basic geometry 1.pdf

School Mathematics Study Group

Geometry

Unit 13

Geometry

Student's Text, Part I

Prepared under the supervision of

the Panel on Sample Textbooks

of the School ~ a t h e k t i c s Study Group:

Frank 3. Alien Lyons Township High School Edwin C. Douglas Tafi School Donald E. Richmond Williams College Charles E. Rickart Yale University

Henry Swain New Trier Township High School Robert J. Walker Cornell University

New Haven and London, Yale University Press

Copyright @ 1960, 1961 by Yale University. Printed in the United States of America.

All rights reserved. This book may not be reproduced, in whole or in pan, in any form, without written permission from the publishers.

Financial support for the School Mathematics Study Group has been provided by the National Science Foundation.

FOREWORD

The Increasing con t r ibu t ion of mathematics t o the culture of the modem world, as well as i t s Importance as a vital part of s c i e n t i f i c and humanistic education, has made it essential tha t the mathematics In our schools be both well selected and well taught.

With t h i s in mind, the various mathematical organizations i n the United States cooperated In t he formation of the School Mathematics Study Group (SMSG) . 3MSG includes college and unlver* s i t y mathematicians, teachers of mathematics a t a l l levels, experts in education, and representatives o f science and technology. The general ob jec t ive of SMSG is the improvement of the teaching of mathematics In the schools of t h i s country. The National Science Foundation has provided substantial funds fo r the support of this . endeavor.

One of the prerequisites for the Improvement of the teaching of mathematics in our schools I s an improved curriculum--one which takes account of the Increasing use of mathematics in science and technology and in other areas of knowledge and at the same time one which r e f l ec t s recent advances In mathematics itself. One of the first projects undertaken by SMSG was to e n l i s t a group of outstanding mathemat ic iana and mathematics teachers t o prepare a series of textbooks which would Illustrate such an improved curriculum.

The professional mathematicians in SMSG believe that t h e mathematics presented l a t h i s t e x t i s valuable for all well-educated citizens i n our s o c i e t y to know and that it is important fo r the precollege student to learn in preparation f o r advanced work In the f i e l d . A t the same time, teachers in SMSG be l ieve that it is presented in such a .form that it can be readily grasped by students.

In most instances t h e material w i l l have a familiar note , bu t the presentation and the p o i n t of view w i l l be d i f f e r e n t . Some material w i l l be ent i re ly new to the traditional curriculum. This I s a8 I t should be, for mathematics is a living and an ever-growing subject, and no t a dead and frozen product of antiquity. T h i s healthy fu s ion of the o l d and the new should lead students to a better understanding of the basic concepts and structure of mathematics and provide a firmer foundation f o r understanding and use o f mathematics in a scientific socie ty .

It is not intended tha t this book be regarded as the only definitive way o f presenting good mathematics t o students a t t h i s level. Instead, It should be thought of as a sample of the kind of improved curriculwn that we need and as a source of suggestions for the authors of commercial textbooks. It I s sincerely hoped tha t these t ex ts will lead the way toward Inspiring a more meaningful teaching of Mathematics, the Queen and Servant of the Sciences.

Below are H a t e d the names of all those who par t ic ipated i n any of the writ ing sessions

at whlch t h e f - o ~ ~ o w l n g SMSG t e x t s were prepared: First Coarse i n Ugebra, Geometry,

IntemedLate Mathematics, Elementary Functions> and ~ n t ~ o d u c t i o n t o Matrix Algebra.

H.W. Alexander, E a ~ l h a n College F.B. Allen, Lyons Township High School, La Grange, I l l i n o i s

Alexander Beck, Oiney High School, Phila- delphia, Pennsylvania

3 3 . Beckenbach, Universi ty o f Cal i fornia a t LOB Angeles

E .G. Begle, School Mathematics Study Group, Yale Vnlversl ty

Paul Berg, Stanford Universi ty Bull Berger, Monroe High School, S t . Paul, Minnesota

Arthur Bemihart, Universi ty of Oklahoma R.H. Bing, Universi ty of Wisconsin A . L . Blalcers, Universi ty of Western

Austra l ia A . A . Blank, New York University Shir ley Boselly, Franklin High School,

Sea t t l e , Washington K.E. -Brown, Department of Health, Eciaca-

t ion , and Welfare, Washington) D . C . 3 .M. Galloway, Carlet on College Hope Chipman, Universi ty High School, Ann

Arbor, Michigan R.R. Christ ian, Universi ty of Br i t i sh

col umbia R.J. Clark, S t . Paul 's School, Concord,

New Hampshire P . H . Daus, Universi ty of Cal i fornia a t Los

Angeles R.B. Davis, Syracuse Universi ty Charles DePrima, California I n s t i t u t e of

Technology Mary Dolclanl, Hunter College Edwin C . Douglas, The Taf t School, Water-

town, Connecticut Floyd Downs, East High School, Denver,

Colorado E .A, Dudley, North Haven High School, Worth

Haven, Connecticut Lincoln Onrst, The Rice Institute Florence Elder, West Hempstead High School,

West Hempstead, Hew York W.E. Ferguaon, Newton High School, Menrton-

v i l l e , Massachusetts N.J. Pine, University of Pennsylvania Joyce 0 , Fontalne, North Haven High School, North Haven, Connecticut

F.L. Friedman, Massachusetts I n s t i t u t e of Technology

Esther Gassett , Claremore High School, Claremore, Oklahoma

R.K. Getoor, Universi ty of Washington V.H. Haag, Franklin and Marshall College R .R . Hartman, Rdina-Morningside Senior H i g h

School, Edina, Minnesota M.H. Heine, Universi ty of I l l i n o i s Edwin Hewitt, University of Washington Martha Hlldebrandt, Proviso Township High

School, Haywood, Ill lnoia

R .C . Jurgensen, Culver Military Academy, Culver, Indiana

Joseph Lehner, Michigan Sta te University Marguerite Lehr, Bryn Maw College Kenneth Leisenrln&, University of Michigan Howard Levl., Columbia University Eunice Lewis, Laboratory High School,

Universi ty o f Oklahoma M.A. Linton. WillSam Penn Charter School,

Philadelphia, Pennsylvania A.E. Livingston, University of Washington L .H. Loomis, Harvard University R .V. Lynch, P h i l l l p s Exeter Academy,

Exeter, New Hampshire W .K . McMabb, Hoeleaday School, Dallas,

Texas K.O. Michaels, North Haven High School, North Haven, Connecticut

E.E. Moise, Universi ty of Michigan E . P . Northrop, University of Chicago O . J . Peterson, Kansas Sta te Teachers

College, h p o r i a , Kansas B . J . Pet t l s , University of North Carolina R . S. Ple te r s , P h l l l i p s Academy, Andover,

Massachusetts H.O. Pollak, Bell Telephone Laboratories Walter Prenowltz, Brooklyn College O.B. Price, Universi ty of Kansas A.L. Putnam, Universi ty of Chicago Pera ls 0. Redgrave, Norwich Free Academy,

Norwich, Comec tiaut Kina Rees, Hunter College D.E. Richmond, Williams College C.E. Rickart , Yale University Harry Rwlerman, Hunter College High School,

New York City J .T . Schwartz, New York University O.E. Stana i t i s , St. Olaf College Robert Starkey, Cubberley H i g h Schools,

Palo Alto, California F h l l l i p Stucky, Roosevelt H i g h School,

Sea t t l e , Washington Henry Swain, Mew Trier Township H i g h

School, Vinnetka, I l l i n o i s Henry Syer, Kent School, Kent, Connecticut Q.3. Thomas, Massachusetts I n s t i t u t e of

Technology A.M. Tucker, Princeton University H.E . Vaughan, Universi ty of ~ l l i n o i s John Wagner, Universi ty of Texas R . J . Walker. Cornell University A.D. Wallace, Tulane University E.L. Halters, Vllliam Penn Senior High School, York, Pennsylvania

Warren White, North Hign School, Sheboygan, Misconaln

D.V. Widder, Harvard University William Wooton, Pierce Junior College, Woodland H i l l s , California

J.H. Zant, Oklahoma S ta t e University

CONTENTS

Chapter 1 . COMMON SENSE AND ORGANIZED KNOWLEDGE . . . . . . . . 1

1.1 . Two Types ofProblems . . . . . . . . . . . 1.2 . An Organized Logical Development of Geometry 8

2. SETS. REALNUMBERSANDLINES , . . , . . . 15 . . . . . . . . . . . . . . . . . . . . . 2.1 Sets 15 . . . . . . . . . . . . . . 2.2 . The Real Numbers 21 . . . . . . . . . . . . . 2-3 . The Absolute Value 27 2-4 . Measurement of Distance . . . . . . . . . . 30 . . . . . . . 2.5 . A Choice of a Unit of Distance 2.6 . An Infinite Ruler . . . . . . . . . . . . . 2.7 . The Ruler Placement Postulate . Befcweeness.

^ Segments and Rays . . . . . . . . . . . . . 39

3 . LINES. PLANES AMD SEPARATION . . . . . . a , . . . , 3.1 Lines and Planes inSpace

53 . . . . . . . . . . 5.2 . Theorems in the Form of Hypothesis and

53

~onclusion . . . . . . . . . . . . . . . . . 60 3-3 . Convex S e t s . . . . . . . . . . . . . . . . 62

4 . ANGLES AND TRIANGLES . . . . . . . . . . . . . . . . 71 4.1 . The Basic Definitions . . . . . . . . . . . 4.2 . Remarks on Angles 77

71 . . . . . . . . . . . . . 4.3 . Measurement of Angles . . . . . . . . . . . 79 4.4 . Perpendicularity. R i g h t Angles and Congruence

of Angles . . . . . . . . . . . . . . . . ; 85

5 . CONGRUENCES . . . . . . . . . . . . . . . . . . . . 97 5-1 . The Idea of a Congruence . . . . . . . . . . 97 5-2 . Congruences Between Triangles . . . . . . . 109 5.3 . The Basic Congruence P o s t u l a t e . . . . . . . 115 5.4 . Writing Your Own Proofs . . . . . . . . . . 117 5.5 . Overlapping Triangles . Using the Figure i n

Statements . . . . . . . . . . . . . . . . . 5.6 . The Isosceles Triangle Theorem . The Angle

123

Bisector Theorem . . . . . . . . . . . . . . 127 5.7 . The Angle Side Angle Theorem . . . . . . . . 132 5.8 . The Side Side Side Theorem . . . . . . . . . 137

Review Exercises. Chapters 1 t o 5 . . . . . . . . . . . 155 6 . A CLOSER LOOK A T PROOF . . . . . . . . . . . . . . . 159

6.1 . How A Deductive System Works . . . . . . . . 159 6.2 . I n d i r e c t Proof . . . . . . . . . . . . . . . 160 6.5 . Theorems About Perpendiculars . . . . . . . . 167 6.4 . Introducing Auxiliary Sets Into Proofs . . . 176 6.5 . Betweenness and Separation . . . . . . . . . 182

Chapter

7 . GEOMETRIC INEQUALITIES . . . . . . . . . . . . . . 189 7.1 . MakineReasonable Con.lectures . . . . . . 189 7.2 . ~ l g e b r a of 1nequaliti6s . . . . . . . . . 191 7.3 . The Basic I n e q u a l i t y Theorems . . . . . . 7.4 . Alt i tudes . . . . . . . . . . . . . . . . 19? 21

8 . PERPENDICULAR LINES AND PLANES IN SPACE . . . . . 219 8-1 . The Basic D e f i n i t i o n . . . . . . . . . . . 219 . . . . . . . . . . . . 8.2 . The Basic Theorem 222 8-3 . Existence and Uniqueness Theorems . . . . 235

9 PARALLEL LINES IN A PLANE . . . . . . . . . . . . 241 9.1 . Conditions Which Guarantee Parallelism . 241 . . . . . . . . . . . . 9.2 Corresponding Angles 251 9.3 . The Parallel Postulate . . . . . . . . . . 252 9.4 . Triangles . . . . . . . . . . . . . . . . 258 9.5 . Quadrilaterals in Plane . . . . . . . . . 265 9.6 . Rhombus. Rectangle and Square . . . . . . 268 9-7* Transversals To Many Parallel Lines . . . 275

10 . PARALLELS IN SPACE . . . . . . . . . . . . . . . . 291 . . . . . . . . . . . . . . 10.1 Parallel Planes 291 10.2 . Dihedral Angles. Perpendicular Planes . . 299 . . . . . . . . . . . . . . . . 10.3 Pro jectlons 306

Appendix I . A Convenient Shorthand . . . . . . . . . . A - 1

Appendix I1 . Pos tu la t e s of Addition and Multiplication A-5 . . . . . Appendix I11 . Rational and I r r a t i o n a l Numbers A-9

Appendix IV . Square Roots . . . . . . . . . . . . . . . A-15 Appendix V . How to Draw Figures in 3-Space . . . . . . A-19 Appendix VI . Proofs of Theorems on Perpendicular i ty . A-25 THE MEANING AND USE OF SYMBOLS . . . . . . . . . . . . . a

. . . . . . . . . . . . . . . . . . . LIST OF POSTULATES e LIST OF THEOREMS AND COROLLARIES . . . . . a . m . . . m 1

INDEX OF DEFINITIONS . . . . . . . . . . . . . following page y

PREFACE

This book is designed for the one-year in t roductory course in geometry which is usually taught in the tenth grade. Students I n t h i s grade normally have learned a fair amount of Informal geometry, including the calculation of areas and volumes for various elementary figures , t h e Pythagorean r e l a tion, and the use of s i m i l a r r i g h t triangles t o c a l c u l a t e unknown heights and dis- t a n c e s . Students who have not learned this mater ia l may have to be given some extra a t t e n t i o n , but the book should still be teachable, a t a su i tab ly adjus ted pace. I n algebra, no special prepara t ion is required beyond the knowledge and skills normally acquired In the ninth grade.

The book is devoted mainly to plane geometry, with a few chapters on solid geometry, and a s h o r t introduction to analytic geometry at the end. It seems natural, in a preface, to give an account of the novel f e a t u r e s i n t h e treatment. We are aware, of course, of a danger In 80 doing. A long string of novelties, offered for the reader's s p e c i a l a t t e n t i o n , may very well convey the impression that the authors have been engaged In an unhealthy pursuit of Innovation for i t s own sake. This is by no means the way in which we have conceived our task. We began and ended our work with t he convict ion t h a t t h e traditional content of Euclidean geometry amply deserves the prominent place which It now holds In high-school study; and we have made changes only when the need for them appeared to be compelling.

The basic scheme i n the postulates I s that of G. D. Birkhoff, In t h i s scheme, it i s assumed that the r e a l numbers are known, and they a r e used freely for measuring both distances and angles. This has two main advantages.

In the first place, the real numbers give us a sort of head s t a r t . It has been correctly pointed out that Euclid's postulates a r e not l o g i c a l l y s u f f i c i e n t f o r geometry, and t h a t t h e treatments based on them do not meet modern standards of rigor. They were Improved and sharpened by Hllbert. But the foundations of geometry, in the sense of Hilbert, a r e not a part of elementary mathematics, and do not belong in the tenth-grade curriculum. If we assume the real numbers, as in the Birkhoff treatment, then the handling of our postu lates becomes a much eas ie r t a s k , and we need n o t f a c e a c r u e l choice between mathematical accuracy and In- telligibility.

In the second place, It seems a good idea i n I t s e l f t o con- nect up geometry with algebra a t every reasonable opportunity, so that knowledge in one of these fields will make its na tu ra l con- t r i b u t i o n to the understanding of both. Some of the topics u s u a l l y studied in geometry are e s s e n t i a l l y algebraic. This i s true, for example, of the propor t ional i ty re la t ions f o r similar triangles. In this book, such topics are treated algebraically, so as to bring out the connections with the work of the ninth and eleventh grades.

We hope t h a t t h e statements of d e f i n i t i o n s and theorems are exact; we have tr ied hard to make them so. Just as a lawyer needs to learn to draw up contracts that say what they are supposed to say, so a mathematics student needs to learn t o write mathemati- c a l statements t h a t can be taken literally. But we are not under the illusion tha t this sort of e x a c t i t u d e I s a substitute for In- tuitlve insight. We have, therefore, based t h e design of both the t e x t and the problems on our be l i e f that Intuition and l o g i c shou ld move forward hand in hand,

Chapter 1

COMMON SENSE ANO ORGANIZED KNOWLEDGE

1-1. Two Types of Problems. - Consider the following problems:

1. A line segment 14 inches long is broken I n t o two segments.

If one of the two smaller segments is 6 inches long, how long is

the other one? 2, In a cer ta in"rectangle , t h e sum of the l e n g t h and the

width is 14 (measured in inches). A second rectangle is three

times as long as the first, and twice as wide. The perimeter of the second rectangle I s 72. What are the dimensions of the first

rectangle?

The answer to Problem 1, of course, is 8 inches, because 6 + 8 = 1 4 . We could solve this problem algebraically, if we wanted to , by se t t ing up the equation

6 + x = 14,

and solving to get x = 8. But this seems a little silly, because it is so unnecessary. If all algebraic equations were as super- fluous as this one, then no serious-minded person would pay any attention to them; in fact, they would probably never have been invented.

Problem 2, however, is quite another matter. If the length

and width of the first rectangle are x and y, then the length and

width of the second rectangle are 3x and 2y. Therefore,

because t he sum of the length and width is hal f the perimeter. We

already know t h a t x + y = 1 4 . Thus we have a sys tern of two l i nea r equations In two unknowns;

x + y = 1 4

3x + 3 = 36.

To solve, ive multiply each term In the first equation by 2, getting 2x + 2y = 28,

and then we subtract this last equation, tern by term, from the

second. This gives

x = 8. Since x + y = 14, we have y = 6, which completes the solution of our problem. It I s easy to check that a length of 8 and a width

of 6 satisfy t h e condi t ions o f the problem.

In a way, .these two problems may seem similar. But in a very important sense, they are different. The first is what you might c a l l a common-sense problem. It is very easy to guess what the

answer ought to be, and it is also very easy to check that the natural guess I s actually the right answer. The second problem

is e n t i r e l y another matter. To solve the second problem, we need to know something about mathematical methods.

There are cases of this kind in geometry. Consider the

following statements: 1, If a triangle has s ides of length 3, 4 and 5, then i t is

a r i g h t t r iangle , with a right angle opposite the longest side.

2. L e t a triangle be given, with s ides a, b and c. If 2 a 2 + b 2 = c ,

then the triangle is a right t r i ang le , with a right angle opposite the longest side.

The first of these facts was known to the ancient Egyptians. They checked it by experiment. You can check it yourself, w i t h

a ruler and compass, by drawing a 3-4-5 t r iangle , and then measuring the angle opposite the longest side with a protractor . You should bear i n mind, of course, that this check is only approximate. For example, if the angle were r e a l l y 89' 5gf 59", instead of 90' exactly, you would hardly expect to tell the difference by drawing your figure and then taking a reading with your protractor. Nevertheless, the "Egyptian method" is a sound

common-sense method of verifying an experimental fact.

The Egyptians were extremely skillful at making physical measurements. The edges of the base of the great pyramid are about 756 feet long; and the lengths of these four edges agree, with an error of only about two-thirds of an inch. Nobody seems to know, today, how the builders g o t such accuracy.

Statement 2 above was not known to the Egyptians; I t was

discovered l a t e r , by t h e Greeks. This second statement is very di f fe ren t from the f irst . The most iJiroortant difference is t h a t

there are i n f i n i t e l y many possibilities f o r a, b and c . For instance, you would have to cons t ruc t tr iangles, and take readings with a pro t rac to r , f o r all of the following cases,

and so on, endlessly, It seems pretty hopeless to try t o verify our general statement by experiment, even approximately. Therefore, a reasonable person would not be. convinced that Statement 2 was t rue in a l l cases until he had seen some logical reason why it should be true in a l l cases.

In fact , this is why it was the Greeks, and no t the Egyptians, who discovered that our second statement is true. The Egyptians

had lots of common-sense knowledge of geometry. But the Greeks found something better , and. much more powerful: they discovered the science of exact geometrical reasoning. By exact reasoning, they learned a great deal that had not been known before t he i r time. The things that they learned were the f i rs t big step toward modern mathematics, and hence, toward modem science in general.

Problem Set 1-1 --

1. Try the following experlaent. Take a piece of string, about six feet long, and put it on the f l o o r in the form of a loop with the ends free:

Then pull the ends of the string apar t , making the loop gradually smaller, and stop when you think that the loop is

the size of your o m waist. Then check the accuracy of your guess by wrapping the string around your waist. After you

have checked, read the remarks at the end of this set of problems.

In this pair of questions, the first can be answered by IT common sense." State only Its answer. The second requires some arithmetic or algebraic process f o r its solut ion. Show

your work f o r it. a. What I s half of 2? b. VJhat is half of 135,790? Answer as in Problem 2:

a. One-third of the distance between two c,it ies is 10 miles. What is the entire distance?

b. The distance between two c i t i e s is 7 miles more than one- third the distance between them. What is the distance between them?

* I ! . . Answer as in Problem 2:

a. If a 5-inch piece of wire I s cut In to two parts so t h a t

one p a r t is 4 times as long as the other , what are the

lengths of the p a r t s ? b. If a 5-inch piece of wire is cu t i n t o two parts such t h a t

a square formed by bending one piece will have four tmes the area of a square formed by bending the other , what are the lengths of the parts?

5. If the s ides of a triangle are 5, 12, 13, is I t a right

t r iangle?

6 . If two s tudents careful ly and Independently measure the width of a classroom with rulers, one measuring from left to r igh t

and the other from right to left, they are likely to get different answers. You may check this with an experiment. Which of the following a r e plausible reasons f o r this?

a. The rulers have different lengths. b. One person may have lost count of the number of feet In

t he width.

c . Things are longer (or shorter) from left to right than right to left.

d. The errors made in changing the position o f the ruler accumulate, and the sum of the small errors makes a

discernable e r r o r . 2 7. S h o w t h a t n - Â£ + 2 = n i f n = 1. Is the equation true

when n = 2? Is it true when n = 3? Is I t always true? 2 2 2 8. a. If 3 , 5 and 7 are divided by 4, what is the remainder

in each case?

b. How many odd integers would you have to square and divide

by 4 t o guarantee tha t the remainder would always be the same?

Number of points connected 2 3 5 6

Number of regions formed 2 4 8 16 ?

Replace the question mark by the number you think belongs there. Verify your answer by m a k i n g a drawing in which six

points on a c i r c l e are connected in all poss ib l e ways.

10. The following opt ical i l lus ions show that you cannot always 11 tniafc appearances. Things are seldom what they seem; sktm

milk masquerades as cream. " From "H.M.s. ina afore" by Gilbert and Sullivan. a . Is CD a continuation of AB?

T e s t your answer with a ruler.

A

b. Are RS and ST equal in length? Compare the lengths with your

ruler or compass.

R

c. Which figure has the greater are a?

d . Which is longer, AB or CD? Check with your ruler.

*11. U s e a ruler to check the accuracy of the measurements of the figure. Show that if these measurements are correct the sum of the areas of the four pieces of the rectangle I s more than

the a r e a of the rectangle. Odd, isn't it?

*12. A trip of 60 miles is to be done at an average speed of 60 m.p .h. The first 30 miles are done a t 30 m.p. h. A t what rate must the remaining 30 miles be covered?

Remarks on Problem 1. Nearly everybody makes a loop about twice as big as it should be. You can get much better results by the following method. The circumference of a circle is equal

to IT times the diameter, and T is approximately equal to 3.

Therefore, the diameter is about one-third of the circumference.

If your waist measure I s , say, 21 Inches, th3s means that the loop on the floor should be about 7 Inches across. This will look unbelievably small, but if you have thought the problem out mathematically, you w i l l have the courage of your convictions.

This is one of a large number of cases in which even a very

crude mathematical approach t o a problem is bet ter than an outright leap in the dark.

1-2. & Organized Logical Development - of Geometry.

If you stop to think, you will realize that by now you know a

great many geometrical facts. For example, you know how to find the area of a rectangle, and of a right triangle, and perhaps of a triangle in general, and you know the Pythagorean relation f o r right triangles. Some o f the things that you know are so simple and obvious that it might never occur to you t o even put them i n t o words, l e t alone t o wonder whether or why they are true. The

following is a statement of this type: Two straight lines cannot cross each other in more than one

point.

But some of them, like the Pythagorean re la t ion , are not obvious a t al l , but rather surprising. We would l i k e to organize our knowledge of geometry, in an orderly way, in such a way that these more complicated statements can be derived from slmpie

statements. This suggests that we ought to be able to make a l i s t of the facts of geometry, with the simplest and easiest statements coming first, and the hard ones coming later . We might try to

arrange the statements in such an order that each statement in the list c a n be derived from the preceding statements by logical re as oning .

Actually, we shall carry out a program that is very much like this. We wil l state definitions, as clear ly and exactly as we can;

and we wi l l establish the facts of geometry by giving logical p~oofs.

The statements that we prove w i l l be called theorems. (E proving of theomms --- Ts not a spectator-sport, more than -- arithmetic is: t h e best 9 to learn about I t 2s & dolng It. - -- ----- - Therefore, in this course, you will have l o t s of opportunities t o

prove lots of theorems for yourself. ) While nearly a l l of the statements that we make about geom-

etry are going to be proved, there will be some exceptions. The simplest and most fundamental statements will be given without proofs. These statements w i l l be called postulates, and will form the foundation on which we will build. In the same way, we wi l l

use the simplest and most fundamental terms of geometry without defining them; these w i l l be ca l led the undefined terms. The

definitions of the other t e rns that we use will be based on them.

A t f irst glance, it might seem better to define every term

t h a t we use, and t o prove every statement that we make. With a l i t t l e r e f l ec t ion , we can convince ourselves that this can't be

done.

Consider f irst the question of the postulates. Most of the time, when we prove a theorem, we do so by showing that It follows

l og ica l ly from theorems t h a t have already been proved. But it

is clear that proofs of theorems cannot always work this way. In particular, the first theorem that we prove cannot possibly be

proved t h i s way, because i n this case there aren't any previously

proved theorems. But we have to s t a r t somewhere. This means that we have to accept some statements without proof. These un-

proved statements are the postulates.

The purpose of stating postulates I s to make it clear just where we are starting, and Just what s o r t of mathematical objects we are studying. We can then build up a solid, organized body of facts about these mathematical objects.

Just as we start with some unproved statements, s o we s t a r t

with some undefined terms. Most of the time, when we g ive a

d e f i n i t i o n of a new geometric term, we define it by m e a n s of o t h e r geometric terms which have already been defined. But i t is clear

t h a t d e f i n i t i o n s cannot always work this way. In particular, the

first definition that we state cannot possibly be stated in this way, because in this case there aren't any previously defined geometric terms. But we have t o start somewhere. T h i s means that we Introduce some geometric terms without defining them, and then

use these basic terms In our first defini t ions. We sha l l use the

simplest and most; fundamental geometric terms without making any attempt to give definitions for them. Three fundamental undefined terns w i l l be point , line and plane .

Postulates, of course, are not made up at random, (if they were, geometry would be of no Interes t or importance.) Postu la tes

describe fundamental proper t ies of space. In the same way, the

mdef tned terns point, l ine and plane a re suggested by physical - objects. To get a reasonably good picture of a point , you make a dot on paper with a pencil. To get a better approximation of the

mathematical Idea of a point , you should first sharped your penci l .

The picture is s t i l l approximate, of course: a dot on paper must cover some area, or you couldn't see it at all. But i f you think

of dots made by sharper and sharper penci ls , you will have a good

idea of what we are driving at when we use the undefined term, point .

lhen we use the tern l lne , we have W m h d the idea of a straight line. A straight line, however, is supposed t o extend infinitely fa r in both directions. Usually, we shall Indicate this in pictures by arrowheads at the ends of the por t ion of the line t h a t we draw, l i k e this:

We shall have mot he^ tern, segment, for a figure that looks l i k e this:

A thin, tightly stretched string I s a good approximation of a

segment. An even thinner and more tightly stretched string I s a

better approximation. And so on.

Think of a perfectly flat surface, extending Infinitely far in every direction, and you have a good idea of a plane.

You should remember that none of the above statements are definitions. They are merely explanations of the Ideas t ha t

people had i n the back of their minds when they wrote the postulates. When we are writing proofs the Information that we claim

to have about points , l ines and planes will be the information given by the postulates.

We have said that theorems are going t o be proved by logical reasoning. We have not explained what logical reasoning is, and

In fact , we doni t know how to expla in this in advance. As the course proceeds, you w i l l get a better and better idea of what log ica l reasoning is, by seeing i t used, and best of a l l by using It yourself. This is the way that all mathematicians have learned

to tell what Is a proof and what Isns t. At the beginning of the next chapter, we s h a l l give a short

account of the Idea of a set, and a short review of the funda- mentals of algebra for real numbers. S e t s and algebra will be used

throughout this course, and our study of geometry will largely be based on them. We s h a l l think of them, however, as things that

we are working with. They w i l l not be an actual p a r t of our system of postulates and theorems. They are supposed to be a v a i l - able a t the start; some of our postulates w i l l invo lve real

numbers; and elementary algebra will be used in proofs. In f ac t , geometry and algebra are very closely connected, and both of them

are easier to learn if the connections between them are brought

out as soon as poss ib l e .

Problem Set 1-2 -- 1. A student wanting to find the meaning of the word "dimension"

went t o a dictionary. This dictionary did not give d e f i n i t i o n s as we have them in geometry but d i d .give synonyms of words.

He made the following chart.

size extent - - longest

dimension dimension - measurement -

a. Point out from the above char t a circular "Mat-, of three terms each of which has its following term a s a synonym.

(1n a circular list, t h e first term is assumed to follow the l a s t .)

b. Make a c i r c u l a r list which contains f o u r such terms. *2. Make a char t similar t o t h a t i n Problem 1, start lnp: with some

word i n your dictionary. 3. John convinced his mother that he did not track mud onto the

living room rug by pointing out that it d i d no t start raining

until 5 o'clock and that he had been in his room studying since 4~30. He mentioned that a person cannot do something if he is not there. The thing he was proving ( t h a t he d i d not

track mud) might be regarded as a theorem and the statement

about a person not being able t o do something If he I s not there might be regarded as a postulate. Make another example of such a convincing argument and point out what corresponds

to the theorem, what t o the proof , and what to postulates.

4. Janie: Mother:

Janie : Mother: Janie : Mother: J an i e : Mother: Janie :

13

What's an architect?

An architect? An architect Is a man who designs

buildings.

?4hatfs "designs"?

Well--plans.

Like we plan a picnic?

Yes, quite like t h a t .

What are buildings? Oh, Janie, you know -- houses, churches, schools. Yes, I see.

Consider the above discussion. What were basic undefined terms as far as Janie was concerned?

The Stuarts have three children. Joe is a senior in high school. Karen is a seventh grader, and Beth I s four. At t h e dinner table :

Joe : We learned a funny rev word in geometry class today -- parallelepiped.

Karen: What in the world Is it? Joe : Well, It's a s o l i d . You know what I mean by a solid

figure -- I t takes up some space. And it's bounded

by planes. You know what a plane i s , don't you? Beth: Like a windowplane? Joe : The word is a windowpane, but thatts the idea. A

parallelepiped is a solid bounded by parallelograms. A candy box is one, but I t ' s a special one because the six faces are all rectangles. If you had a candy box and could shove it at one comer you'd get a parallelepiped. Got the idea?

In the above discussion what basic, undefined terms did Joe

use in his descript ion?

6. What do you think is wrong with the following faulty definitions?

a. A square is something t h a t is n o t round. b. A right triangle is a triangle each of whose angles has

a measure of 90'. c . An equilateral triangle is when a triangle has three sides

the s m e length. d. The perimeter of a rectangle is where you f ind the sum of

the lengths of the s ides of the rectangle. e . The circumference of a circle is found by m u l t i p l y i n g the

diameter by IT. *7. Indicate whether the fol1orri.ng are true or false;

a . It is possible to define each geometric term by using

simpler geometric terms. b. Exact geometric reasoning leads us to geometric truths

tha t cannot be deduced from measurement. c . Theorems are proved only on t h e basis of d e f i n i t i o n s and

undefined terms. d . If you are willing to write In a l l the steps, each

theorem can be deduced from postulates without making recourse to previous theorems.

Chapter 2

SETS, REAL NUMBERS AND LINES

2-1. Sets - You may not have heard the word s e t used In mathematics be- -

fo re , but the idea l a a very familiar one. Your family I s a s e t

of people, consisting of you, your parents, and y o u r bro thers and s i s t e r s (if any). These people are t h e members of the s e t . Your geometry class I s a s e t of students; I t s members are you and your classmates. A school a th l e t i c team l a a set of students. A

member of a s e t is said t o belong t o the set. For example, you belong to your family and to your geometry class, and so on. The members of a s e t are of ten called i t s elements; the two terms, members and elements, mean exactly the same thing. We say t h a t a set contains each of I t s elements. For example, both your family and your geometry class contain you. If one set contains every element of another set, then we say that the first s e t

contains the second, and we say that the second s e t is a subset

of the f irst . For example, the student body of your school conta ins your geometry class, and your geometry class I s a subset of the student body. We say that t h e subset lies In the set that -- contains it. For example, the s e t of a l l violinists lies In the set of a l l musicians.

Throughout this book, lines and planes will be regarded as sets of points. In fact, a l l the geometric figures that we t a l k about are sets of points. (YOU may regard this, if you l i k e , as a postulate. )

When we say that two sets are equal, or when we write an

equality A = B between two sets A and 3, we mean merely t ha t t h e two sets have exactly the same elements. For example, let A be

1 1 the set of all whole numbers between and %, and l e t B toe the

1 1 set of all whole numbers between 7 and %. Then A = B, because each of the seta A and 3 has precisely the elements 1, 2, 3, 4

and 5. In fact , it very of ten happens that the same s e t can be

described In two different ways; and If the descriptions look d i f f erent , this doesn't necessarily mean that the sets are different.

TWO sets intersect if there are one or more elements t h a t belong to both of them. For example, your family and your geometry class must intersect, because you yourself belong t o both of them. But two different classes meeting at the same hour

do not intersect. The intersection of t w o sets l a the set of a l l objects that belong to both of them. For example, the intersection of the set of all men and the set of all musicians is the

set of all men musicians. Passing t o mathematical topics, we see that t h e set of all

odd numbers I s the set whoae members are 1, 3, 5 , 7 , 9, 11, 13, 15, * = *

and so on. The set of a l l multiples of 3 I s the se t whose members are

3, 6, 9, 12, 15, . and so on. The intersection of these two s e t s is

3, 9, 15, 21, . and s o on; its members are the odd multiples of 3.

In the figure below, each of the two rectangles l a a set

of points, and their intersect ion contains exactly two points.

S i m i l a r l y , each of the corresponding rectangular regions is a s e t of points, and t h e i r intersection is the small rectangular region

in the middle of the f igure . In t h e next figure, each of the two lines is a set of points, and t h e i r Intersection consists of a

Below, we see two sets of points, each of which is a f la t rectangular surface. The Intersection of these two s e t s o f points

is a part of a s t r a i gh t l i n e .

The union of the two sets 1s the set of all objects that belong to one or both of them. For example, the union of the set

of a l l men and t h e s e t of a l l women I s the s e t of all adults. The intersection, or the union, of three or more sets is defined

s i m i l a r l y . Thus a triangle is t h e union of three sets, each of which i s a subset of a line.

The figure below is the union o f five sets, each

In some s i t u a t i o n s , it is convenient t o use

of which is a

t h e Idea of the

empty se t . The empty set Ls the set that h&s no members at al l . T h i s i d e a

very much following

(1)

(2)

( 3 )

Once

may seem a little peculiar at f i rs t , but it I s really

like the idea of the number 0. For example, t h e three statements all say the same th ing :

There are no married bachelors In the world. The number of married bachelors In the world I s zero. The set of a l l married bachelors in the world I s the empty set.

we have Introduced the empty set, then we can speak of t h e Intersection of any two s e t s , remembering that the inter- s e c t i o n may tu rn out to be the empty se t .

F o r example, the Intersection of t h e set of a l l odd numbers

and the set of a l l even numbers is the empty s e t .

A word of warnine: If you cornmre the definitions of t h e --- terms Intersect and Intersection, you w i l l see that these two terms are not related In qui te the simple way that you m i g h t ex- pec t . When we apeak of the i n t e r s e c t i o n of two sets, we allow

the possibility that the intersection may be empty. But If we say t h a t the two sets intersect, this always means that they have an element In common.

Another word of wamirg: Statements ( 2 ) and (3) above mean -- the same thing. But t h i s does not mean that a set that contains

only the number 0 is empty. For example, the equation x + 3 = 3 has 0 as I t s only root, and so the s e t of roots is no t the empty s e t ; the s e t of roots has exactly one element, namely, the number 0. On the other hand, the set of all roots of the equa-

tlon x + 3 = x + 4 really I s the empty'set, because the equation

x + 3 = x + 4 has no roots at a l l ,

Problem Set 2-1 -- 1. Let A be t h e s e t ( 3 , 5, 6 , 9, 11, 12) ( t h a t Is, t h e set whose

members are 3, 5, 6 , 9, 11, 12) and B be the s e t

(4, 5 , 7 , 9, 10, 111. What is the i n t e r s e c t i o n of s e t s A and B? What is t h e union of A and B?

2. Consider the fo l lowing sets: S , is t h e set of a l l students In your school .

Sy is the set of all boys in your student body.

S is the s e t of a l l girls In your student body. 3 Sj, Is the set of all members of the faculty of your school. S is the set whose only member I s your se l f , a student in 5 your school . a. Which pairs, of s e t s Intersect?

b, Which set I s the union of Sn and S ?

c . Which set I s the union of S., and S ? 5

d. Describe the union of Si and S 4 .

e . Which of the sets are sub-sets of Si?

3. In the following figures, consider the l i n e and the c i r c l e as

two sets of points. In each case, what Is t h e i r in tersec t ion?

Caae I. Case 11. Case 111.

4 . Consider a se t of t h ree boys, ( A , 3, Any set of boys

selected f r o m these t h r e e will be called a committee. , a . How many different two-member committees can be formed

from the t h r e e boys?

b. Show t h a t any two of t h e committees in (a) intersect. - What does the word " Intersect" mean?

5. Consider t h e set of all posi t ive even Integers and the s e t of a l l positive odd integers. Describe t h e set which l a the

union of these two s e t s .

6 . Describe the Intersection o f the two se ts g i v e n in Problem 5. 7 . In the figure, what l a the in te r sec t ion of the triangle ABC

and the segment EC? What is the i r union?

8. L e t A be the set of pairs of numbers (m,n) which satisfy the equation 4m + n = 9. Let B be t h e set of pairs of numbers (m,n) which satisfy the equation 2m + n = 5. Find the intersection of the sets A and B.

9. Let A be the s e t of pairs (x ,y) for which x + y = 7 . L e t B be the s e t of pairs (x ,y) for which x - y = 1.

What is the intersection o f A and B?

10. Let A be t h e set of pairs (x,y) for which x + y = 3 .

Let B be the set of pairs (x,y) for which 2x + 2y = 7 . What la the Intersection of A and B?

11. Consider the s e t of a l l positive Integers d i v i s i b l e by 2 . Consider the s e t of all posi t ive integers d i v i s i b l e by 3. a . Describe the intersection of these two sets. Give i t a

first four members. b. Write an algebraic expression for the Intersection. c. Describe the union of the two sets. Give its first

eight members.

12. a . How many s t r a i g h t lines can be drawn through 2 points? b. If three points do not lie in a s t r a i g h t l i n e , how many

straight l i n e s can be drawn through pairs of t h e points?

c. If four points are given and no s e t of three of them lie in a s t r a i g h t l i n e , how many s t ra ight lines can be drawn containing sets of two of t h e points? Answer the same question if five points are given.

*d. Answer Question c if n points are given.

2-2. The Real Numbers -- The f i r s t numbers t h a t you learned about were the "counting

numbers" o r " na t u r n 1 numtoe rs" , 1, 2, 3, 4, 5, - 4

and so on. (YOU knew about these before you learned to read or write. And ancient roan learned to count long before the inven- tion of writing.) The counting numbers never end, because starting with any one of them, w e can always add 1, and get another

one. We may think of the counting numbers as arranged on a line,

s t a r t i n g at some point and continuing t o the right, like t h i s :

To t h e left of 1, we put In the number 0, like t h i s :

And the next step I s to put I n the negative whole numbers, l i k e t h i s :

The numbers that we have ao far are cal led the Integers or whole - numbers (pos i t ive , negative and zero). The counting numbers are the posi t ive integers, and are often referred to by t h i s name.

Of course, there are l o t s of points of the line that have no

numbers attached to them so far. Our next step is to put in the 1 1 2 fractions n, 3, -T, 1 1 2 - Â¥x- - -5, - 7 and so on. The new numbers

that we want to put In include a l l numbers that can be expressed as t h e r a t i o 2 of any two integers (w i th q not equal t o zero) .

4 We can Indicate a few of these, as samples:

The numbers t h a t we have so far are called the ra t ional numbers* h his term I s not supposed t o mean that they are in a

better state of mental hea l th than o t h e r and less fortunate numbers. It merely refers t o the f a c t that they are r a t i o s of whole numbers).

The rational numbers form a very large s e t . Between any two uf them there I s a third one; and there are Infinitely many of them between any two whole numbers. It is a fact, however, t h a t the rational numbers still do not fill up the line completely.

For example, F i s not rational; it cannot be expressed as the ratio of any two integers ye t it does correspond t o a po int on the line. (For a proof, see Appendix 111. ) The same I s true :for i/T" and fl, and also for such "peculiar1' numbers a s T . Such non- r a t i o n a l numbers are cal led irrational. If we Insert all these extra numbers, in such a way t h a t every po in t of the line has a number attached t o it, then we have the - real numbers. We I n d i c a t e some samples, l i k e this:

You should check that these numbers appear on the scale In approximately the positions where they belong. ( / IS

approximately 1.41 . HOW would you find.,/^?)

The real numbers w i l l form part o f the foundation of almost a l l that we are going to do in geometry. And it will be important throughout f o r us to think of the real numbers as being arranged on a l i n e .

A number x I s less than a number y if x l i e s t o the left of y.

We abbreviate t h i s by writ ing x < y . We notice that every negative number lies t o t h e left of every posit ive number. Therefore, every

negative number is less than every p o s i t i v e number. For example, -1,000,000 < 1,

even though the number -1,000,000 may in a way look "bigger".

Expressions of the form x < y are called inequalities. Any

inequality can be w r i t t e n in reverse. For example, 1 > -1,000,000;

and In general, y > x means t h a t x < y .

The expression

x i y means that x is less than or equal to y . For example, 3 A 5 because 3 < 5, and 5 < 5, because 5 = 5.

In your study of algebra, you have by now learned quite a lot about how the real numbers behave under addit ion and multiplication. All the algebra that you know can be d e r i v e d f r o m a few tr iv ia l - looking statements. These statements are the postu- '

lates f o r addition and multiplication of real numbers. You will find them listed in Appendix 11. You may not have studied algebra on the basis of the postulates; and we are not going to s tar t such a proceeding now. In t h i s course, we are simply going t o use the methods of elementary algebra, without comment.

We should be a little more careful, however, about inequal- i t i e s and square roots. The r e l a t i o n < defines an order f o r t h e - real numbers. The fundamental properties of this order relation are the following:

0-1, (uniqueness of order) For every x and y, one and only one of t h e following conditions holds: x < y, x = y, x > y.

0-2. (Transitivity of Order) If x < y, and y < z, then x < 2 .

0-3. ( ~ d d i t i o n for ~ne~ualities) If x < y, then

x + 2 < y + z for every z . 0-4. (~ultiplication for Inequalities) If x < y and z > 0,

then xz < yz .

The statements 0-2. and 0-3. have an Important consequence, which is worth mentioning separately:

0-5. If a < b and x < y , then a + x < b + y .

T h i s is true f o r the following reason: By 0-3, we know that

a + x < b + x

and also t h a t

b + x < b + y .

hat IS, an i n e q u a l i t y is preserved if we add the same number on each side.) By 0-2, these last two Inequalities fit together t o

g i v e us a + x < b + y ,

which is what we wanted.

F i n a l l y , we are going to need the following proper ty of the

real numbers : R-1. (Existence of Square Roots.) Every p o s i t i v e number has

exactly one positive square root. There is one rather tricky point In connection with square

roots. When we say, In words, that x I s a square root of a , t h i s 2 means merely t h a t x = a. F o r example, 3 Is a square root of 9,

and - 3 is a square root of 9. But when we write, in symbols,

t h a t x =& we mean that x is t h e p o s i t i v e square root of&. Thus, t h e following statements are t rue or false, as Indicated:

T r u e : - -3 I s a square root of 9. False: -3 = i/ ". - True: fl= 3. - False: fl= + 3. -

The reason for t h i s usage is simple, once you t h i n k of it. ~ f c w e r e allowed to denote either the positive root or the neg-

ative roo t , then we would have no way at a l l to write t h e posi-

t i v e square root of 7. (Pu t t ing a plus sign in front of the

express!oni/T'gets us nowhere, because a-p lus sign never changes the va lue of an expression. If y^were negative, then + f l would a l s o be negative),

Problem -- S e t 2-2

1. Indicate whether each of the following l a true or false. a. The real number scale has no end points. b. There exists a point on the real number scale which

repreaenta -i/^exactly. 6

c . The poin t corresponding to 7 on the real number scale 7 lies between the points corresponding t o 3 and 3.

d m Negative numbers are real numbers. 2. Restate the following in words:

a. AB < CD. e. 0 < 1 < 2 .

b. X > Y . f* 5 1 x > -5- c . X Y > YZ. g. x > 0. (3. n < 3.

3. Write as an inequality: a. k is a posi t ive number. b. r l a a negative number. c . t Is a number which I s not pos i t ive . d. s is a non-negative number. e. g has a value between 2 and 3.

f, w has a value between 2 and 3 inc lus ive . g, w has a value between a and b.

4 . For which of the following I s I t true tha t 1/t2- 5 x?

5. How would the points corresponding to the following sets of

numbers be arranged from left to r i g h t on a number scale in which the positive numbers are to the right of O?

*6. If r and s are real numbers, other than zero, and r > s, indicate whether the following are always true (T), sometimes

true (s), o r never true (N), a. s < r.

*7. Follow the I n s t r u c t i o n s of Problem 6 for the following:

2-3. - The Absolute Value The idea of the absolute value of a number I s easily under- -

stood from a few examples:

( 1) The absolute value of 5 I s 5.

(2) The absolute value of -5 13 5. (3) The absolute value of TT Is TT.

(4) The absolute value of -T is ir, and so on,

Graphically speaking, the absolute value of x is simply the distance between 0 and x on the number scale, regardless of whether x lies to the left or to the right of 0. The absolute

I Ã ‘ l x l Ã Ã‘IÃ‘Ã‘Ã‘I

x 0 x<o

The two p o s s i b i l i t i e s for

each of the two cases, ] x

value of x is written as 1x1 . *

1Ã‘Ã‘lX1 Ã‘Ã‘Ã‘Ã‘Ã‘I

0 x X > O

x are indicated In t h e figures. In

1 is the distance between 0 and x .

If a particular number is written down arithmetically, it l a

easy t o see how we should write I t s absolute value. The reason l a t h a t In arithmetic, the p o s i t i v e numbers are writ ten as I, 2, 3, 4, and so on. A way to wri t e negative numbers I s to put minus signs i n front of t h e positive numbers. This gives -1, -2, -3, -4, and so on. Therefore, In arithmetic, If we. want to" write the absolute value of a negative number, we merely omit the minus

sign, thua, - 1 = 1 - 2 = 2, and so on.

We would like t o give an algebraic def in i t ion f o r 1x1, and we would like the definition to apply both when x is p o s i t i v e and when x is negative. In algebra, of course, the l e t t e r x can represent a negative number. In working algebra problems, you

have probably written x = -2 nearly as often as you have written x = 2. If x is negative, then we can't write the corresponding

positive number by omitting the minus sign, because there i s n t t

any minus sign t o omit. There I s a simple device, however, that gets around our d i f f i c u l t y : if x I s negative, then t h e correspond-

ing p o s i t i v e number I s -x. Here are some examples: x = -1, -x = -(-1) = 1; that l a , if x a -1, then -x = 1.

x = -2, -x = -(-2) = 2; that is, if x = -2, then -x = 2. x = -5, - x = -(-5) = 5; that la, If x = - 5 , then -x = 5. In each of these cases, x I s negat ive and - x la the corre-

sponding p o s i t i v e number. And In fact , this l a what always

happens. Since we knew a l l along t h a t 1x1 = x when x Is positive or zero, It follows that the absolute value is described by the

following two statements:

(1) If x I s positive or zero, then 1x1 = x, (2) If x is negative, then 1x1 = -x . If t h i s s t i l l looks doubtful to you, try substituting various

numbers for x. No matter what number x you pick, one of the con- d i t i o n s above w i l l apply, and will give you the right answer f o r

t h e absolute value.

Problem Set 2-3 1. Indicate which of the following are always true:

a. 1-31 = 3 . b* 131 = -3.

C . 12 - 71 = 1 7 - 2 1 , d. 10- 51 = 1 5 - 01.

e. i n = n . * 2 . Indicate which of the following are always true: ,

e. ! d l + 1 = I d + 11. 3. Complete these statements:

a. If 0 < r, then = - b . If 0 > r, then lrl =-.

c . If 0 = r , t h e n r ! = - . 4. The following three examples give a geometric Interpretation

t o algebraic statements. x < 2.

All p o i n t s of the scale t o the l e f t of 2. - I I i I

-2 - 1 0 1

I 2

1 The set of points between 2 and -2. 1 -

T w o p o i n t s . - - A f I I I I 1 f -

-3 -2 -1 0 1 2 3

[sec . 2-31

Continue as above f o r the following problems:

a. x < 0. e. 1x1 = 1.

b. x = 1. f. I x l < l . c. x > 1. g . 1x1 > 1. d , x A 1. h. 1x1 2 0 .

5. a. How would the set of points represented by x 0

d i f f e r from the set represented by x -> O? b. HOW +would t he s e t of points represented by 0 <Â x <' 1

differ from the set represented by 0 <- x <- I?

2-4. Measurement of Distance - The f i r s t step in measuring the distance between two points

P and Q la to lay down a ruler between them, like thiat P 0

RULER

Of course we want to use a straight ruler, since we cannot expect to get consistent results if ou r rulers are curved or notched. A s t r a i g h t ruler has the property that however it is placed between P and Q the line drawn along I t s edge I s always the

same. In other words, this line Is completely determined by the

two given points . We express this basic property of lines as our f i rs t geometric postulate:

Postulate 1. Given any two di f fe ren t points,

there Is exactly one line which conta ins both of them.

We shall of ten refer to t h i s postulate, b r i e f l y , by saying that every two points determine a line. This Is simply an

abbreviated way of stating Postulate 1. To designate the l i n e determined by two points P and Q we use

the notation s. (The double arrow will recall our picture of the line. ) Of course we can always abbreviate by Introducing

a new le t ter and calling t h e line L, or W, or any th ing else. Now let us consider the marks on the ruler* and the actual

diatance between P and Q. The easiest way to measure the distance is to place t h e ruler l i k e t h i s :

P Q

This gives 7". Of course, there I s no need t o put one end of the ruler at P. We might put it like this:

P Q

In t h i s case, the distance between P and Q, measured in Inches, l a 9 - 2 = 7, as before.

On many rulers that are sold now, one edge I s l a i d o f f in

inches, and the other edge In centimeters. Using the centimeter scale, we can measure the distance between P and Q like t h i s :

This gives t h e distance as approximately 18 cm., where cm. stands f o r centimeters.

A foot I s , of course, 12", and a yard I s 36". A meter I s

a hundred centimeters; m, stands f o r meters* A millimeter I s i a t e n t h of a centimeter ( or =of a meter) ; m. stands for

millimeters. We can therefore measure the distance between

P and Q I n at least t h i s many ways: 18 em., 180 mm., .18 m , , 7 in., ft-, & y d s -

T h a t I s , t h e number we get, as a measure of t h e distance, depends

on the unit of measure. We can use any unit we like, as long as -- we use it consistently, and as long as we say what unit we are us ing .

Problem Set 2-4 -- 1. What common f ract ions (or integers) are needed to

complete the following table?

a. 2 in. = - ft. = yd. - b.

1 - In. = 4 -, ft. = yd. - in. = 2 =. - - ft. = 3 yd.

2 . What numbers are needed to complete the following table? a. 500 mm. = - cm. = - m. b. - mrn. = 32.5 cm. = m. - - mm. = - cm. = 7 . 3 2 m.

1 3. a. Suppose you dec ide to use the w i d t h of an 8 in. by 11 in.

sheet of paper as a u n i t of length. What I s the length and the w i d t h of the sheet in terms of t h i s u n i t ?

b. Repeat the problem with the length of the sheet as your

new u n i t .

4. I f the lengths of the sides of a t r iangle are 3 ft., 4 ft., 2 and 5 ft., I t Is a right triangle because 3 + 4 = 52,

V e r i f y t h a t t h e Pythagorean relationship still holds i f the

lengths above are expressed I n Inches.

5. If the length of each side of a square is 4 ft. I t s perimeter

is 16 ft. and I t s area I s 16 sq. ft. Observe that the numerical value of the perimeter is equal to t h e numerical

value of the area. a. Show that the numerical values of the perimeter and

area w i l l no longer be equal t o each other if t h e length

of t h e side is expressed In Inches.

b, In yards. *6. Generalize Problem 4. Given that the numbers a, b and c are

the number of u n i t s I n t h e sides of a triangle If some 2 2 par t i cu la r uni t of length I s used and t h a t a + b2 = c .

Show that t h e Pythagorean relat ionship k i l l s t i l l ho ld If the u n i t of length i s multiplied by n. (Hint: The lengths of

a b c the sides will become ;, Ã and . If a, b and c seem too abstract use 3, 4 and 5 at f i r s t . )

*7. Generalize Problem 5. Show that if the numerical values of the area and per imete r of a square are equal for some particular unit of measure, then they w i l l not be equal for any other unit. (Hint: Start by letting t h e number s be

the length of the s i d e of the square f o r some u n i t and equating the area and perimeter formulas.)

2-5. Choice of a Unit of Distance ---- We have noticed t h a t t he choice of a unit of distance I s

merely a matter of convenience. Logically speaking, one unit works as well as another, for measuring distances. Let us there- f o r e choose a unit, and agree t o talk i n terms of t h i s unit In a l l of our theorems, (it w i l l do no harm to think of our u n i t aa being anything we like. If you happen to like inches, feet, yards, centimeters, cubits, or furlongs, you are free to consider t h a t these are the u n i t s that we are using. A l l of our theorems will hold t r u e for 9 unit.) ---- -

Thus, to every pair of points, P, Q there w i l l correspond a number which l a t h e measure of the distance between P and Q In terms of our u n i t . Such numbers will be used ex tens ive l y In o u r work, and it would be very inconvenient to have to be continually

repeating the long phrase "measure of the distance between P and

Q in terms of ou r unit". We shall therefore shor ten t h i s phrase to "distance between P and Q", trusting that you will be able to fill In the remaining words if I t should ever be necessary.

We can now describe t h i s s i t ua t i on in the follow^ precise f om:

I Postulate 2. h he Distance Postulate. ) To every 1 1 pair of different points t h e r e corresponds a unique positive^ 1 number. 1 Definition. The distance between two points l a the posit ive

number given by the Distance Postulate. If the points are P and Q, then the distance I s denoted by PQ.

It w i l l sometimes be convenient t o allow the possibility

P = Q, that is, P and Q are the same point; in t h i s case, of

course, the distance is equal to zero. Notice that distance I s

def ined simply for a pair of points, and does not depend on the order in which the points are mentioned. Therefore PQ is always the same as QP.

Some of the problems you w i l l be asked to do w i l l Involve various units of distance, such as feet, miles, meters, e t c . As

noted above, our theorems will be applicable to any of these uni ts , p ~ o v i d e d you consistently yse Jus t one un i t throughout any one theorem. You can use Inches in one theorem and feet In another , If you wish, but not both in the same theorem.

2-6. - An I n f i n i t e Ruler A t the beginning ,o f this chapter we l a i d o f f a number-scale

on a line, like t h i s :

We could, of course, have compressed the scale, like t h i s :

or s t r e t c h e d It, l i k e t h i s :

But l e t us agree, f r o m now on, that every number-scale t h a t we

l a y off on a line is to be chosen In such a way t h a t the point labeled x lies at a distance 1x1 from the point labeled 0. For

example, consider the points P, Q, R, S, and T, labeled with the numbers 0, 2, -2, -3, and 4, as in the figure below:

Then PQ = 2, PR = 2, PS = 3 and PT = 4. If we examine various pairs of points on the number-scale,

it seems reasonable to f i n d the distance between two points by

t ak ing the difference of the corresponding numbers. For

36

example,

PQ = 2, and 2 = 2 - 0; Q T = 2, and 2 3 4 - 2; S Q = 5, and 5 = 2 - ( - 3 ) ; RT = 6, and 6 = 4 - (-2).

Notice, however, that if we look at the pairs of points in reverse order, and perform the subtractions In reverse order, we w i l l get the wrong answer every time: instead of get t ing the distance (which I s always pos i t ive ) , we w i l l g e t the corresponding negative number. This d i f f i c u l t y , however, is easy to get around. A i l we need t o do I s t o take the absolute value of the difference of the - numbers. If we do t h i s , then a l l of our positive right answers will still be right, and a l l of our negative wrong answers will

become right. Thus we aee that t h e distance between two p i n t s IEI the ----- -

absolute value of the difference of the corresponding -7- -- numbers.

Surety all thia seems reasonable. But surely we have not

proved it on the basis of the only postulatesthat we have written down so far. (And, In fac t , it cannot be proved on the basis of the Distance Postulate.) We therefore aum up the above discussion In the form of a new postulate, like this:

I Postulate 3 . h he Ruler Postulate. ) The po in t s of - a line can be placed i n correspondence with the real

numbers In such a nay that (1) To every point of the line there corresponds

exactly one real number,

(2) To every real number there corresponds exact ly

one point of the line, and

( 3 ) The distance between two points Is the abso-

lute value of t he difference of the corresponding numbers .

We c a l l t h i s t he Ruler Pos tu l a t e because In effect it fur -

nishes us with an I n f i n i t e ruler, wi th a number-scale on It, with

which we can measure distances on any line. Definitions. A correspondence of t h e sor t d e s c r i b e d In

Pos tu la te 3 is called a coordinate system f o r the l i n e , The

number corresponding to a given point I s cal led the coordinate

of the point. Problem Set 2-6 --

1. Simplify:

a. 1 3 - 61. d. 1-4 - (-2)l. b. 16- 31. e. la - (-a)l. c. 1-2 - 1 f. Ja l - I-a(.

2. Using the kind of coordinate system discussed in the text, f ind the distance between poin t pairs w i t h the following coordinates : a. 0 and 12.

b. 12 and 0.

c. 0and -12.

d. -12 and 0.

f. -5.1 and 5.1.

g. -/"? and 0. h . x, and x2. i. 2a and -2a.

1 e. -3 3 and -5 . j. r - s and r + s .

3.

The lower numbering on this scale was put there by Jim.

Pete began the upper numbering but qui t . a. Copy the scale and write in the rest of Pete I s numbering.

b. Show how to f i n d the distance from P to Q, f irst by using Jim's scale and then by using Pete's scale.

c . Do the same f o r the distance from ^f to P. 4. Suppose in measuring the distance between two points P and Q

you intended to place t h e zero of t h e number-scale at P and

read a posi t ive value at Q. However, you happen to place the 1 number-scale so that P is at -n- and Q is farther to t he r i g h t .

How is it still possible to measure the distance PQ? *5. Consider a coord ina te system of a line. Suppose 2 I s added

t o the coordinate of each point and t h i s new sum 13 assigned

to the point. a. W i l l each point then correspond to a number and each

number t o a point? to. If two points of t h e line had coordinates p and q in the

coordinate system given, what numbers are assigned to them In the new numbering?

c. Show t h a t the formula (Number assigned to one point)-(Plumber assigned to"

o t h e r point) [ gives the distance between the two points. d . Does the new correspondence between points and numbers

satisfy each of the three conditions of Postulate 3? (if it does it may- be called a coordinate system.)

* 6 . Suppose a coordinate system is set up on a l ine 30 t h a t each po in t P corresponds to a real number n, If we replace each n by -n, then the point P will correspond to a number -n. Show t h a t t h i s correspondence Is a l s o a coordinate system for the line. ( H I N T ; It is apparent that each point will have a number associated w i t h It and each number a point . You must show In addition that the absolute value of the difference

of the numbers assigned to the two points w i l l , remain unchanged when the numbering l a changed. )

7 . In a certain county the towns of Alpha, Beta and Gamma are colllnear ( on a l ine) but not necessarily In that order. It is 16 miles from Alpha to Beta and 25 miles from Beta to

Gamma. a . Is it possible to tell which town l a between t h e o the r

two? Which town l a not between the other two? - b. There migh t be two different values f o r the distance

from Alpha to Gamma. Use a sketch to determine what

these are.

c . If you are given the a d d i t i o n a l informat ion that t h e d i s -

tance from Alpha t o Gamma i s 9 miles, then which town is

between the other two?

d . If t h e distance between Alpha and Beta were r miles, the distance from Alpha t o Gamma s miles, and the d i s t a n c e from B e t a t o Gamma r + s miles, which city would be


A, B, C are three c o l l l n e a r points. A and B are 10" apart, and C I s 15" from B. Is there just one way to arrange these po ints? Explain. Three different coordinate sys tems are assigned t o the same line. Three f i x e d points A, B, C of the l i n e are assigned values as follows:

With the f i r s t system the coordinate of A Is -6 and t h a t of B I s -2.

With the second t h e coordinates of A and C are 4 and -3 respectively.

With the third the respective coordinates of C and B are 7 and 4. W h a t point l a between the o the r two?

Evaluate AB + BC + AC.

The Rule r Placement Postulate - Betweenness- Segments 2-7 And Rays - The Ruler P o s t u l a t e (Pos tu la t e 3) te l l s us that on any line,

we can set up a coordinate system by laying off a number-scale. Thia can be done i n l o t s of di f f erent ways. For example, given

a point P of the l i n e , we can start by making P t h e zero-point . And we can then l a y o f f t h e scale In either d i rec t ion , like t h i s :

P - t I I I 1 I I

-3 -2 -1 0 1 2 3

This means that given another point Q of the line, we can always choose the coordinate system in such a way that Q corresponds t o

a p o s i t i v e number, like t h i s :

P Q f I I I I 1 I

1 -3 -2 -1 0 1 2 3

or t h i s :

Let us w r i t e t h i s down, for future reference, in the form of a

postulate . Postulate - 4. he Ruler Placement Postulate.)

Given two points P and Q of a line, the coordinate

system can be chosen in such a way that the coordinate of P is zero and the coordinate of Q is positive.

Everybody knows what It means to say that a point B I s between two points A and C. It means that A, B and C l i e on

the same line, and that they are arranged on the line like t h i s :

# I I I I

A B C or like this:

If we are going to use betweenness as a mathematical Idea, however, we had better give a mathematical definition t h a t state8 exactly what we mean, because t h e feelings that we have In our bones are not necessarily reliable. To see this, let us look at the corresponding situation on a circle. In the figure on the left,

it seems reasonable to say that B is between A and C. But C can be moved around the c i r c l e In easy stages, without passing over A or B, so as to l i e Jus t to the left of A, as in the right- hand figure. In the final position, I n d i c a t e d by the exclamation point, It looka as if A is between B and C. In this respect,

circles are tricky. Given any three points of a circle, It I s

quite reasonable t o consider that each of them I s between the

other two. Betweenness on a line is not at a l l tricky. It I s easy to

say exactly what it means for one point of a l ine to be between two o thers . We can do t h i s In the following way:

Definition. 3 l a between A and C If (1) A, 3 and C are distinct points on the same line and (2) AB -I- BC = AC.

It is easy to check that t h i s definit ion really expresses our common-sense Idea of w h a t betweenness ought to mean. It may be a good idea t o explain, however, the way In which language is ordinarily used In mathematical definitions. In the d e f i n i t i o n of betweenness, two statements are connected by the word if. What we really mean I s that the statements before and after the word - If are completely equivalent. Whenever, In some theorem, or problem, we are given or can prove that conditions (1) and ( 2 )

both hold, then we can conclude that 3 l a between A and C. And

whenever we find that B is between A and C then we can conclude that (1) and ( 2 ) both hold. T h i s I s not a s t r i c t ly logical use of the word - if, and parthular the word never uaed In t h i s way In postulates, theorems or problems. In definitions, however, i t l a common.

The following theorem describes betweenness in terms of coordinates on a line.

Theorem 2-1. L e t A, B, C be three points of a l ine , with - coordinates x, y, z . If

X < Y < z , then B is between A and C.

Proof: Since x < y < z, we know that the numbers y - x, z - y, and z - x are a l l posi t ive . Therefore, by def ini t ion of the absolute value,

Therefore, by the Ruler Postulate,

AB = y - X ,

m = z - y , AC 3 z - X.

Theref o r e A B + BC = ( y - X ) + ( Z - y)

- x + 2

= z - x = AC.

Therefore , by the definition of betweemeas, B i s between A and C,

which was to be proved.

Problem Set 2-78 -.- - A number-scale I s placed on a line with -3 falling at R and 4 at S. - If the Ruler Placement Postulate is applied with 0 placed on R and a positive number on S , what will this number be?

Same question 1f -4 f a l l s at R and -10 a t S.

Same question if 8 fa l l s at R and -2 at S. Same question if -4 Ã fa l l s at Rand 4 at S. Same question if 5.2 f a l l s a t R and 6.1 a t S. Same question if x-, f a l l s a t R and 5 at S .

2. Explain br ie f ly how t h e Ruler Placement Postulate simplifies the procedure given by the Ruler Postulate f o r computation of distance between two points,

Suppose R, 3 and T are collinear points. What must be true of the lengths RS, ST and RT If S is to be between R and T? ( See definit ion of between. )

A C B AC and BC each equals 8.

The coordinate of C is 6. The coordinate of B I s greater than the coordinate of C. What are the coordinates of A and B? If a, b and c are coordinates of collinear points, and If

a - c l + l c - b) = la - bl, what I s the coordinate of t h e po in t which l i es between the other two? Be able to juatify y o u r answer.

If x,, x2 and x3 are coordinates of points on a l ine such that x3 > x, and x2 < xl, which point is between the other

two? Which theorem would be used to prove your answer?

Consider a coordinate system in which A is assigned the number 0, B is assigned the positive number ry E the number 1 2 r, and F the number 3 r. Prove that: a. A E = E P = F B b. E is between A and F.

Prove: If A, B and C are three points of a line with

coordinates x, y and z respectively and If x > y > z, then B is between A and C.

Theorem=. Of any three di f ferent points on the same line,

one I s between the o the r two.

Proof: L e t t h e points be A, B and C. By the Ruler Postulatey there I s a coordinate system f o r the l ine . L e t the coordinates of A, B, and C be x, y, and z . There are now six possibilities:

(1) x < Y < 2, (2) x < < Y, (3) Y < x < 2,

( 4 ) Y < < x,

( 5 ) Z < X < Y ,

( 6 ) < Y < x. In each of these cases, Theorem 2-2 fol lows by Theorem 2-1.

In cases (1) and (6), B l a between A and C. In cases (2) and

f 4 ) , C Is-between A and B. In cases ( 3) and (5 ) , A I s between B and C.

Theorem u, Of three different polntg on the a m 0 lac, only

one la between the other two.

Restatement. If A, B and C are three different points on the same line, and 3 l a between A and C , then A I s not between B and C, and C is not between A and B.

I t o f t e n happens that a theorem Is easier t o read, and

easier to refer to, If It l a s t a t e d In words. But t o prove

theorems, we usually need t o s e t up a notation, giving names to

the objects that we w i l l be talking about. For th i s reason, we shall o f t e n give restatements of theorems, I n the s t y l e that we have just used f o r Theorem 2-3. The restatement gives us a s o r t of head-start In the proof. )

Proof: If 3 l a between A and C, then

AB + BC = AC. ,

If A Is between B and C, t h en BA + AC = BC.

What we need to prove is t ha t these two equations cannot both

hold at the same time. If the f irst equation holds , then

AC - EC = AB.

If the second equation holds, then AC - BC = -BA e -AB.

Now A 3 is positive, and -AB is negative. Therefore, these equations cannot both be true, because the number AC - BC cannot be both positive and negative.

[sec. 2-71

In an ent i re ly similar manner we can show t h a t C is not between A and B.

Definitions. For any two points A and B the segment is

the set whose points are A and B, together with a l l points that

are between A and 3. The points A and B are called the - end- points of AB. -

Notice that there is a big difference between the segment and the distance AB. The segment l a a geometrical figure, that Is, a s e t of points. The dis tance is a number, which t e l l s us how far A Is from 3.

Deflnltion. The distance AB is called t h e length of the segment m.

A ray Is a f igure that looks like t h i s :

A B The arrow-head on the right Is meant t o Indicate that the ray in-

cludes all points on the l i n e t o the r ight of the point A, plus - the point A Itself. The ray la denoted by a. Notice that when

we write a, we simply mean the ray that starts at A, goes through

3, and then goes on In the same direction forever. The ray might

look l i k e any of ths following:

1, That is, the arrow In the symbol AB always goes from left to right, regardless of how the ray is pointed In space.

Having explained informally what we are d r i v i n g at, we proceed

t o give an exact definition.

Definltlona. Let 4 and 3 be points of a line L, The 2 Is the set which is the union of ( 1) t h e segment and (2) the

s e t of all points C for which It Is true that B ig between A and 0. The point A Is called the end-point of f i .

These two parts of the ray are as indicated: . a

(I) yy /-Â ' -S f -

A 0 C If A is between B and C on L, then the two rays A and &S

' 'go in opposite direction, like this : 3 AB

- - \_/^ - - - -

C A 6 Beflnltton. If A I s between B and C, then ($ and 8 are

c a l l e d opposite mys, Note that a pair of points A, B determines six geoometrlc

figures : 0

The line A3,

The segment E, Â¥Ã‘Ã‘Ã‘Ã‘Ã

The ray &, -Ã

The ray BA,

The ray opposite to 3, - -Ã

The ray opposite to BA.

The Ruler Placement Postulate has three more simple and

useful consequences. Theorem 2-4, h he Point Plotting heo or em) L e t SB be a ray,

and let x be a p o s i t i v e number. Then there Is exactly one ->

poin t P of AB such that AP = x .

Proof: By the Ruler Placement Postulate, we can choose the

coordinate system on the line In such a way that the coordinate of A I s equal to 0 and the coordinate of B Is a posi t ive number r:

----------- A B P - 1 r

0 r x

(In the figure, the labels above the line represent points, and the labels below the l i n e represent the corresponding numbers.)

Let P be the point whose coordinate I s x. Then P belongs to i6, and AP = !x - 01 = 1x1 = x, because x la positive. Since only one point of the ray has coordinate equal t o x, only one point of the ray l i es at a distance x from A.

Definition. A point B I s called a mid-point of a segment - AC if B l a between A and C, and AB = BC.

T h e o r e m s . Every segment has exactly one mld-point..

Proof, On the segment we want a point B such t h a t AB = BC. We know, by definition of a segment, that 3 is between A and C. Therefore, AB + BC = AC. From these two equations we conclude

1 t h a t 2AB = AC, or AB a - AC. Since B is to lie on segment AC 2.

it must also lie on ray AC, and Theorem 2-4 te l l s us tha t there l a exactly one such point B.

Definition. The mid-point of a segment 1s s a i d to bisect the

segment. More generally, any figure whose Intersection w i t h . a segment is the mid-point of the segment I s s a i d t o bisect the

segment.

Problem Set 2-7b - 1. If three poin ts are on a line, how many of them are not


2. Each of the following la a particular case o f what definition or theorem?

If three colllnear points R, S and T have coordinates respectively 4, 5 and 8:

a. S I s between R and T because 4 < 5 and 5 < 8, b. R cannot be between S and T aince S I s between

R and T. c. S I s between R and T because RS + ST = RT.

4 3, Describe in mathematical language what points are included In:

- 4 a. XY b. XY

*4. Show that the restriction "between A and C" in the def in i t i on of the midpoint of AC is unnecessary by proving the

following theorem: If B i a any point on the line 8 such that AB = BC, then B 3.5 between A and C. (~int: Show t h a t A cannot be between B and C nor 0 between A and 3. Use algebra In

showing this. Use Theorem 2-2 to f i n i s h the proof.)

*5. Suppose that P is a point on a line M and r I s a posi t ive number. Which of the previous theorems shows that there are exactly two points onM whose distance from P l a the

given number r?

*6. Prove that If B is between A and C, then AC > AB.

7 . a. Copy the following paragraph. Supply the appropriate missing symbol, if any, over each letter pair.

XZ contains pointa Y and R, but XZ contains neither points Y nor R. . R belongs to XZ but Y does not.

YZ + ZR = YR.

b. Make a d r a w i n g showing the relative position of the four points.

Review Problems 1. Consider the following sets:

S, 18 the set of a l l boys in the 10th grade. Sg is the set of all girls I n the 10th grade.

3, I s the set of a l l 10th grade geometry students. SL Is the set of all students In high school. S I s the set of all 10th grade students. 5

a. What l a the intersection of S,and S ? 5 b. What I s the union of Sa and 3h?

c. What l a the Intersection of Sg and Sh?

d . What i s the union of S., and Sg?

e. What is t h e intersection of S-, and Sp?

2 . a. How many squares does a given pos i t ive number have? b. How many square roots? c . Is i/^ever negative?

3. a. Draw a line and locate the following points on It.

(The coordinate of each point is given in parentheses. ) Use any unit of measure you choose, but use the same unit throughout.

P (21, Q (-11, R (01, s ( -31 , T ( 4 ) . b. Find PQ, RT, TR, FT, QS.

4 . a. If a > b, then a - b is - b. If 0 < k and k < 4, then k I s . c. If a < b then a - b is .

a. Write a n equation t h a t describes the relative positions of these three points.

b. Under what condi t ion would 3 be the midpoint of z? 6 Pour points A, 3, C, D are arranged along a l i n e so that

AC > AB and BD < BC. Picture the line with the four points in place. Is t h e r e more than one possible order? Explain.

7. The l e t t e r pairs contained i n the following paragraph are

either numbers, lines, line segments, or rays. Indicate which each l a by placing the proper missing symbol, If any, above each letter pair.

'AB + BC = AC. DB contains points A and C, but DB con- taina neither point A nor point C, A belongs to DB but C does not." D r a w a picture t h a t i l lustrates your response ,

8. A Is the s e t of a l l integers x and y whose sum is 13. B l a

the set of a l l integers whose difference is 5 . What is the

Intersection of A and B?

9. John s a i d , "My house I s on West Street halfway between Bfll*~ house and Joe's house." P e t e said, "So l a mine!" What can you conclude concerning John, and Pete?

10. N men a i t on a straight bench. Of how many may I t be said, 'He a i t s between two people?"

11. Use the figure below t o answer questlona a. through e.:

a. Describe the intersection of triangle AEP and rectangle ABCD.

b. Describe the intersection of segment and

rectangle ABCD. - - c. Describe the union of segments AF, EP, and E. d. Describe the Intersection of segments and E. e. Describe the union of triangle AEP and segment z.

12. Given a group of five men (Meaara . Andrews, Brown, Crawford, Douglas, and ~ v a n s ) . a. ~ r o m t h e group, how many different 4-man committees can be formed? b. 2-man? c. 3-man?

13. Given that A, B and C are co l l inea r and that A 3 = 3 and EC = 10, can AC = 6? U v e a picture to explain your answer.

14. Indicate which of the following statements are true and

which are fa lse . For any that are falae, give a correct

Looking at t h i s number-scale, Jack said, h he length of RQ is 1 x-y 1 .'I S a m maintained that when giving the length of

it would be Just as correct to use simply y-x. Do you agree with Sam? Explain.

16. The first numbering of the points on the l ine below represents a coordinate system. Which of the other numberings are not coordinate systems according to Postulates 2 and 3?

- . . -7

-3 -2 -1 0 1 2 3 4 5 6 7 a. -7 -6 -5 -4 -3 -2 -1 o 1 2 3

b. 0 1 2 3 4 5 4 3 2 1 0

c. 11 12 13 14 15 16 17 18 19 20 21

d . -11 -12 -13 -14 -15 -16 -17 -18 -19 -20 -21

e . -3 -2 1 0 -1 2 3 4 5 6 7 *17. Consider the points of a line whose coordinates are descr ibed

as follows: a* x < 3. e . x = - 3 . b. x - 1 . f. 1x1 < 3. c. x > 2. 6. 1x1 > 2. d. x < 1. h. I x l > O .

Whlch of the above sets l a a ray? A polnt? A line? A segment?

Chapter 3

LIKES, PLANES AND SEPARATION

3-1. Lines and Planes in Space, -- In t h e l a s t chapter, we were t a l k i n g only about l i n e s and the

measurement of distance. We shall now proceed t o the study of

planes and space. We recall -that our basic undefined terms are point, line and plane. Every line I s a set of points , and e v e ~ y plane Is a set of poin ts .

D e f i n i t i o n . The s e t of all po in t s Ts called space. In t h i s section we will explain some of the terms we are go-

ing t o use in talking about points, lines and planes, and s t a t e

some of t h e basic facts about them. Most o f these basic facts w i l l be stated as pos tu l a t e s . Some of them w i l l be s ta ted as theorems.

These theorems w i l l be so simple that it would be reasonable t o

accept them w i t h o u t p roof , and call them postulates. We do not do this, however; the first of them is going to be proved In t h i s sec-

tlon, and the r e s t of them will be proved, on the basis of t h e

postulates, in a later chapter. For the present, however, let us

not worry about this question, one way or another; l e t us simply t r y t o get these basic facts straight.

Problem Set 3-la -- 1. On a piece of paper, or on t h e blackboard, place two marks to

represent poin ts A and B. How many dif ferent lines can you draw through both A and B? What happens i f you consider "linel1 In a sense other than "straight"?

2. Take a piece of st iff cardboard or your book. Can you support it in a fixed position on the ends of two pencils? What is

the minimum number of penc i l s needed to support it in this

way? 3. Think of one cover of your book as p a r t of a plane. How many

points are needed to determine t h i s plane?

4. How many end-points does a line have? How many end-points does

a line segment have?

Definition, A set of poin ts is colllnear if there Is a line which contains all the points of the s e t .

Definition, A set of points is coplanar if there I s a plane

which contains a l l t h e points of the s e t ,

For example, In the above figure of a triangular pyramid, A, E and B are colllnear, and A, F and C are colllnear, but A, B and

C are not colllnear. A, B, C and E are coplanar, and A, C, D, F and G are coplanar, but A , B, C and D are not coplanar.

One of the properties we desi re for the s e t s of points which

we c a l l lines, planes and space is that they should contain lots of points Also, a plane should in some sense tie "biggertf than a

l i ne and space should be 'b igger than any plane. The existence of plenty of po in t s on a l i n e is insured by the Ruler Postulate; for planes and space the following postulate w i l l give us the pro-

perties we want:

Pos tu la t e 5. ( a ) Every plane contains at least three non-colllnear points.

(b) Space contains at least four non-coplanar points,

For convenience in reference we repeat Postulate 1.

Postulate 1. Given any two different points, there Is exactly one l ine which contains t h e m .

Theorem 3-1, Two different lines Intersect in a t most one point.

The proof of t h i s follows from Postulate 1. It is Impossible for two different lines to intersect in two different po in t s F and

Q because by Postulate 1 there is only - one l i n e t h a t contains P and

Q

Problem S e t 3- lb -- 1. Given: 1. Li and Lp are different lines.

2. Poin t P lies on L, and Lp. 3 . Point Q lies on L, and Lp.

What can you say must be true about P and Q? 2. How many lines can contain one given point? two given points?

any three given points? Â¥Ã‡Ã‘> w

3. The diagram shows three different lines AB, CD, and EF, whose Â¥Ã‡

view is partially obstructed by a barn. If B and CD intersect t o the left of t h e barn, which postulate says t h a t they cannot a l so intersect to the right of t h e barn?

11. Draw a diagram to illustrate each part of t h i s problem and jus-

t i f y your answers in terms of Postulate 1. a. How many l ines can be drawn through both of two f ixed

points?

b. How many lines can be drawn through three points taken two

at a t i m e ? 5 , a. How many lines can be drawn through four coplanar points,

taken two at a time, if no three of the points are collinear? ( ~ i n t : Call the points A, 5, C, D. )

b. How many lines would there be if p o i n t s A , B, and C were collinear?

c . Draw a diagram f o r (a) and (b) . *6 . "A point lies on a line" and "a line conta ins a point" a re two

forms of saying the same thing. a . The d e f i n i t i o n s : ~ ? collinear and coplanar are phrased using

the second form. Rephrase these def in i t ions using the first form.

b. The first part of Postu la te 5 I s phrased u s ing the second form. Rephrase t h i s part of Postulate 5 using the F i r s t

form . *7. As in Problem 6, Postulate 1 Is written in one of the two forms

Which form? Restate Postulate 1 in the other form.

By Pos tu la t e 5 a plane contains a t least three poin ts . Does it contain any more? On the b a s i s of our present postulates we cannot conclude that it does, so we introduce

Postulate 6 . If two points l i e in a plane, then t h e l i n e containing these p o i n t s l i e s in the same plane.

This postulate essentially says that a plane is flat, that is, that if it contains part of a line it contains the whole line.

Theorem 3-2. If a line Intersects a plane not conta in ing it, then t h e Intersection i s a single point.

This follows from Postulate 6 i n the same way that Theorem 3-1 follows from Postu la te 1.

The figure shows line L, intersecting a plane E in a point P. You are going to see lots of drawings like t h i s , of figures in space, and to learn to draw them yourself. You should examine them carefully t o see how they work. We usually Indicate a plane E by d r a w i n g a rectangle in E. Seen Ln perspective, the rectangle looks somewhat like a parallelogram. The l ine L punctures E at I?. Part of L I s dotted. This I s the part t h a t you "can't seett, because the

rectangular piece of E gets In the way. (For a discussion on drawing 3-dimensional figures see Appendix V . )

We have seen that two points determine a l ine . The next postulate specif ies a similar determination of a plane.

Postulate 7 . Any three points l i e in at least one

plane, and any three nun-collinear points l i e in exactly 1 one plane. More brief ly , any three points are coplanar, 1 1 and any three non-colllnear points determine a plane.

Theorem 3-3. Given a l ine and a point not on the line, there is exactly one plane containing both of them.

The figure shows a plane E determined by the line L and the point P.

Theorem 3-4. Given two intersecting lines, there is exactly

one plane containiry them.

The figure shows two lines L, and Lp, intersecting in a point P. E I s the plane that contains both lines.

Finally, we state one more postulate;

Postulate 8. If two di f fe ren t planes intersect , i then their inter~ectlon is a lme.

Problem Set 3-lc

1. How many planes can contain one given point? two given points?

t h ree given points?

On a level floor, why will a four-legged table sometimes rock, while a three-legged table is always steady?

Complete; Two different l ines may intersect in a 9

and two d i f f e r e n t planes may Intersect in a Can two poin ts be non-colllnear? three points? fou r points?

n points?

Write a careful definition of a set of non-colllnear points. Given: 1. Points A, 3, C lie in plane E.

2. Points A, B, C lie i n plane F. Can you conclude that plane E is the same as plane F? Explain.

Complete the following statements using the accompanying diagram.

a. Points D, C , and - are colllnear. b . Points E, F, and - are col linear, c. Points B, -, and A are collinear . d . P o i n t s A , B , C , D , E , F , a r e A AB E

Examine the following figure of a rectangular s o l i d until you see how it looks as a three-dimensional drawing. Then close

the book and draw a figure l i k e this for yourself. Practice until you are sa t i s f ied w i t h the results.

After doing Problem 8, draw a figure that represents a cube. Draw a plane E, using a pa-rallelogram to indicate the plane.

Draw a l ine segment which lies in the plane E. Draw a l ine that Intersects the plane E but does not intersect the line seg-

ment. Use clashes to represent the part of the line hidden by the plane.

11. The accompanying figure is a triangular

pyramid, or tetrahedron. It has four vertices: A, B, C, D, no three o f

which are colllnear. a. Make a definit ion of an edge

of th is tetrahedron. Use the I deas of t h e t e x t to h e l p you

form the defini t ion. b. How many edges does the tetrahedron El

have? Name them. c . Are there any pairs of edges t h a t do not Intersect? d . A face is the triangular surface determined by any three

vertices. There are four faces; ABC, AED, ACD, BCD, Are . there any pairs of faces that do not Intersect? Explain.

12. How many different planes (determined by D tr ip le ts of labeled p o i n t s ) are there in the pyramid shown? Make a complete list, D> (YOU should have seven planes. ) ^"^

# ^

A 8 3-2. Theorems ---- in the Form of Hypothesis and Conclusion. -

Nearly every theorem is a statement that - If a certain thing is true, then something else is also true. For example, Theorem 3-1 states that - if Ll and L2 are two different lines, then Ll Intersects Lp in at most one po in t . The - if part of a theorem i s cal led the hypothesis, OF t h e ~ 5 v e n data, and the - then part 1s called the E- elusion, or the thing to be proved. Thus we can write Theorem 3-1 in this way:

Theorem 3-1. Hypothesis: Li and L2 are two different lines. Conclusion: L., intersects L2 in at most one

point.

Postulates, of course, are like theorems, except t ha t they are

not going to be proved. Most of them can be put in the same if then form as theorems. Postulate 1 can be stated th i s way: -

Hypothesis; F and Q are two different po in t s . Conclusion: There is exactly one line containing P and Q. There are cases in which the hypothesis-conclusion form does

not seem natural or useful. For example, the second part of Postu-

late 5, expressed in this form, looks awkward: Hypothesis: S is space. Conclusion: Not a l l points of S are coplanar.

Such c a s e s , however, are very rare. It is not necessary, of course, that all theorems be stated I n

the hypothesis-conclusion form. It ought to be clear, regardless

of the form in which t h e theorem is stated, what part of it is t h e

hypothesis and what part is the conclusion. It is very important, however, that we be able to s ta te a theorem i n t h i s form if we want to, because if we cannot, the chances are tha t we do not understand exactly what the theorem says.

Problem Set 3-2

1. Indicate which part of each of the following statements is the hypothesis and which part is the conclusion. If necessary, rewrite in if-then form first. a . If John is ill, he should see a doctor . b. A person with red hair is nice to know. c. Four points are coll inear if they lie on one line. d. If I do my homework well, I will get a good grade.

e . If a s e t of poin ts lies inone plane, the points are coplanar.

f. Two intersecting lines determine a plane. 2. Write the following statements in conditional, or if-then, form:

a. Two different l ines have at most one point In common. b. Every geometry student knows how t o add integers. c. When it rains, it pours. d . A l ine and a point not on the l ine are contained in

exactly one plane.

e . A dishonest practice is unethical.

f. Two parallel lines determine a plane.

3. Using the words "lf" and "then", write Postulate 1 and Theorem 3-1 in conditional form. Indicate the hypothesis and the conclusion for each case.

4. a. Does the following statement mean the samething a s Theorem

3-4? " ~ w o lines always intersect in a point, and there I s exactly one plane containing them." Why or why not?

b. Write Theorem 3-4 in the "hypothesis and conclusion" form.

3-3. Convex Sets. Definition. A set A is called convex if f o r every two points

P and Q of A, the entire segment l i e s In A .

For example, the three sets pictured below are convex.

Here each of the sets A, B and C consists of a region of the plane. We have Illustrated their convexity by showing a f e w segments F. None of t h e sets D, E and F below is convex:

D E F

We have shown why not, by showing pairs of points P, Q for which t he segment ^5 does not lie entirely in the given set.

A convex set may be very large. For example, take a line L in a plane E and let H, and H2 be the sets lying on t h e two sides of L, l i k e t h i s :

The two sets Hi and Hp are called half-planes or sides of L, and

the l i ne L is called an edge of each of them. (Notice that L does

not l i e in either of the two half-planes; L is not on either s ide

of Itself . ) If two points P and Q are in the same half-plane, say HI, then

the segment 7S[ a l s o lies in Hi, and so does not intersect L.

Thus Hz is convex. And in the same way, Hg is convex; t h i s is

illustrated by the points R and S in the figure. We notice, however, that if T and U are points belonging to

different half-planes, then the segment intersects L, because you cannot get from one side of L to the o ther side without crossing

the edge. We express t h i s

H2 in the plane, o r that L Hl and Hz.

fact by saying that I. separates H, from separates the plane in to two half-planes

T h i s discussion I s a fair account of the facts, but it is not very good mathematical form, because It is based on a postulate that w e haven't even stated so far. We shall therefore state t he

postulate t h a t is needed, and then s t a t e the definitions that are

based on it.

Postulate 9 . (The Plane Separation Postulate. ) Given a line and a plane containing It. The points of

the plane that do not l i e on the l i n e form two sets such

t h a t (1) each of t h e s e t s is convex and (2) if P is in one set and Q is In the other then the segment T intersects t h e line.

Definitions. Given a line L and a plane E containing It, the two sets determined by Postulate 9 are called half-planes, and L I s called an edge of each of them. We say that L separates I3 In to the two half-planes. If two points P and Q of E l i e In the same

half-plane, we say that they lie ---- on the same side of L; if P lies In one of t h e half-planes and Q in the other they l i e on opposite sides of L. -

We see that the Plane Separation Postulate says two things

about the two half-planes into which a l i n e separates a plane:

(1) If two points l i e in t h e same half-plane, then the seg-

ment between them lies in the same half-plane, and so - never Inter- s e c t s the line.

(2) If two points l i e in different half-planes, then the segment between them always Intersects the line.

If we do not restrict our a t t en t i on t o a single plane we can have many half-planes with t h e same edge. The picture

f i v e o f the infinitely many poss ib l e half-planes having line L for edge. Note that points P and Q, although they l i e in d i f f e ren t half-planes, cannot be said to be on opposite sides of L. This can only be said of points l i k e P and R which are coplanar with L.

A plane separates space, in exactly the same way, i n t o two convex sets cal led half-spaces.

P

In the figure, Hi is the half-space above E and H p i s t h e half-

space below E. P and Q l i e in H,, and so also does the segment 7^. P and S are in different half-spaces, so tha t the segment 7S ' intersects E in a point X. R and S are in the same half-space H2, and so also I s the segment E.

This situation is described in the following postulate.

Postulate 10. (The Space Separation Postulate . ) The points of space t h a t do not lie In a given plane form two s e t s such that (1) each of the sets I s convex

and (2) if P is In one set and Q is in t h e other, then

t h e segment P^ intersects t h e plane.

Definitions. The two sets determined by Postulate 10 are called half-spaces, and t he given plane I s called the face of each - - of them.

Note that while a line is an edge of i n f in i t e ly many half-

planes , a plane is a face of cn ly two half-spaces,

Problem Set 3-3

In answering the following quest ions use your intuitional understanding of planes and space in situations not covered by our pos tu la t iona l structure. 1. Be prepared to discuss the following questions orally.

Is a l ine a convex set? Explain. Is a s e t consisting of only two points convex? Why?

Is a ray a convex set? If one point I s removed from a line, do the remaining po in t s form a convex set? Why?

Is the set of points on the surface of a sphere convex?

Why? Is the space enclosed by a sphere a convex set?

Does a point separate a plane? space? a line? Does a ray separate a plane? Does a line? Does a line

segment?

Can two lines in a plane separate the plane i n t o two regions? Three regions? Four regions? F i v e regions? In to how many parts does a plane separate space? What are these parts called?

Every point on ?Q is contained In the set shown. Does th i s mean that t h e s e t I s

convex? Explain.

Which of the regions indicated by

Roman numerals are convex sets?

Give reasons f o r your choice.

Is every plane a convex set? Explain. Which postulate I s

essential in your explanation? The interiors of circles A and B

are each convex sets. a. Is the i r Int&*secfcion a convex

set? Illustrate . '3D b. Is their union a convex set? Illustrate.

If one point is removed from a plane, is the s e t formed convex? Why?

If L I s a l ine in a plane E, i s the set of a l l points of E on one side of L a convex set?

Draw a plane quadrilateral ( a figure with four sides ) whose i n t e r io r 1s convex. D r a w one whose in te r ior I s not convex. Is the set of points containing all points on the surface and a l l paints i n ?!be i n t e r i o r of a sphere convex? Is the set of points in a to rus ( a doughnut shaped figure) convex? Is the union of t w o half -planes which are contained in a plane the whole plane if a. the half-planes have the same edge? Explain.

b. the edge of one half-plane intersects t h e edge o f the

other half-plane In exactly one point? Explain, using a diagram if necessary.

a. Into how many parts does a point on a line separate the

line? What name would you suggest giving to each of these parts?

b. Using the terminology you developed In part ( a ) , write out a Line Separation Statement similar to Postulates9 and 10.

13. How does a ray di f fer from a half-line? 14. Can three lines in a plane ever separate the plane into three

regions? four regions? f i v e regions? s ix regions? seven regions?

15. Into how many parts do two intersecting planes separate space?

Two parallel planes?

16. What is the greatest number of parts i n t o which space can be separated by three distinct planes? What I s the least number?

*17. Wri-te a careful explanation of why the following statement is true. The intersection of any two convex sets which have at

least two points i n common is convex. (Hint: L e t P and Q be

any two points belonging to the intersection.) *l8. Sketch any geometrical solid bounded by plane surfaces such

that t h e set of points in the interior of the figure is not

convex.

Review Problems

1. Each of 3 planes intersects each of the others. May they

intersect i n one line? Must all three intersect in one l ine?

Explain.

2. How many planes will contain the three given points A, B, and

C if no l i n e contains t h e m ?

3. Write each of the following statements In the "if -then" form.

a. Zebras with polka dots are dangerous.

b. Rectangles whose sides have equal lengths are squares. c . There will be a celebrat ion if Oklahoma wins.

d. A plane is determined by any two intersecting lines. e. Cocker spaniel dogs are sweet tempered.

4. Supply the following information about the postulates in the chapter.

What prope r ty of each of the half-planes is mentioned in the Plane Separation Postulate?

Do the half-spaces of the Space Separation Postulate have the same property?

Cri t i c ize the following statement: 'The t o p of the table is a plane."

List a l l the situations we have studied which determine a single plane. A s e t is convex If f o r every p a i r of points in It, all points

of the segment Joining the two po in t s l i e

Given that plane E separates space into half-spaces R and S, and that point A is in R and point B is in S, does have to Intersect E? L, intersects plane E in P but does not l i e In E. LA lies.in plane E but does not contain point P . Is It possible for L, and Lp to intersect? Explain.

a . A s e t o f points is coll lnear if

b. A set of points is coplanar i f

c . May 5 po in t s be col l lnear? d . Must 2 points be co l l lnea r? e . May n points be colllnear? f . Must 5 points be coplanar? g . May n points be coplanar? Points P and Q l i e In both planes E am d F which i n t e r s e c t d n

# line AB. Would It be correct to say that P and Q l i e on AB? Explain.

Is the union of a half-plane and a ray on its edge convex?

Chapter 4

ANGLES AND TRIANGLES

4-1. The Basic Definitions. -- An angle is a figure that looks l i k e one of these:

To be m o r e exact: Definitions, An angle 1s the unlon of two mys whlch have

t h e same end-point but do not l i e in the same line. The two ray3 are called the sides of the angle, and their common end- point I s called the vertex. + +

The angle which is the union of A3 and AC is denoted by

LMC, or by LCAB, or aimply by LA If I* l a clear which rays

are meant. Notice that/^^^ can be equally well descr ibed by means of A and any two points on different sides of the angle.

A C E +

In the above f i g u r e LDAE l a the same as /_BAC, because AD is the same as Afi and a 13 the same as 3.

Notice that an angle goes out Infinitely far In two direc- t lons, because I t s sides are rays, rather than segments. The figure on the left, below, determines an angle uniquely, but I s

not all of the angle; to get a l l of the angle, we have to extend -> ->Â

the segments and AC getting rays AB and AC, as on t h e r ight .

Definitions. If A, B, and C are any three non-collinear

points, then the union of the segments a and AC I s called

a triangle,

and is denoted by AABC; the points A, B and C are called I t s - - vertices, and the segments AB, BC and AC are called Its - sides.

Every triangle determines three angles; AABC determines the

angles LING, LABC and LACB, whlch are called the angles

of AABC. For shor t , we w i l l often write them simply aa - - /A, /B, and LC.

Note that while AABC determines these three angles, I t does

not actually contain them. J u s t as a school does n o t contain I t s own graduates, so a triangle does n o t contain Its own angles,

because t h e s ides of a t r iangle are segments, and the sides of an angle are r a y s . To draw the angles of a t r iangle , we would have t o extend the sides of the tr iangle t o g e t rays, like t h i s :

There usually Is not much point in doing th i s , however,

because it is plain what the angles of a triangle are supposed

to be.

[sec. 4-11

The interior of an angle consists of all points that l i e Inside the angle; and the exterior of an angle consists of all the points t h a t l ie outside, l i k e t h i s :

Exterior / Interior

- - #'

Exterior *s

We can sta te t h i s more exactly as follows:

Definitions. Let /BAC be an angle lying in plane E. A point P of E lies In the interior of /BAC If (1) P and B are on the same s i d e of the line and a l s o (2) P and C are

0 on the same s i d e of the line AB. The exterior of /BAG is the

s e t of all points of E that do not l i e I n the In t e r io r and do no t l i e on the angle i t se l f .

You should check carefully t o make sure that this really says w h a t we want it to say. In the figure, P l a In the i n t e r i o r ,

Â¥ because P and B are on the same s i d e of AC and also P and C are on the same side of %. Q is in the exterior, because Q and C

<Â¥ are not on the aame a i d e of A3. R l a In the exterior, because

< - -fr> R l a on the "wrong side" of - both o f the lines A 3 and AC. S is in

0 the exterior because I t is on the "wrong s ideH of AC.

Notice that we have d e f i n e d the interior of an angle as the intersection of two half-planes. The half-planes look l i k e this:

Here one of the half-planes is cross-hatched horizontally, the other I s cross-hatched vertically, and the Interior of /BAc is

cross-hatched both ways.

The interior of a triangle consists of the points t h a t lie

ins ide it, like this:

More precisely: Definitions. A point lies In the i n t e r i o r of a triangle if

It l ies in the In te r ior of each of t h e angles of t h e triangle. A point lies In the exterior of a triangle If I t lies in the plane of the triangle but Is not a point of t h e triangle or of I t s In t e r i o r .

You should check ca re fu l ly to make sure t h a t t h i s rea l ly says w h a t we want it to say.

Problem Set 4-1 -- Complete this definition of angle: An angle I s the

of two which have the same end-

point, but do no t l i e in the same . Complete this definition of triangle: A triangle I s the

of the three joining each pa i r of three points . C Are the sides AC and of

AABC the same as the sides of LA? Explain,

A B Is t h e union of two of t h e angles of a tr iangle the same

as the triangle I t s e l f ? Why?

Into how many regions d o the angles of a triangle separate the plane of the triangle?

M,' Complete:

Name the angles In the figure.

HOW many angles are determined by the figure? Name them. How many may be named using the vertex l e t te r only?

A 8

9. Name the angles in the figure. here are more than six. )

10. Name a l l the triangles in the figure.(There are more than e igh t ) .

11. a. Name the points of the *D figure which are in the In-

eF terlor of /CBA. OM

b. Name the points of the N - figure In the exterior B

- - G

A Of /B. 'H

12. Is the vertex of an angle in the Interior of t h e angle?

in the exterior? Explain.

13. Is the Interior of an angle a convex set? l a the exter ior? 14. Is a tr iangle a convex set?

15. I s the interior o f a t r i ang le a convex set? is the exterior? 16. a. Can a point be in the exterior of a triangle and In the

i n t e r io r of an angle of the triangle? Illustrate. b. Can a point be In the exterior of a t r i ang le and not

In the i n t e r i o r of any angle of t h e triangle? Illustrate. l7+ Given AAEC, and a po in t P. P is in t h e i n t e r i o r of LBAC

and also in the I n t e r i o r of /ACB. What can you conclude about point P?

18. GivenAABC and a poin t P. P and A are on the same side of

8. P and 3 are on the same a i d e of E.

a* 1 s P in the in te r ior of LACB? b. Is P in the interior of AABC?

19. Carefully explain w h y the following statement 18 true: If a line m Intersects two s i d e s C of a t r iangle ABC in points D and E, not the vertices of the

triangle, then line rn does not i n t e r s e c t the third side. ( H i n t : Show t h a t A and B are I n the same half -plane. )

4-2. Remarks On Angles. - What we have presented in t h i s chapter is the simplest form

of the idea of an angle. According t o our d e f i n i t i o n , a n angle i s simply a s e t which is t h e union o f two non-collinear my&

Angles, I n t h i s aense, w i l l be quite good enough for the purposes of t h i s course . Later, you w i l l see the Idea of an angle I n various o ther forms. Here we explain these other forms br ie f ly , merely I n order to a v o i d confusion In case you may have heard o f

them already.

(1) In the first place, we sometimes t h i n k of an angle as being obta ined by rotating a ray from one position to another . In t h i s case, one r a y i s the i n i t i a l s i d e , and the other is t h e tenalnal side. Thus we would cons ider the two angles below as being d i f f e r e n t , because t h e ro ta t ions are i n two d l f ferent

The first Is called a posi t ive angle; the rotation l a counter-

clockwise. The second I s a n e g a t i v e angle; the ro ta t ion la clockwise.

(2) People sometimes speak of s t r a i g h t angles, which look

l i k e this:

Here the rays A and 8 are considered t o form an angle, even though A, B, and C are collinear.

(3) F i n a l l y , we sometimes distinguish between an ordinary angle and a reflex angle having the same rays as its sides. The

double-headed arrow below is supposed t o Indica te a reflex angle;

These complications, and various others of the same s o r t , w i l l not be used in t h i s book, because they w i l l not be needed. For example, the angles o f a triangle are never reflex angles, and there I s no reasonable way to d e c i d e in which direction they should be considered to go. Not until we get to trigonometry do these fancy angles become necessary and Important .

4-3. Measurement Of' Angles. - Angles are usually measured In degrees, with a protractor .

With t h e protractor placed as in the f igure below, with its edge

on the edge of the half-plane H, we can read of f the measures of a large number of angles.

Figure A.

The number of degrees In an angle l a cal led i t 8 measure. If there are r degrees In t he angle /XAY, then we write

m - / XAY = r. For example, In the figure we read off that

m /PAB = 10,

m / W = 40, m - / RAB = 75, m LSAB = 90,

m / TAB = 105, - and so on. Of course, the rays t h a t are drawn form more angles than t h i s . By subtraction, we can see t h a t

m /W = 40 - 10 = 30,

m /SAR = 90 - 75 = 15,

and so on.

Since m - / QAB = 40, we speak of /QAB as a ~ 1 0 angle, and we indicate Its measure in a figure l i k e this:

But we don't need t o uae the degree sign when we write ~ / Q A B = 40, because as we explained at the outset, rn /QAB means the number of degrees in the angle.

Notice that in Figure A there is no such t h i n g as the angle

/CAB, because the r a y s A? and Afi are collinear. But we notice ->

that the ray AC corresponds to the number 180 on the number-scale

of the pro t rac to r , and the ray 3 corresponds to t h e number 0. Therefore we can find m LCAU by wri t ing

m /CAU 180 - 130, = 50.

Similarly,

ITI LCAQ = 180 - 40, = 130.

The following postulates merely summarize t h e facts about protractors that we have Just been discussing. Each of them I s

I l l u s t r a t e d by a f igure .

P o s t u l a t e 11. h he Angle Measurement Pos tu la te . ) To every angle /BAG t h e r e corresponds a real number between 0 and 180.

Definition. The number specified by Pos tu l a t e 11 is

called the measure of the awle, and written as m L B A C . --

f Postulate 12. (The Angle Construction Postulate . ) 1 Let A? be a ray on the edge of the half-plane H. For

1 every number r between 0 and 180 there is exactly one 1 ray A?, with P In H, such t h a t m /PAB = r.

Postulate 13. (The Angle Addition postulate . ) If D is a point in the i n t e r io r of D A C , then m @ G = m@D + ~ / _ D A c .

In was on this basis that we computed the measures of angles by subt rac t ion , with a p r o t r a c t o r placed with I t s edge on t h e

-> ray AB. (~LDAC = ~ ~ B A C - BAD.)

Two angles form a l inear pair i f they look l i k e this:

That is : ->

Definition. If A and A are opposite rays , and AD Is

ano the r ray, then /-BAD and ~ A C form a linear pair. Definition. If the sum of t h e measures of two angles is 180,

then t h e angles are called supplementary, and each is ca l led a

supplement of the o t h e r .

Hence the name of the following postula te*

Pos tu la te 14, h he Supplement Postu la te . ) If two angles f o m a l inear pair, then they are

1 supplementary.

Problem -- Set 4-3

\

1. Using t h e figure, find <,he value of each of t h e following:

a, m L F A B . g. m L E A D . b. m L E A B . h. FAG + KI/_GAH.

C . m /MAC. 1. RI /GAF + m /FAE. d. ~ / F A E . j. m LMAB - m LFAE e . m LGAE. k . rn/-mB - m/-DAB. f. MAN. 1. m LMAB - n~ LNAH.

2. With continued practice you should "be able t o estimate t h e

s ize of angles fairly accura te ly without using a protractor. Do - not use a pro t rac to r to dec ide which of the angles shown

have measures within t h e indicated ranges.

Match the corresponding pairs :

3. Using only a straightedge and - not a prot rac tor , sketch angles

whose measures are approxlJnately 30, 150, 45, 60, 135, 90, Then use your pro t rac to r to check your sketches.

4. On the edge of a half-plane, take a segment about 3 Inches

long. A t A draw ray A?S In the half-plane f o r m i n g / ~ ~ ~ of

58'. A t B draw ray B& in the same half-plane forming /ABD

of 72O. Measure the remaining angle of the triangle fonned.

5. In the figure, a. m / _ B H F + m / J 3 V P a g i / _ ?

to. r n / G F H + m / . B P H = r n ^ _ ?

8

6. In the figure,

a. ~/XZK + ~ ^ _ K z R + IR/_YZR a m/_ ?

b. m / X Z R - m / R Z K = m/ ? - - c . m/ X Z Y - rn/XzK ss m i - ?

d . If Y, R, K and X are co l l inea r ,

- then m/ - YRZ + rn JSRx = ?

In the figure, and C6 Intersect forming four angles. Using t h e Indicated measure, find a, b and c .

Determine the supplement of each of',the following:

lloO, goo, 36') 15.5', no, (180 - n)', ( 9 0 - n)Â¡ If one of two supplementary angles has a measure 30 more t h a n t h e measure of t h e o ther , what is the measure of each angle?

If the measure of an angle is twice t h e measure of its supplement, find the measure of the angle. The measure of an angle is f o u r times t h e measure o f i t s

supplement. Find t h e measure of each angle. ->

a. Given a ray AC lying on the edge of a half-plane H, and a number r between 0 and 180. In how many ways can you

-> construct a ray AB In H such that r n 1 3 A ~ = r? Why?

-> b. Given a ray AC l y i n g in a plane EJ and a number r between

0 and 180. In how many ways can you construct a ray 3 in E such that ~/BAC = r? Why?

4-4. Perpendicularit& R i a t Angles and Con~ruence of Angles. - Definitions. If the two angles of a l inear pair have the

aame measureJ then each of the angles is a right angle.

Since r + r = 180, by the Supplement Postulate, we see t h a t a ri~$~t angle is an a n ~ l e of &. Th33 can be regarded as an - -- alternative definition of a right angle; It i s equivalent t o our first def in i t ion .

In terms of right angles it is easy to define perpendicularity

of any combination of line, ray or segment. In applylng the following definition remember t h a t a ray or a segment determines a unique line which contains i t .

Definition. Two intersecting sets, each of which i s either

a l i ne , a ray or a segment, are perpendicular if the two l ines which contain them determine a right angle.

Definition. If t h e sum of the measures of two angles is 90, then the angles are called complementary, and each of them i s called a complement of the other. (Compare t h i s with the def ln l t ion of supplementary amleg, Jus t before the statement of the Supplement Postulate . )

An angle with measure leas than 90 I s called acute and an -' angle with measure greater than 90 i s called obtuse.

I Obtuse , /

Definition. Angles with t h e same measure are called congruent angles.

Tha t Zs, - / BAC and LPQR are congment if ~LBAC = ~LFQR. In t h i s case we write

/ BAC 2 /PQR. - Notice t h a t the equa t ion m/ BAG = ~ / P Q R and t h e congruence - / BAC 2 / PQR are completely equivalent; we can replace one by L - the other any time we want to.

The following theorems are easy to prove, if we remember

clearly what the words mean: Theorem 4-1. If two angles are complementary, then both

of them are a c u t e . Theorem 4-2. Every angle is congruent to i t s e l f . Theorem 4-3. Any two right angles are congruent. Theorem 4-4. If two angles are both congruent and supple-

mentary, then each of them is a right angle.

(~int: L e t r be the number which is the measure of each of the

t w o angles, and then f ind out what r must be. ) Theorem 4-5. Supplements of congruent angles are congruent. Restatement: If (1) LB 2 LD, ( 2 ) LA and L 3 are

~upplementary and [ 3) LC and LD are supplementary, then

?roof: The statement t ha t LEi 2 L D means that m L B and m/ D are the same number r, as in the figure. Since LA and - L 3 are supplementary, it follows that

For t h e same reason, m/C = 180 - m/D = 180 - r.

Therefore rnL~ = m k C , which means that LA 2 LC. You must not conclude from t h e above picture that supplement-

ary angles must necessarily be placed beside one another* In a way that makes it evident t h a t their measures add up to 180.

[see. 4-41

The following pic tu re also serves to illustrate Theorem 4-5.

In drawing pictures to Illustrate theorems or problems you should real ize that the figure in the book is not t he only correct one, and you should try t o make your picture dif ferent from t h e one given In t h e book.

Theorem G. complement^ of congruent angle3 are congruent. The proof of t he theorem I s exactly analogous to the pre-

ceding proof, and you should write I t out for yourself.

When two lines Intersect, they form four angles, l i k e t h i s :

Ll and are c a l l e d v e ~ t l - 1 angles, and L2 and L4 are a l so called vertical angles. More precisely:

Defini t ion. Two angles are vertical angle^ if their sides form two pairs of opposite rays .

It looks as if these pairs of congruent, and In fact t h i a l a w h a t

Theorem 4-7. V e r t i c a l angles

Proof:

vertical angles always happens : are congruent.

Given t h a t and 3 are opposite rays, and

ought to be

& and A i are opposlte ray3, so m a t Ll and L2 are vertical angles. Then

Ll and L 3 are supplementary, and L2 and ,/-3 are supplementary.

Since L 3 is congruent to itself, t h i s means that and L 2

have congruent supplements. By Theorem 4-5, Ll 2 L 2 , whlch

was to be proved.

Theorem - 4-8. If two intersecting lines form one right

angle, then they form f o u r r i g h t angles.

t You should be able to supply the proof.

Problem Set 4-4

1. a. In a plane, how many perpendiculars can be drawn to a line at a given point on the line?

b. In space, how many perpendiculars can be d r a w n to a line

at a given point on the l i n e ? +' 4 2 . If OR and OS are opposite rays and 6% is a r a y such t h a t

<-Ã

m /RON = m /SON, what can you conclude about 8k and RS?

Explain. In half-plane H, % and 3 are opposite rays, m L R X B = 35 and

m /RXS = 90. a. Name a pair of perpendicular

rays, If any occur in the

figure . b. Name a pair of complementary

angles, if any occur in the figure.

c. Name a pair of v e r t i c a l angles, If any occur in t h e figure.

d . Name two pairs of supplementary

angles In the figure.

Determine the measure of a complementary angle f o r each of the following:

a. 10'. 0 d . x *

b. 80'. e. (90 - XI'. c. 44.5'. f. (180 - x)', a. If two angles with t h e same measure are supplementary,

what is the measure of each? b. If two angles w i t h the aame measure are complementary,

what is the measure of each? a If two lines intersect, how many pairs of vert ical angles

are formed? b. If t h e measure of any one of the angles in (a) is 70,

what is the measure of e&ch of t h e others? c. If all of the angles In (a) are congruent, what i s the

measure of each? If one of a pair of vertical angles has a measure of r, write t he formulas f o r the measures of the other three angles formed . In half-plane H, & and & are opposite rays ,

Prove Theorem 4-1. A G E Prove Theorem 4-4, Given: In the figure f o r Problem 8, & -L tf& and Ok and & are opposite rays . Prove; /AGB and /DOE are complementary.

Given: In plane E, lines 8, S, %, % Intersect at 0.

Prove: b + g + d = a.

-Ã

13. If and & and OC are three different r a y s in a plane,

no two of them opposite, indicate true or false f o r each

of the fol lowing statements and explain your answer. a. ~ / _ A O B + m/JBOC = ~ / _ A o c . b. rn LAOB + m LBOC + rn /AOC = 360.

14. The measure of an angle is nine times t h a t of I t s supplement.

What is the measure of the angle?

15. A l ayou t drawing l a a plane drawing which can be folded to form t h e boundary of a given solid. Below Is pictured a cube and a layout drawing for it.

- --

u

(~otted lines Indica te f old3 . ) Use your Imagination, your ruler and your p r o t r a c t o r to make a l ayou t drawing f o r each o f the figures below. Then

cut out your drawing, f o l d on d o t t e d l i n e s , and tape t o -

gether. Use cardboard o r heavy paper f o r a rigid figure. a, A pyramid whose base is

8 square with 2" sides and whose o t h e r faces are isosceles triangles w i t h 60Â base angles.

2"

( Problem 15 continued) b. A prism whose bases are

pentagons w i t h 1 inch s i d e s and 108' angles,

and whose height I s

2 Inches.

Review Problems

1. What tool is used to measure angles? 2. To every angle there corresponds a real number between

and , called the measure of t h e angle. 3. An angle w i t h a measure of less than 90 is

4. Two angles formed by the union of two opposite rays and a t h i r d ray a l l with the same end point are a of angles.

5 . If the sum of the measure of two angles is 90, then each is called a of the other.

6 . A n angle with a measure greater than 90 is called 7 . Angles with t h e same measure are 8. If two angles are both congruent and supplementary,then each

of them is a 9. Supplements of congruent angles are 10. If two angles are complementary, then each of them Is

11. An angle is t h e of two which have a common

end point.

12. If X, Y, Z are t h r e e points, t h e union of the t h r ee

segments connecting them In pairs l a a 13. A point X is In the inter ior* of RST if points R and

lie on the same s i d e of and if points X and

l i e on the same side of . 14- If the sum of t h e measures of two angles l a they are

cal led complementary and If t h e sum is they are ca l led

15. Two opposite angles formed by two Intersecting lines a r e angles. They are always congruent.

16, AS andm opposite rays. The points E, F, and H are on t h e same side of AS* Poin ts E and H are on opposite

aides of %. Points A and H are on t h e same side of

. 8 1 8 1 % and 313. B I F B E = 20. D r a w t h e

figure and f ind:

a. rn - / E B A . b. m /FBH. c. m - /EX. 17. Given: Find :

m /BCD = 90, a. m /-DOC.

III /BOC = 50, b. m /KO. m /DCO = 25, C. m LDOA. m /DAO = 45. d . ~LAOB.

If one of two supplementary angles has a measure of 50 more than the measure of t h e o ther , what is the measure of each angle?

The measure of an angle is five times t h a t of i ts complement. Find the measure of each angle. Under what condi t ions are the angles of a l i nea r pa i r

congruent ?

Is there a poin t i n the plane of a t r iangle such that the

point I s neither In t h e ex te r io r nor the I n t e r i o r of a triangle and n e i t h e r i n the i n t e r i o r nor the exterior of any of I t s angles?

Is t he measure of an angle added to the measure of an angle the measure of an angle? Explain.

23. Could the interior of a triangle be considered as the intersection of three half planes? Illustrate.

c 24. How many triangles are in t h i s figure?

25, Does m LBAC = m /^BAE?

26. Does LBAC = /BAE?

27, Is LABE aupplernentary to LEBC? 28. How many angles are ind icated in the

drawing?

Problems 24 - 28.

29. Explain ca re fu l ly why the following statement I s true: If a line m in te r sec t s 2 s i d e s of a, triangle ARST In

points U and V, not the vertices of the triangle, then line m does no t intersect the t h i r d s ide .

< zAs B C

31. 1f' you were glven t h a t L a 2 / b and that / x 1s ~ ~ P P l e m e n t a ~ - - t o L a and that L y is supplementary to L b , what theorem or postulate would you use to prove that L x 2 / y? -

32, The Angle Measurement Postu la te places what limitation on angle measures?

33. Is the following a correct restatement of the Angle Construction Postulate: Given a ray X? and a number k

-> between 0 and 180 there i s exactly one ray XP such that m /FXY = k? Explain.

34. By giving its name, or by stat ing I t in f u l l , give t h e postulate which seems to you to be most appropriate In each

of the following cases, a s reason f o r the statement.

35. Is the following statement always true?

If !S and 8 Intersect at 0, then /AOC 2 /BOD.

Chapter 5

C ONGRUENCES

5-1. The Idea of a Congruence. - - - - Roughly speaking, two geometric figures are congruent if

they have exactly the same s i z e and shape. F o r example, in the

figure below, all three triangles are congruent.

One way of descrialnp; t h e situation is to say that any one of

these triangles can be moved onto any other one, in such a way

that it fits exactly. Thus, t o show what we mean by saying that

two triangles are congruent, we have to explain what points are

supposed t o go where. For example, t o move AABC onto A DFE,

we should pu t A on E, B on F, and C on D. We can write down the pairs of corresponding vertices l i k e t h i s :

A-E

B-F C - D.

To describe the congruence of the first triangle and t h e th ird , we should match up the ver t ices like t h i s ;

A- G

3-H . !-

c -I.

How would you match up the v e r t i c e s to describe the congruence

of the second triangle w i t h t h e t h i r d ?

A matching-up scheme of thls kind is called a one-to-one

correspondence between the vertices of the two triangles. If t h e matching-up scheme can be made to work - - t h a t I s , i f t h e

triangles can be made to f i t when the vertices are matched up in the prescribed way - - then the one-to-one correspondence is called a congruence between the two t r i ang les . For example, the

correspondences tha t we have just given are congruences. On the other hand, if we write

A - F B-D

c -E,

this does give us a one-to-one correspondence, but does n& give

us a congruence, because the first and second triangles cannot be made to coincide by this particular matching-up scheme.

We can write down one-to-one correspondences more b r i e f l y , in

one line. For example, the correspondence

A-E

B-I'

c -D, which i s t h e first example that we gave, can be wri t t en i n one

l i n e l i k e this: ABC - EFD.

Here it should be understood that t h e f i r s t letter on the l e f t corresponds to the first letter on the r igh t , the second corresponds t o the second, and the t h i r d to t h e t h i r d , l i k e t h i s :

ABC EFD L e t us take one more example.

These two f igures are of t h e same size and. shape. To show how one can be moved onto the o t h e r , we should match up t he v e r t i c e s l i k e

this:

A - H

B-G

C-F

D - E.

These two f igures are congruent, because the correspondence t h a t we have w r i t t e n down is a congruence, t h a t is, t h e f igures can be made t o f i t If t h e ve r t i ces are matched in the given way. For short, we can write the congruence in one l i n e , like this:

Notlce that t he order in which the matching pairs are written

does not matter. We could have wri t t en our list of matching pairs this way:

D-E B-G

C - F

A- H;

and we could have described our one-to-one correspondence In one

l i n e , l i k e this: DBCA - EGFH .

A l l that matters Is which poin t is matched w i t h which.

It i s q u i t e pos s ib l e f o r t w o figures t o be congruent i n more than one way.

Here t h e correspondence

ABC - FDE is a congruence, and the correspondence

ABC - FED

is a d i f f e r e n t congruence between the same two figures . Obviously AABC coinc ides with itself. If we agree t o match

every ver tex w i t h i t s e l f , w e get t h e congruence

ABC - ABC. This is called the i d e n t i t y congruence. There i s another way of matching up the vertices of t i l l s t r iangle , however. We can use

t h e correspondence

A B C - ACB. Under t h i s correspondence, t h e figure I s made to coincide with itself, with the vertices 3 and C interchanged. This is not

possible for all triangles by any means; it won't work unless at least two sides of the triangle are of the same l e n g t h .

Problem -- Set 5-1

I n t h e problems of t h i s s ec t ion , t h e r e are no t r i c k s in the

way that t h e figures are drawn. That is, correspondences t h a t

look l i k e congruences when the f i g u r e s are measured with reasonable care really are supposed to be congruences. In t h i s s e c t i o n

we are n o t trying to prove things. We are merely trying to learn,

informally, what the idea of a congruence Is all about.

1. Below there are s i x figures. Write down as many congruences as you can, between these figures. ( ~ o not count the iden-

t i t y congruence between a figure and itself but recall that there is a congruence between a triangle having two congruent

s ides and itself that is not the identity.) You should get 6 congruences in all. {One congruence is DEF -SUT.)

4 . Answer as in Problem 1:

5. Name the f igures t h a t do not have a matching figure.

6. Which pa i r s of the following figures are congruent?

7. The t r iangle below is equilateral. That is, AB = AC = BC.

For the triangle on the preceding page, write down all congruences between the triangle and i t s e l f , starting with

t he identity congruence ABC - A B C , (you should get more than four congruences.)

8. Write down a l l of the congruences between a square and itself.

9. a . I f two f igures are each congruent t o a t h i r d , are they congruent to each o the r?

b. Is a figure congruent to i t s e l f ? c . Can a tr iangle be congruent t o a square?

d. A r e t he t o p and bottom faces of a cube congruent?

e . Are two adjacent faces of a cube congruent? f . Are t h e top and bottom f a c e s of a rectangular block, such

as a brick, congruent? g. Are two adjacent faces of a brick congruent?

10. Pick out the pairs o f congruent figures,

106

1.1. Write down t h e four congruences of t h i s figure with itself.

12. Suppose A, B, and C are three points of a l i n e as shown with

AB = EC.

a. Describe a motion of the line that takes A t o where B was. Does it necessarily take B to C?

b. Describe a motion of the line that interchanges A and C. Under what conditions can the following pairs of f igures be made to coincide by moving one in space without changing i t s size and shape? (it Is understood that this moving is done abstractly In the mind. One figure can move through another

so t h a t a s o l i d can be moved onto another solid of the same size and shape. For example, one segment can be moved to

coincide w i t h another if they have the same length. One sphere can be moved to coincide w i t h another if their radii

are the same length. ) a. Two segments. b. Two angles. c. Two rays. d . Two c irc les . e . Two cubes.

f. Two p o i n t s . g. Two l ines .

1 4 . Given a c i r c l e containing three p o i n t s A, B, C as shown, with

the arc from A to B the same length as t he arc from B

to C.

a. Describe how the c i r c l e may be moved to take A to where

B was and 3 t o where C was. b. Describe how t h e circle may be moved to leave B fixed

but to Interchange A and C. 15. Suppose t h a t the following ornamental f r i e z e extends infin-

itely In both directions, as a l i n e does.

a. Describe motions of two different types that Induce congruences of t h e frieze with I t s e l f . How many such congruences are there altogether?

b. Do the same f o r this frieze.

16. Which of the following figures can be f i t t e d onto each other? For each matched p a i r , tell whether you must tu rn the f igure over in space as well as s l i d e and rotate it in a plane to

make it f i t on the other so that a l l segments f i t .

17. The f igure below is a f i ve -po in t ed s tar .

Write down a l l of the congruences between the star and itself. To save t i m e and paper, l e t us agree t h a t a congruence f o r

this figure is sufficiently described if we say where the

points A, B, C, D, E of the star are supposed to go. For ex-

ample, one of the congruences that we are looking for can be written as ABODE -BCDEA.

5-2. Congruences between Triangles. In the preceding sec t ion , we have explained the basic idea of

what a congruence is. Let us now give some mathematical definitions so t h a t we can talk about congruence in a care fu l way, in

terms of distance and angular measure, instead of having to talk loose ly about things falling on each other.

For angles and segments, it is easy to say exactly what we mean :

Definitions. Angles are congruent If they have the same measure. Segments are congruent If they have the same length. The first definition above is merely a repetition from Section 4 - 3 .

Analogous to Theorem 4-2 for angles we have a theorem f o r

segments : Theorem 5-1. Every segment is congruent to i t se l f . We sometimes refer to these two theorems by the term Identity

congruence.

Just as we indicate tha t L A and C B are congruent, by

writing Z. A 2 LB, so we may write m2^S -

t o indicate that the segments AB and CD are congruent. In t he

table below, t h e equation on the left and the congruence on the right in each l i n e may be used interchangeably:

1. m L A = m LB. 1. L A 2 LB. 2. AB = CD. 2. "S5^^5 . Each of the equations on t he left is an equation between numbers. The first says t h a t m L A and rn L 3 are exactly the same number. The second says that the distance A 3 and the distance CD are exactly the same number.

Each of the congruences on the right I s a congruence between geometric fi~ures. We do not write = between two geometric fig-

ures unless we mean that the figures are exactly the same, and

occasions when we mean t h i s are rare. One example Is this:

Here it is correct to write L B A C = LEAD,

because L 3 A C and LEAD are not merely congruent, they are - ex-

,actly -- the same angle. Similarly, and are always exactly - the same segment, and so it is correc t to write AB = x.

Consider now a correspondence ABC - DEF

between the vertices of two triangles AABC and ADEF.

This automatically gives us a correspondence betww.n t h e s ides of

t h e t r iangles , like this: - AB - "EE

and it gives us a correspondence between the angles of the two triangles, l i k e this:

LA - Z-D

LB- LE

LC - LF.

We can now state the d e f i n i t i o n of a congruence between two

triangles.

Definition. Given a correspondence ABC - DEF

between t h e vertices of two triangles. If every pair of corresponding sides are congruent, and every pair of corresponding

angles are congruent, then the correspondence AX-DEF is a congruence between the -- two triangles .

You should read t h i s de f in i t ion a t least twice, very care-

fu l ly , t o make sure that it says what a d e f i n i t i o n of the idea of a congruence between triangles ought to say.

There is a shorthand f o r writing congruences between triangles. When we write

L A 2 ZD, t h i s means that the two angles LA and LD are congruent. (That

is, m L A = m L D.) Similarly, when we write

AABC 2 ADEF,

this means that the correspondence ABC - DEF

is a congruence. Notice that this is a very efficient shorthand: f\s

the single expression AAEC = ADEF tells us six things at once; namely ,

AB = DE ~~~ AC = DF I ^ ^ F BC = EP S?3?^

mZA = m Z D L A 2 Z D m L B = m L E Z B 2 LE

" m Z C = m d F . L C = LF.

In each of these s i x l i n e s , the equations on the left and the congruences on the r i gh t mean the same thing, and we can choose either notation at any time, according to convenience, Usually

we will write AB = DE, instead of 2 'GE, simply because it is easier to write. For the same reason, we will usually write

L A ~ Z D instead of m L A = m L D ,

It is sometimes convenient to indica te a congruence graph- 1ca l ly by malting marks on the corresponding sides and angles, like t h i s :

AABC 2 ADEF

We can also use t h i s method t o i nd ica te t h a t c e r t a i n corresponding parts of two figures are congruent, whether or not we know about

o the r parts.

The marks in t h e figure Ind ica te t h a t (1) A 3 = DE, (2) AC = DF and ( 3 ) m L A = m Z D .

Question: Would It be correct to write A 3 2 DE, or LA = AD? Why o r why n o t ?

It seems pretty clear, in the above f igure, that the congruences we have indica ted are enough t o guarantee that the correspondence ABC-DEF Is a congruence. That is, if these three pa i r s of corresponding parts are congruent, the triangles must also be congruent. In f a c t , t h i s is the content of t he basic con-

gruence postulate, to be stated in the next section.

Problem Set -

A ABF a AKiRQ. Complete the following list by telling what

should go In the blanks. A 6 - M-Q

AABR 2 AFBR. L i s t the six pains of corresponding, congruent part s of these two triangles.

3 . Ar"M 2 AFHW. L i s t the s i x pairs of corresponding, congruent parts of the-ie triangles. (it is not necessary t o have

a picture but you may make a sketch if you wish. ) 1;. . ARQP 2 AABX. List t he s ix pairs of corresponding congruent

p a r t s of these triangles. Do not use a f i g u r e .

AAZW 2 ABZW. L i s t the six pairs of corresponding, congruent parts of these t r i ang les , Here is a list of the six pairs of corresponding parts of two congruent triangles. Give the names of the two triangles t h a t

would fit In the blanks below. - m , - AB = MK. L A 2 LM. --" BW = T3?. Z B Z L K .

m ^ m. Z W S Z F . 2 -

A. A-â If AABC 2 AXYZ and ADEF 2 AXYZ, what can be said about the relationship of AAEC to ADEF? State a theorem gener-

a l i z i n g t h i s s i t u a t i o n .

Using ruler and protractor, draw a triangle ABC in which

A 3 Is 3 inches long, EC Is 2 Inches long and angle B Is 50'. Compare your triangle with those of other members of the class. Draw AABC In which AC is 3 inches long, BC is 4 inches long and angle C I s 70'. Compare t r iangles . Draw AABC with A B 3 inches long and EC 2 inches long. Make dB any size t h a t su i t s your fancy. Compare triangles.

If these three exercises suggest t o you an Idea concerning a congruence between two triangles, try to sta te or write

t h i s idea for tr iangles In general.

9 a . Given that AABC and ADEF do not in te r sec t , and t h a t

X is a point between B and C, Tell which of t h e syrn- 1\f

bols =, = may be i i l l e d in t he blanks to make t h e state-

ments meaningful and possibly true.

1. AABC - ADEF. 2. m L A - m LD. - v

3 . AB - DE . 4. EC - EF . 5. ^3- L C .

6 . L A E X - L A X .

7 , m Z ABX - rn .L EDF, b. Which of t h e .blanks could have been filled with either =

or 2 ? - c . If had been the same segment as DE but if C were

a d i f f e r e n t point than F, which blank could have been f i l l e d by = t h a t should otherwise have been filled by 2 ?

5-3, B s I c Congruence Postulate.

To get a t the facts on congruences of triangles, we need one

new postulate. In t he name of t h i s postulate, S.A.S. stands for Side Angle S i d e .

Postulate s. h he S , A .S, P o s t u l a t e . 1 Given a

correspondence between two triangles ( o r between a triangle *and i t se l f ) . If two sides and t he inc luded

angle of t he first triangle are congruent to the corresponding parts of t he second triangle, then the

correspondence is a congruence.

To Illustrate this, we repeat the previous f igure .

The postulate means t h a t if - - - AB = DE, - AC 2

and

L A 2 ZS,

as indicated in the figure, then

AABC 2 ADEF;

t h a t is, t h e correspondence AX-DEF is a congruence.

It is very important t o no t i ce t h a t in the S.A.S. Pos tu la te ,

t h e given angle is the angle included between the two given s ides, l i k e t h i s :

Under these conditions, the S.A.S. Postulate says t h a t t h e corres-

pondence ABC-DEF is a congruence. If we knew merely t h a t some one angle and some two s ides of t h e first t r i ang l e were congruent t o the corresponding p a r t s of the second t r iangle , then it would not necessarily follow that the correspondence was a congruence. - For example, consider t h i s figure;

Here AB = DE, L A '= fD, BC = EF, Note that LA and LD are

not included by t he pairs of congruent sides. This correspondence - is ce r t a i n ly n o t a congruence, because it matches AC w i t h E,

L B with ZE, and LC with LF. Since these are not congruences, the definition of congruence between triangles I s not sat isf ied.

5-4. Writin6 -- Your Own P roo f s ,

You now have enough basic material to be able t o write real geometric proofs of your own. From now on, writing your own proofs will be a very important part of your work, and t h e chances

are that It will be more fun than reading other people's proofs. L e t us take a couple of examples, to suggest how we go about

f i n d i n g proofs and writing them up. Example 1. If two segments b i s e c t each o the r , the segments join- i ng the ends of t h e given segments are congruent. -

' Given; AR and bisect each o t h e r at F.

B S t a r t i n g t o work on a problem l i k e this, we should first draw

a f igure and l e t t e r I t , using a capital letter f o r each ver tex .

Then, s t a t e t h e hypothesis and conclusion In terms of t h e l e t t e r - ing of t h e figure .

Next, we divide t h e page into t w o columns as shown, and write In t h e headings Statements and Reasons,

A l l t h i s , of course , Isn't going t o do us a b i t of good unless we can t h i n k of a proof to wri t e down.

Since our o b j e c t is t o prove two segments congruent, we must reca l l what we know about congruent segments. Looking back we can

f ind the definition of congruent segments, of congruent triangles, and the S.A.S. Pos tu l a t e . These are t h e available weapons about congruent segments in our arsenal, and a t t h i s po in t the search i s sho r t , because our arsenal I s small.

To apply the postulate, we have to set up a correspondence between two t r iangles, in such a way that t w o sides and t h e In-

cluded angle of the first triangle are congruent to t h e corresponding parts of the second triangle. From the figure, this correspondence looks as if I t ought to be

AFB-RFH.

Two pairs of sides are congruent, because we have f r o m the given data and the def in i t ion of bisec t that'

AF = RF and J3.F' = HF.

How about the included angles , LAFB and ZRPH? We need t o know that they are congruent, too. And they are, because they are vertical angles. Therefore, by t h e S . A . S . Postu la te , ou r correspondence I s a congruence. The sides and are corresponding sides,

and so they are congruent. This I s what we wanted t o prove. Written down in tine double-column form, our proof would l ook

l i k e this: - Given: AR and bisect each o t h e r

at F. --"- A

TO prove: AB = m. +R 0

Statements Reasons

1. A F = RF

2. B F = H F .

3. ZAFB 2 LRF'H.

~ t . A A F B ~ ARPH. - * - 5. AB = RH.

1. Defin i t ion of b i s e c t .

2. Definition of bisect. 3. Vertical angles are

congruent.

4 . The S.A.S. Postulate.

5. Definition of a congruence between triangles.

This i s given merely as a sample of how your work might look. There is a limit to how standard1 we can expect the form of a

proof to be. For example, in t h i s proof we have Indicated congruences between segments by writing AF = HP and BF = HF, and

- + - - - - so on. Ve cou ld j u s t as well have written AF = RF, BF = HF, and

so on, because in each case the congruence between the segments

and t h e equation between the distances mean the same thing.

There are only two really important t h ings In writing proofs , F i r s t , what you write should say what you real ly mean. Second,

the things tha t you really mean should form a complete logical explanation of why t h e theorem I s true,

By now, you should have the idea, and so we give we second example in an incomplete form. Your problem I s to f i l l in t h e

blank: spaces in such a way as to get a proof. H Example 2.

- * - Given: AH = FH.

- -

HB bisec t s LAW.

To prove: L A 2 LF.

2. Definition of the bisector of an angle.

Statements Reasons

3 . Every segment is congruent to Itself.

4.

5.

- * - 1. A H = P H . .

A mistake often made.In proofs is that the student assumes as

true the very thing he is t rying t o prove to be true. Another

1. Given.

common mistake is to use a s a reason In his proof a theorem which

is actually a consequence of the fact that he is trying to prove.

Such arguments are called circular arguments, and are worthless as logical proofs.

A p a r t i c u l a r l y bad kind of c i r cu l a r argument is the use of the theorem we are trying t o prove as a reason fo r one of the steps In its p r o o f " .

Problem Set 5-k -- (Note: - I n some of the following problems we make use of a square. A square ABCD is a plane figure that is t h e union of f o u r con- - gruent segments AB, 3, E, DA such that L A E C , ZED, ZCDA,

^ D A B are right angles. Tile square will be discussed in a later

chapter of the t ex t . ) 1. In each pair of t r i ang l e s , if l i k e markings ind ica te congruent

parts, which triangles could be proved congruent by S.A.S.?

2. In the figure I t is given

t h a t intersects "Â¤ at C, that AC = DC and BC = EC.

Show (i .e . , prove) that ZB 2 LE. Copy the follow- B E

ing proof and supply t he missing reasons.

Statements Reasons

1. A C = C D .

2. 3C = EC. 3. ZACB 2 LEE.

4 . AACB 2' ADZE.

Suppose In t h i s figure mzm, L x 2 f y and - 3 is the midpoint of AF. show that ZR ZH.

1. Given.

3 . angles are congruent .

4. . [ ~ o t e that Statement 3 refers to angles and Statement 4 to triangles, so that your reason here should refer to triangles],

5. Corresponding partsof congruent t r i ang les are .

Copy and complete t he following proof . A B F

Statements Reasons

1.

2. Given. 3. From the definition of rnid-

p o i n t . 4. I " - S.A.S.

a . If ABCD I s a square and - R is the midpoint of AB,

prove that RC = RD. (see

note preceding Problem 1,)

b. What pairs of congruent

acute angles appear i n the A figure? Prove your answer. In this figure A 3 = FH and

m A x = m L g .

Show that m Z A = m L F.

In this f igure it is given that m Z AEH = m .L FBH, A 3 = FB. Prove AH = F'H.

- A

then A F A B Z AFHR.

Prove: If the line segments

Prove that if segments AH, 1?B bisec t each other at poin t F,

- AD and "S5" bisect each other, then A 3 = DC and AC = DB. A

a. Given: Square ABCD, R I s

t he midpoint of m, F I s a

po in t between A and D, Q

is a po in t between C and 3,

DF = CQ. To prove: RF = RQ.

b. A r e there two o the r po in t s F', Q1 of square ABCD n o t on - AD or such t h a t RF' = RQ'? Where are they?

Suppose in this figure t h a t

AB = AH and that A? bisec t s ZHAB. Prove t h a t FH = FB.

5-5. Overlapping Triangles. Using the Figure & Statements. Frequently In geometric figures, the triangles that we need

t o work with are not entirely separate but overlap, like' AAFM

and AFAH in the figure below. R

The easiest way to avoid getting mixed up, and making mistakes, In dealing with such cases, is to write down congruences i n a standard form, l i ke t h i s ,

AAFM 2 AFAH.

Check t h a t the correspondence AFM-FAH real ly is a congruence, and then later refer back to AAFM AFAH when we want to conclude that two corresponding sides (or corresponding angles) are congruent.

Of course, if you don't see the congruences between the overlapping triangles, you will have nothing to check and nothing to apply later. To practice up, write all t h e congruences t h a t you

can between triangles contained in the figure above, if it is given that AR = FR and M, H, B are the midpoints o f the res-

pective s ides . L e t us now look at a case In which this s o r t of thing comes

up in t h e proof of a theorem. Given: HA = HI?.

HM = HQ. To prove; PM = AQ.

A F

A very common way to prove that two segments are congruent is to show t h a t the segments are corresponding s i d e s of congruent triangles! If t h i s way can be used successfully here, then t h e - f irst thing to do is locate the triangles which contain FM and - AQ. These are AHMF and AHQA, and these triangles overlap quite a bit. Now the problem becomes one of proving the triangles congruent. The proof i n the double-column form goes l i k e this:

St a t ement s Reasons

1. HA = HP. 2. LH 2 ZH.

3. K M = H Q .

4. A H M F ~ AHQA. 5. FM = AQ.

1 1. Given. 2. An angle is congruent to

itself. 3 . Why?

4. why?

5. Why?

A s t r i c t l y log ica l proof must not depend on a figure but must

follow from the postulates, the def in i t ions , and the previously proved theorems. But geometers In practice use f igures as a mat ter o f convenience, and readily accept many observable facts without a tedious restatement in words, unless such a restatement i s essen-

tial t o c l a r i f y i n g the problem a t hand. To i l l u s t r a t e , let us look at a restatement of Example - 1 used

previously.

Example 1. Le.t A, B, F, H and R be f i v e non-collinear poin t s lying in a plane. If (1) F is between A and R, (2) F I s between 3 and H, (3) AF = FR, and ( 4 ) BF = FH, then (5) AB = RH.

T h i s conveys all the information conveyed by the figure on the lef t and t h e notation on t h e right below.

- Given: AR and b i s e c t each

other at F. To prove: A 2 T3T.

Notice t h a t (1) t e l l s us t h a t and $ are opposite rays, and (2) t e l l s us that % and FH are oppos i te rays. These two

things, taken together , m e a n that LAFB and LRFH are v e r t i c a l angles. (See definition of vertical angles . ) This I s t h e sort of information that we normally read from a figure.

In stating problems In t h i s tex t we will frequently avoid tedious repe t i t ion by referring to a figure. You - can use the f i g - ure to give the colllnearlty of points, t h e order of points on a l i ne , t he location of a point in the Interior or e x t e r i o r of an angle or in a certain half-plane, and, i n general, t h e r e l a t i v e position of points, l i n e s , and planes. Things you cannot assume because "they look that way" to you are the congruence of segments

or angles, that a cer ta in point is a midpoint of a segment, that two l ines are perpendicular, nor that two angles are complementary.

1.. If In this figure AC = DB,

LACF 2 ZDBE and PC = EB, A prove tha t AF = DE.

2. In t h i s f igure BC = ED B

AC = AD and C

LACE 2 LADB.

Prove AACE ~ D B . A

Proof: (~111 in the blanks. ) D

1. Given. E

Statements

1. BC = ED. 2. CD = DC. 3 . BD = EC.

4. AC = AD.

5 LACE 2 LADB.

2.

3 . Addition, from statements 1 and 2.

Prove t h a t the diagonals of a

square are of equal l eng th .

(see no te preceding Problem 1

of Problem Set 5-4 .) Given: ABFH is a square. To prove: A F = BH.

In t h i s figure LAW 2 LRHQ and F is the midpoint of m. Can you prove AWBP 2 AQHF?

Explain. -

If A B F H is a square and AX,

"B? are congruent segments on -> ->

the rays AH, BF respectively, show t h a t m, are congruent. Restatement :

Given: ABFH is a square. + +

X, Y are points of AH, BF, respect ive ly . ^ Z W

- * - To prove: AY = EX.

In the f igure , X is between A and H, and Y is between B and F. Would the proof be affected if H were between

A and X, and F were between B and Y?

Suppose I t 1

f igure that and x = y.

s given In this AH = BF, r = m Prove that HB = FA.

7 a If In t he figure 1 T5?, - B R L ~ , AR = RX and ER = RY, prove t h a t A 3 = XY.

me I&osceles T r i a l @ Theorem. The A n ~ l e Bisector Theorem. 5-6. - A t t he end of Section 5-1 we mentioned the case of matching

up the vert ices of a tr iangle AABC in which at least two sides of t he triangle are of t h e same length. This, i n f a c t , i s the

case that we deal with In our first formally stated congruence theorem:

Theorem 5-2. If two sides o f a triangle are congruent, then

the angles opposite these s ides are congruent. Restatement: Given a t r iangle AABC. If AB = AC, then

LB 2 L C .

Proof : Consider the correspondence ABC - ACB,

between AABC and Itself. Under this correspondence, we see that ^B-AC, A5'-m, LA - L A .

Thus two sides and the Included angle of AABC are congruent to

the parts t h a t correspond to them. By the S.A.S. Postulate, this

means that AABC 2 AACB,

that is, t he correspondence ABC-ACB I s a congruence. By the

d e f i n i t i o n of a congruence between triangles all pairs of corresponding parts are congruent. Therefore

ZB 2 L C ,

because these angles are corresponding parts. We now show how t he above proof looks in t w o - c o l m form.

The same figure Is used.

Theorem 5-2, If two s i d e s of a triangle are congruent, then

the angles opposite these sides are congruent. - * -

Given : AABC with AB = AC.

To prove: L B Z L C .

Proof: Statements

- .%- 1. AB = AC. - -

AC 2 AB. f\f

2. L A = L A .

3 . AABC 2 AACB.

4. ZB 2 L C .

Reasons 1. Given

2. I d e n t i t y congruence.

3 . Steps 1 and 2 and the S.A.S. Postulate.

4. Definition of a congruence between triangles.

Usually, we will state theorems in words, as we have stated Theorem 5-2, and then restate them, using notation which will be the nota t ion of t h e proof ,

Definitions. A triangle w i t h two congruent sides is called isosceles. The remaining s ide is the - base. The two angles t h a t

include the base are base angles. In these terms, we can state Theorem 5-2 in this form: 'The base angles of an isosceles triangle are congruent. "

Definitions. A triangle whose three s ides are congruent is called equi la te ra l . A t r iangle no two of whose s ides are congruent is called scalene.

Definition. A triangle is equiangular if a l l three of its angles are congruent.

Using the term equiangular we s t a t e a theorem which r e a d i l y fol lows from Theorem 5-2. We denote t h i s theorem as Corollary 5-2-1. A corollary Is a theorem which is an easy consequence of another theorem. The proof of Corollary 5-2-1 is le f t f o r you t o do.

Corollary 5-2-1. Every equilateral triangle I s equiangular,

In proving theorems for yourself, you will need to make your own figures. It is important to draw figures in such a way that they remind you of what you know, without suggesting more than you

know. For example, the figure given in the proof of Theorem 5-2

looks l i k e an Isosceles triangle, and this I s as it should be,

because t h e hypothesis of the theorem says t h a t t h e triangle h a s

two congruent s ides . In the f igure for the S.A.S. Pos tu la te , i t f\f

looks a s if AABC = ADEF, and this I s as It should be, because

this is t h e situation dealt with In the postulate. But it would not have been good t o draw isosceles triangles to illustrate t h e

S.A.S. Postulate , because this would suggest t h i n g s t h a t the postulate doesn't say. +

Definition. A ray AD bisects , or is a bisector of, an angle ZBAC if D is in the i n t e r i o r of LBAC, and Z3AD 2 LDAC.

Ã‘

Note that if AD bisects LBAC, then m LBAD = m LDAC

Theorem 5-3. Every angle has exactly one b i s e c t o r .

Proof: Given L A . By the Point P l o t t i n g Theorem we can find B and C, points on the sides of L A , such that (1) AB = AC.

L e t D be the mid-point of BC, so that (2) DB = DC. Since AB = AC, I t follows by Theorem 5-2 that (3) LB 2' L C , h his follows even though the Isosceles triangle AABC is " ly ing on i t s s i d e . " ) From (l), (2) and ( 3 ) , and the S.A.S. Postulate it

follows that AABD 2 AACD.

Therefore , ZBAD 2 ZCAD, and so m ^BAD = m Z C A D , By t h e def ' l -

n i t i o n of bisector of an angle, this means t h a t & b i s e c t s LBAC.

To jus t i fy our use of t h e word "exactly" we must prove that

3 1s t h e only ray having t h l s property. Suppose there is a ray

A^ a l s o a b i s e c t o r of LA. Then m Z C A D = m ZCAE, s ince each of J. these equals Ã rn LBAC . Applying the Angle Construction Postu- c 0

l a t e to t h e half-plane with AC as edge shows that we must have 3 ~3, t h a t I s , 3 and 8 stand f o r t h e same ray. Hence, there is exactly one bisector .

The following definitions are useful I n discussing proper t ies of triangles.

D e f i n i t i o n . A median - - of a triangle is a segment whose end-

p o i n t s are one vertex of t h e triangle and the mid-point of the

opposite s ide . Ikf3ni t ion . An angle b i sec to r of a triangle is a segnent - -

whose end-points are one vertex of the t r i ang le and a po in t of the opposite side which lies in the ray bisecting the angle at the given vertex.

Note t h a t every triangle has three medians and three angle

bisec to r s . The figure shows one median and one angle b i sec to r of

t h e median from B, and is the angle b i sec to r from B.

Problem Set 5-6 A

1. I n t h e f i g u r e A B = A C . We s tar t t h e proof t h a t

f\f Zrn = Ln. complete t h i s

proof supplying reasons.

Proof : Statements Reasons

2. Z m is supplementary to ^ A = .

l n Is supplementary to Z ACB.

2. Given: In t h e figure FA = FD c

and AB = DC.

Prove : AAFB 2 ADFC,

AFBC 2 AFCB.

- 3 . If in the figure EB 2 m,

prove t h a t LEEA 2 ZECD.

5 . If AC = AB and CD = BD In the plane figure, show

LACD 2 ZABD.

A B C D 4 . If AB = AC and DB = DC in

the plane f igure , show that LABD 2 LACD.

c

6. Give a paragraph proof rather C than a two-column proof of the

following:

Given: X and Y are the mid-

points of the congruent sides - AC and of the i sosce les triangle ABC. A 8 To prove: ZCXY 2 LCYX-

7. Prove Corollary 5-2-1. v very equilateral triangle Is equiangular. ) A

8. Given equilateral triangle ABC w i t h Q, R and P, t h e mid-

points of the s ides as shown.

Prove that APQR is equilateral.

C R B 9 Prove the following: If median 1 of AFAB is perpendicular -

to side AB, then AFAB is Isosceles.

5-7 , The Angle Side Angle Theo~em, - Theorem 5-4. h he A.S.A. Theorem.) Given a correspondence

between two t r i a n g l e s , (or between a triangle and itself). If two

angles and the included side of the first t r i ang le are congruent t o the corresponding parts of the second triangle, then t h e correspondence is a congruence.

Restatement: L e t ABC-DSF be a correspondence between two triangles. If L A 2 ZD,

AB = DE,

then

c

2 . A 3 = DE and m L A = mLD. 3 . A A E Z ADEF'.

4 . L A B C 2 LDEFt.

Proof: Statements Reasons

Ã‘ Ã‘ 7 . EF and EF1 are the same 8. F = F ~ . ray.

4 1. On the ray DF there is a

point Ft such that DF' = AC.

9 . A A B C Z ADEF,

1. The Point P l o t t i n g Theorem.

2 . Given.

1 3 * The S.A.S. Postulate. 4. Definition of a congruence

I between triangles. 5. Given. 6. Steps k and 5, and the

definition of congruent angles .

7 . Step 6 and pos tu la te 12. # -^*

8. Two lines (EF and DF) intersect in at most one p o i n t .

9. Statements 3 and 8.

The proofs of t h e following theorem and corollary are left to the s t u d e n t . The proofs are analogous to those of Theorem 5-2

and Corollary 5-2-1.

Theorem 5-5- IF two angles of a t r i a n g l e are congruent, the sides oppos i te these angles are congruent.

Corollary 5-5-1. An equiangular triangle I s equilateral.

Problem Set 5-7

1. In some parts of this exercise there is not enough information to enable you to prove the two triangles are congruent even if you use all other facts that you know, f o r example, that "vertical angles are congruent". If it can be proved that the two triangles

are congruent, name the statement {A.S .A. or S.A.S.) supporting

your conclusion; if there is not enough information given to prove

the t r iangles are congruent, name another pa i r of congruent parts t h a t would enable you to prove them congruent. If there are two possibilities, name both.

134 - * -

a . Given only that AH = AB.

b. Given only that i^c 2 Ld. I= c . Given only t h a t L a 2 Z b

and LC 2 ^.d.

- - - d. Given only that AR = MR.

A e . Given only that L A 2 Z M .

f. Given only that LXFY 2 d KJ?Y.

g . Given only t h a t L X Y F 2 LKYF.

2. In accordance with the specifications at the left, list the

data which would correctly f i l l the blanks.

a. Side , angle, side of AABH: - m. -, HB. b. Angle, side, angle of AABH:

-' s, -- c . Angle, s ide , angle of ABFH:

LF, -9 L HBF. A A F d. S ide , angle, side of ABFH:

7

mj -9 -*

3 . Follow the di rec t ions of Problem 2. a. Angle, side, angle of AABP: -

-' BF, -- b. Side, angle, s ide of ARAF:

-' ZR, c . Side, angle, side of ARAB:

- , LB, -*

d. Side, angle, side of ARAB: - m, - 9 m.

e . Angle, side, angle of ARAF:

L R , -9 L. RFA. f . Angle, side, angle of AAFB:

ZFAB, 3, -.

Follow t h e d i rec t ions of Problem 2. a. Side, angle, side of AHFB:

-' ZHW, -. b. Angle, side, angle o f AABH: - - , HB, - m

c . Side, angle, side of AHFB: - HE, __> w.

d. Angle, side, angle o f AHFB: - -' 3F, -*

e . Side, angle, s i d e of AABH: - - AH9 -3 AB.

I f bisects and L a 2 Lb in the f igure, prove that cS? G bisec t s z.

A

pLV; Prove Theorem 5-5.(If two angles of a triangle are congruent, the sides opposite these angles are congruent.) Restatement: If I n AABC,

L B 2 &, then AB = AC.

Hint: Use congruency of the triangle with I t se l f .

B C Prove Corollary 5-5-1. v very equiangular triangle is equilateral . ) U s e a paragraph proof.

If AABC is equilateral, prove AAEC 2 ACAB.

If the b isector of Z G In A FGH Is perpendicular to the opposi te side at K, then t r iangle FGH is isosceles.

Given: The figure with f^

Z x = Z y and - H3 m. -

Prove; HP 2 HR.

136

11. I n t h e figure, ffi bisec t s LRMS and LRWK 2 LSWK. R

Can it be proved that Z R 2 ZS? If so, do so,

12. Prove that AN 2 if 2 m, '\1

L A = LR and Zx d y in

t h e f igure .

"13. a . I f , in t he figure, X is R

- the midpoint of AB,

LA 2 ZB and LAXR 2 LBXF,

show that 2 m. ,

b. -Do you need as a part of the

hypothesis that the figure A x B lies in a plane?

<Vf 14. Given: La = Z b and

Z w 2 Zs in the figure. - w - Prove: GR = KH.

G H

Can the following be proved on t he basis of the informa-

t i o n given?

Given: LAOB with OA = OB

and P, Q, points on rays a, 66 with AQ = BP. 0 A: 0 B

Prove: OP = 00.

Prove t h a t RX = RY if it is

given t ha t In the figure: BQ = TS, m d B = m LT and m -^Q = m L S .

S Y T

The Side Side Side Theorem. 5-8- ---- Theorem 5-6. ( ~ h e S. S . S, Theorem. ) Given a correspondence

between two triangles (or between a triangle and i t se l f ) . If all

three pairs of corresponding s i d e s are congruent, then the correspondence is a congruence.

Restatement : Let ABC -DEF be a correspondence between two

triangles. If A 3 = DE, AC = DF, s

BG = EF, 8 then AABC 2 ADEF.

Proof: Statements

4 1. There I s a ray AG such that

L C A G s- LFDE, and such t h a t

a^. G are on opposl'te side

of 4

2. There is a point E1 on AG such t h a t AEt = DE.

Reasons

1. The Angle Construction Postulate.

2. The Point Plotting Theorem.

3 . The S.A.St Postulate .

What we have done, so far, is to duplicate ADEF on the

under side of AABC, using the S.A.S. Postulate.

6 . The segment 'E1?1 intersects t h e line 3' In a point H.

4 . AB = DE by hypothesis; and DE = AE1, from State- ment 2 ,

5. BC = EF, by hypothesis; and EF = EtC from Sta te - ment 3.

6 . By Statement 1, 3 and E 1 are on opp I t e aides of the line a.

We shall now complete the proof for t he case in which H I s

between A and C, as in the figure. The other possible cases

will be discussed later.

7 . LABH 2 LAEIH. 8. ZCBH 2 LCEtH. 9 . m L A B H + mLCBH = m L A B C .

10. m f A E ' H + m^CETH = m LAETC. 11. . L A X 2 Z A E I C .

12. LABC ZDEF.

13. AABG 2 ADEF.

7 . Statement 4 and Theorem 5-2. 8. Statement 5 and Theorem 5-2. 9. The Angle Addition Postulate. 10. The Angle Addition Postulate. 11. Statements 7, 8, 9 and 10. 12. Statements 3 and 11,

13. Statement 12, the hypothesis, and the S.A.S. Postulate.

This completes the proof for the case in which H is between <->

A and C. We recall that H is t h e point in which the l i n e BE1 <-Ã

intersects the l i n e AC, If H = A, then B, A and E t are co l -

linear, and the f igure looks l i k e t h i s :

In this case Z B 2 Z E r because the base angles of an isos-

celes triangle are congruent. Therefore dB 2 LE, "because

LE 2 LET. The S.A.S. Pos tu l a t e applies, as before, to show that A A X 2 ADEF.

If A is between H and C, then the figure looks l i k e this:

and we show that ZABG 2 LE by subtracting the measures of angles, instead of by adding them. That is,

m L A E = m L W - mLHI3A

and m LAErC = m ZHEIC - m ZHEtA,

so t h a t LAW, LAEtC 2 LDEF, as before . The rest of the proof I s exactly the same as in the

fir st case . The two remaining cases, H = C and C between A and H,

are similar to the two above.

Pmblem Set 5-8 . A

1. Given: AABF and AAHP with AHSm and H?zm.

Prove: LHAFZ ZBAF.

2. In the figure, 2 i% and - + - f\i AH = FB. Show that .P = Ls .

3. In the figure, z m and ^? 2 E. Prove t h a t Z H 2 ZR.

Ã 4. Consider the pairs of triangles pictured below. If on the basis of our information t o date they can be proved congruent, t e l l which congruency statement you would use.

b

too" m A 4 B F

d .

A 7 B xa A s B ew5K

Consider ARWM and Am. AW = XM, AB = XR, /A 2 /x.

h. Consider: i. ARMW and AQMH.

J . AWMX and AHMK.

F W R W xQ X

5. A suppl ier wishes to telegraph a manufacturer f o r some parts in the form of triangular metal shee ts . In a d d i t i o n to the thickness, kind of metal, and number of pieces wanted, what

I s the least he can say in order to speci fy the s ize and shape of the t r i a n g l e s ? (consider the possibility of more than one cho ice . )

6 . Prove t h e following theorem: If the bisector of the angle opposite the base in an isosceles triangle in tersects the base, It is perpendicular to

the base,

Restatement : A H B

Given : & A X with AC = 3C and

H a point on such that ZACH 2 ZBCH.

To prove: mLm. 7 . Prove the theorem that t h e

median from t h e vertex of an isosceles triangle is the b i - sector of the ve r tex angle. A

D B

8. Prove the theorem: The b i sec to r of the ver tex angle o f an

i sosce les triangle is t h e perpendicular bisector of the base. Restatement:

Given: A ABF w i t h AF = BF and - H a point on AB such t h a t bisects LAFB. - To prove : AH 2 and

mL'.

A A H B

9. a. Given: In the figure, - 4 . - AF = BR and --"- AR = BF.

Prove : LAW 2 ZBFR. - h he gap in RB was left there

so t h a t the figure would not reveal whether or not - intersects A F . )

b. Do you need as part of the hypothesis that the figure

l i es in a plane?

10. a. Given: In the f igure , AH = FB, A 3 = FH, and - RQ bisects in K.

Prove: OK = RK. Is the figure necessarily planar?

11. Given: square ABGD with P, Q, R, S the midpoints of m, - PC, m, m, respectively.

Prove: A P Q S Z ARQS.

12. Point out why the following argument Is circular, and thereby Invalid.

Theorem: The base angles of an i sosce les triangleare congruent. - + - Given: A ABC with AB = AC.

To prove: 0 2 L C .

Proof: Statements Reasons - * -

1. AB = AC. - * - 2. AC = AB. - 2 - 3. BC - CB.

4. A A B C S AACB.

5. d B ^ L C .

---

1. Given. 2. Given. 3. Identity. 4. S.S.S.

5. Defini t ion of congruent tr iangles.

Point out why the following argument is c i r c u l a r .

Theorem: Given a correspondence between two tr iangles (or be-

tween a triangle and ~ t s e l f ) . If two s i d e s and the Included angle of the f i r s t triangle are congruent to the corresponding parts of the second t r i a n g l e , then t he correspondence is a congruence.

Given: ABC-DEF, B z m , KT=-, LABC 2 LDEP. Prove: A A E Z ADEF.

Proof: Statement Reasons

Let 3 be the ray on <Â¥

t h e same s ide of A 3 as 1 I-

AABC ^ ADEF. 112.

!& such that LBAC t % ADF, Â¥<-

Angle-Construction

Postulate.

intersecting BC in C t .

L A E ' 2 LABC. ->

C t is on ray BC, from s tep 1.

Given.

S t e p s 2 and 3. Given.

Step 1.

A . S . A .

Corresponding parts.

Given . Steps 8 and 9. Step 10 and the reason f o r Step 2.

Steps 7 and 11.

2 .

'Vi If La = Z b and Lm 2 Lw in the figure, prove that

A K F ' T B .

If in t h e figure mJJl# at F, Elz at A, and m / l a = m z; b y can you prove that FB = AB? If so, do so.

*

In A HAP, 'points 3 and W are on

sides T^ and AH, respect ive ly , and mlm, = L A ^ , and

AW = AB.

Prove: rn = HB. A A 8 F

If In he f igure X L m , "S i b i s e c t s LAW,

bisects LAQF and

bisec t s LF'QR, - * -

prove t h a t BQ = HQ.

1 8 . In AABC and AHRW, AB = HR,

AC = HW and median 2 median ' S. On t h e basis of theoreins you have

had so far, can yov show that AABC 2 AHRIJ? H ,.

If so, do so.

Â¥i

19. Use the diagram f o r Problem 18 and suppose now it is given - *

that AB = HR, EC = RW, and median AF = median m. Can you

prove t h a t AABC 2 AHRW? If so, do so. * 20. Given: Points A, R, S, and C lie on Line L.

R l i e s b e t w e e n A and S.

S lies between R and C.

3 and D do no t l i e on L.

AR = CS,

AB = CD. BS = DR. a. P r o v e t h a t : L B S A Z LDRC.

b. Need the poin ts A, R, S, C, B, D be coplanar? * 21, In this figure D is the midpoint -

of AG, BE, and m. Prove tha t AEFG 2 A K A . C F

-

:22. Does t h e proof for Problem 21 hold even if the segments m; m, ^5 are not coplanar?

23, Gtven: I n the figure, m l m . - .%- RQ = SQ. - RC 2 SC". Prove that: ZRCA 2 LSCA.

v 24. A t r i p o d wi th three legs of

equal lengths VA, VB, VC

stands on a plane. a. What can you say, if anything, about the distances

AB, AC, BC? About the six angles O A B , LVAC, LVBA, e tc?

b. Answer part (a) if you are given also t h a t the tripod legs make congruent angles with each o t h e r ; t h a t is, L A V B 2 LBVC 2 LAVC.

K- 25. a . L e t and bisect each o the r at M. Prove t h a t

A 3 = RQ and AQ = RB.

b. Now l e t ^Z also be bisected at M. How many pairs of

congruent segments, as in (a) can you f i n d ?

c . You probably thought of as l y ing in the same plane as - AR and S . Is t h i s necessary, or do your conclusions In (b)

hold even if s t i c k s out of the plane of and B^? T r y t o visualize t h e figure in t h e l a t t e r case, and either

draw a p ic tu re or make a model. 26. L e t A A E be any t~tangle and D a point In the plane of

t h i s triangle. The s e t consisting of the union of s i x seg- - merits AB, my E, AD, m, CD we shall call a skeleton o f a

tetrahedron. Each of t h e six segments I s called an edge of t he tetrahedron, each of t h e four points A, B, C, D is a vertex, each tr iangle formed by three ver t i ces is a face, each angle of a face is a face angle. Edges and faces of a tetrahedron were considered In Problem 11 of Problem Set 3-lc, a. How many f ace s are there? How many face angles?

b. Two edges of a tetrahedron are opposite edges if they do

not intersect. They are adjacent if they do Intersect. If each pa i r of opposite edges are congruent, are any of

the faces congruent? If each pair of adjacent edges are

congruent, what k ind of triangles are t h e faces? c . Construct an equilateral skeleton of a tetrahedron w i t h

toothpicks and quick-drying glue or w i t h soda straws by

threading string through them.

[sec. 5-81

Review Problems f o r Chapter 2

Complete:

If the ver t ices of two triangles correspond so that every pair of corresponding angles are and every pair of cor-

responding are congruent, then the correspondence is a between the two triangles. Consider t he set of abbreviations A.S.A., S.S.A., S.A.S.,

S.S.S., A .A.A. a . Which subsets are abbreviations of postulates in this

chapter? b. Which subsets are abbreviations of theorems proved in

this chapter? If ARST I s I sosce les with RT = ST, what correspondences are congruences be tween

the triangle and itself?

Given A F = B F and D F = E F ,

what would be t h e final reason in the most direct proof that

AAFD 2 A m ? That AAEC A m ?

1 A 6

Given: In the f igure AR = RH and A F = BH. Prove; R 3 = RF.

In the figure f o r Problem 5, if RB = Ry and AB = HF, prove tha t AR = HR.

7. A person wishes to find the distance - - - - - --- across a river. He -- 7

does t h i s by sighting River - a tree, T, on the other -- side opposite a point P,

- perpendicular to PQ at Q until \ he determines the point X where his

<-> path meets line TM. What o t h e r segment - In the figure has the same length as TP? \ What Is the principal theorem used in show- \ ing that: ATPM 2 AXQM ?

such that 7S?Im. - * - - -- Marking the midpoint, M, +

of m, he paces a path

8. Napoleon's forces, marching into enemy territory, came upon a

stream whose width they d i d not know. Although the engineers

were in t h e rear, nevertheless, the impetuous commander de- manded of h i s off ice rs the width of the river. A young officer Immediately stood erect on the bank and pulled the

v i so r of his cap down over his eyes u n t i l h i s line of vis ion was on the opposite shore, He then turned and sighted along t he shore and noted the point where h i s visor r e s t ed . He then paced off t h i s distance along the-shore. Was this d i s t ance the width of the river? What t w o triangles were congruent?

Why?

In ARST: Point X lies between S and T, and SX = SR. Point Q lies between R and T, and - SQ bisects L S , ^X I s drawn Find an angle congruent t o

LR, and establish the con- S x

Given: The figure with A R BLTR, m - m , Lx '='Ly, QB = WH and F,

the midpoint of m. Prove: ABF& 2 A m .

Gtven: In the f igure ,

A% = AR and LBAH 2 ZRAH.

Prove: F3 = FR.

R 12. In this figure given that:

A 3 = KF and +

RB = RJ?. Prove: AAFR 2 AHBR.

13. Given: In t h e figure,

AB = FB and Iffi = RB.

Prove : A AQR 2 A FQM.

14. In t h i s figure given R that B and F trisect* %B,

A Z L - H and A R = H Q .

Prove: BW = FW. * Trisect means to separate Into - three congruent parts. A F

15. In th is figure, given that HA = KB, + AF bisects L HAB and * BF bisects L HBA.

Prove: AF = BF.

16. A polygon ABODE has five sides

of equal length and f ive angles of equal measure. Prove that

LDAB 2 LDBA.

17. Prove: If two medians of a triangle are perpendicular to their respective s ides , then the triangle is equilateral.

18. I n t h i s f igu re "SB 2 fSS and

^ '\i m* Prove : L A 2 ZH and m 2 m.

19. Prove that the b i s e c t o r s of a pair of corresponding angles of two congruent t r iangles are congruent.

20. In t h i s figure it X is given that:

m = QR,

~a Lb, Z X 2 ZQ.

Prove: KA = KM.

21. In t h i s figure I t is given that Zl 2 Z2, Z3 = L4,

and JT = JB.

Prove: ~5 2 L 6 .

22. If PA = PB and QA = QB then LAP& 2 ZBPQ. Will the same

proof hold regardless of whether

A is in the same plane as P, Q, and 3 ?

a. Prove: If PA = FB, QA = QB and R I s on ?$ as shown I n

the figure, then RA = RB.

b. Must the five points be

coplanar? Will the proof hold whether or not A is i n the

same plane as B, R, P, and Q?

In t h i s f igure, points F and H t r l s e c t AT, and points F

and 3 trisect MR. If AF = FB, I s A ABT A MHR? Prove your answer.

<-Ã If RS I s perpendicular to each of three different rays, & & A

RA, RB, RC at R and RA = RB = RC, prove that SA = SB = SC.

(Draw your own f igure. )

Q Let A PAB and A &B lie in

di f fe ren t planes but have the - common side AB. L e t A PAB 2 A QAB. B

Prove that If X is any po in t in

A 3 then A PQX is isosceles. P A

Complete Euclidls proof of the theorem that the base angles of an isosceles triangle are congruent. Given: A 3 = AC . Prove: L ACB 2 L AX. Construction: Take a point F

with B between A and F, and a point H w i t h C between A and H so t h a t AH = A F . Draw

CT and BtT.

*28. Given: The p l a n e figure ABCD AB = CD, AD = BC. - Prove: AC and bisect

each other.

*29. Given: The figure ABCD w i t h

AB = AC, DB = DC, and

LBAX 2 LXAY 2 ^CAY.

Prove: AX = AY.

Review

Chapters 1 to 5 REVIEW EXERCISES

Write the numbers from 1 to 80. Follow each with a 11+" or a "-" to i n d i c a t e whether you cons ide r the statement t r u e or false . True will mean "true under all conditionsi1.

Every two rays Intersect. - AB designates a l i n e . If m Â£ Q = 100, then .L Q has no complement. A l ine and a point not on it determine a plane. If a poin t I s In the Interior of two angles of a triangle It

is in the i n t e r i o r of the triangle. If a line Intersects a plane n o t containing it, then the i n t e r - sect ion is one point. The union of two h a l f planes is a Whole plane. A poin t which belongs to t he in te r ior of an angle belongs to

the angle, If = m, then either A = C or A = D. The intersection of every two half planes Is the i n t e r io r of an angle. The i n t e r i o r of every triangle is convex. It Is poss ib le t o f i n d two s e t s , neither of them convex, which

have a union which is convex. A ray has two end-points,

Experimentation Is always the b e s t way of reaching a v a l i d

conclusion. Given four different points , no three of which are collinear, there are exactly six d i f f e r e n t lines determined by pairs of

these points . - If m L R S T = m LXYZ, then^ R S T ~ Z XYZ.

In the f igu re the best way fco name + +a the angle formed by DA and DC is Â£ D.

In this text b e t w e e n for points on a l i n e I s an undefined

term.

The vert ices of a triangle are non-collinear . The intersection of two s e t s I s the s e t of all elements that

belong to one o r b o t h of them.

Every statement about geometric f igu res which is not a def in i -

tion can be proved.

If A X Y Z 2 A CAB, then L A 2 L X. It is possible f o r two lines to intersect in such a way that three of the angles formed have measures 20, 70, and 20. . Each side of an angle Is a ray. All nouns which the t e x t uses that relate to geometry are def ined in the t e x t . The i n t e r i o r of an angle is a convex s e t ,

If rn L A B C = 37 and m L DEF = 63, t h e n L- A E and i- DEF are

complementary.

If A is not between B and C, "then C Is between A and B.

m is never a negative number. If poin t Q Is in the exterior o f t - A N , then Q and C are

@ on the same s i d e of AB.

The distance between two points is the absolute value of t h e

sum of the i r coordinates.

The longest side of any triangle I s called I t s hypotenuse. # Â¥<-

If A B ~ C D at p o h t P [different from po in t s A, B, C, D), then m L APC + m L C P B + m LBPD + m L DPA = 360. Given a line, there is one and only one plane containing it. A rational number is one which is t h e r a t io of two integers.

Given two po in t s on a l i n e , a coordinate system can be chosen so that the coordinate of one point is zero and the coordinate

of the other one Is negative. Two triangles are congruent i f two s ides and an angle of one are congruent to two sides and an angle of the other. A collinear s e t of po in t s is a line.

The absolute value of every real number except zero I s

p o s i t i v e .

If CD + CE = DE, then D is between C And E. If in A A E , m^. A = m ^.B = m f C, then AB = BC = AC. + + If, in a plane 2, P T ~ l ine L, PQL line L, and P 1s on

0 <&Ã L, then PT = PQ. From the statements (1) If q is false, then p is false,

and (2) p is t rue , we can conclude that q is true. The Ruler P o s t u l a t e s t a t e s t h a t any unit can be reduced t o

Inches.

If R I s a po in t in the i n t e r i o r of L X Y Z , then m L XYR + m L Z Y R = rn L XYZ. There are certain points on a number scale which are not in correspondence with any number. Every line is a colllnear s e t of points. - n = n.

The distance between two points is a positive number. From the facts that m L AOB = 20 and m L BOC - 30 it can

be concluded that rn L AOC = 50.

A p o i n t on the edge of a half-plane belongs to t h a t half-plane.

A line L in a plane E separates the plane i n t o two convex s e t s .

The median of a triangte b i sec t s the s ide to which it is drawn.

If two points l i e In the same half-plane, then the line de te r - mined by them does no t in tersect t h e edge of that half -plane. If two supplementary angles are congruent, each is a right angle.

The Interior of an angle Includes t h e angle itself.

Vertical angles have equal measures. The sides of an angle are rays whose intersection is the vertex of the angle. If L C I s supplementary t o LA and m L A = 67, then m L C = 113,

If two lines intersect, there are exactly two points of each

which are contained by t h e other.

If two angles have equal measures the angles must be congruent. From the statement (1) If p is t rue , then q Is t r u e , and (2) p is not true, we can conclude that q is false.

It h a s been proved In t h e first five chapters of t h i s text t h a t the sum of the measures of the angles of a triangle is

180.

The s i d e s of a triangle are lines. The midpoint of a segment separates it into two rays. If two l i n e s Intersect so t h a t the ve r t i c a l angles formed are supplementary, then the measure of each angle is 90.

If m L B = 93, then L I3 is acute.

For all numbers x, 1x1 = x. - + The intersection of AB and BA is E, - In A ABC all points of BC are in the interior of L A.

If A ABC 2 A BCA, then A A X is equilateral. 2 2 If 1x1 = jy l , then x = y .

A ABC and A RFH which are In different planes are congruent if AB = RF, PC = FH and AC = RH.

A ABC 2 A MQT if AB = QM, BC = TQ and A Q '= L B . Median a in AACE bisects L A .

2 2 IS x = y , then 1x1 = y l .

If three points are on three d i f fe ren t l i n e s , the points are

There is no A ABC in which L A = L B.

Two po in t s not on a plane are in opposite half-spaces deter-

mined by the plane i f and only if t h e segment joining them

intersects the plane.

Chapter 6

A CLOSER LOOK AT PROOF

6-1. How A Deductive System Works. -d

In Chapter 1 we t r i e d to explain in general terms how our study of geometry was going to work. After the experience that

you have had since then, you ought to be In a much better position

to understand the explanation. The idea of a set , the methods of algebra, and t h e process

of logical reasoning, are things that we have been working with.

The geometry Itself is what we have been working - on. We started with point, line and plane as undefined terns; and so f a r , w e

have used f ifteen postulates. Sometimes, new terms have been defined by an appeal to postulates. (FOP example, t h e distance

PQ was defined to be the positive number given by the Distance Postulate. ) Sometimes definitions have been based only on the undefined terms. o or example a set of points is c o l l l n e a r if a l l points of the set l i e on the same line. 1 But at every polnt we have b u i l t our definitions with terms that were, in some way, previously known. By now we have p i l e d d e f i n i t i o n s on top of each other so often that t h e list is very long. And in f ac t , the length of the list I s one of the main reasons why we had to be careful, a t the outset, to keep the record straight.

In the same way, a l l the statements t h a t we make about geometry are based ultimately on the postulates. Sometimes we

have proved theorems directly from postulates, and sometimes we have based our proofs on theorems that were already proved. But

i n every case, the chain of reasoning can be traced backward t o

the postulates . You might find it a good idea, a t th i s point, t o reread the

second half of Chapter 1. It will seem much c learer to you now

than i t d i d the first time. It is much easier to look back, and understand what you have done, than to understand an explanation of what you are about to do.

6-2. Indirect Proof.

We remarked in Chapter 1 t h a t the best way to learn about logical reasoning is by doing some of it. There is one kind of

proof, however, that may require some additional discussion. For

!heorern 3-1, we used what I s called an Indirect proof. !The theorem and i t s proof were as follows;

II Theorem 5-1. Two different lines intersect in at most one point.

Proof: It is Impossible for two different lines to intersect i n two d i f f e r e n t points P and Q. This is impossible

because by Postulate 1 there is only one l i ne that contains both P and Q."

Probably this was the first time that you had seen this kind

of reasoning used in mathematics, but you must have encountered t h e same s o r t of thing, many times, in ordinary conversation.

Both of the following remarks are examples of indirect proofs;

(1) "1t must be raining outside. If it were not raining, then those people corning in the door would be dry, but they

a r e soaking wet. "

(2) "Today must not be the r ight day for the f o o t b a l l game.

If the game were today, then the stadium would be full of people, but you and I are the only ones here. It

In each of these cases, the speaker wants t o show t h a t his first statement is true. He starts his proof by supposing t h a t

the thing he wants t o prove is wrong; and then he observes that this leads to a conclusion which contradicts a known fact. In

t h e first case, the supposition is that it is not raining; this leads to t h e conclusion tha t the people coming in would be dry,

which contradicts the known f a c t that these people are w e t ; and

therefore it is raining, after all. Similarly, in the second case the assumption that the game Is today leads t o a contradiction

of t he known fact of the empty stadium.

In the proof of Theorem 3-1, the supposition is that some

two di f fe ren t lines i n t e r s ec t in two points. By Postulate 1,

t h i s leads to the conclusion that the l i n e s aren't different

af te r a l l . Therefore t h e supposition is wrong, and this means

that the theorem I s right.

Problem S e t 6-2a - ---.

1. For the sake of argument accept each of the following assump- t i o n s and then give a logical completion for each conclusion. a. Assumption: Only men are co lo r b l i n d .

Conclusion: My mother --------. b. Assumption: A l l men are left-handed.

Conclusion: My brother --------. c . Assumption: The only th ing that makes Jane ill i s hot

chocolate. Jane I s ill. Conclusion: Jane --------.

2, Which of the following arguments are indirect? The temperature outside must be above 32' F. If t he

temperature were not above 32', then the snow would not be melting. But it I s melting. Therefore, the

temperature must be above 32O.

That movie must be very entertaining. If It were not very e n t e r t a i n i n g , then only a few people would go to see It. But large crowds are going to see it. There-

fore, It must be very entertaining. The air-conditioning in this building must not be working correc t ly . If it were working correctly, then the temperature would not be so high. But the

temperature is uncomfortably high. Therefore, the

air-conditioning i s not working correctly. 3 . Mrs. Adams purchased a set of knives, forks, and spoons

advertised as a s t a i n l e s s steel product. After u s ing the set for several months, she found that the set was beginning

to rust. She thereupon decided that the set was not stain-

less s t ee l and returned it for refund. In this example of Indirect proof identify (1) the

statement to be proved, (2) the supposition made, (3) the conclusion resulting from the supposition, and ( 4 ) the Icnown fact contradictory to ( 5 ) .

What conclusions can you draw from the following hypothesis in which x, y and z stand for di f fe ren t statements? If x is true, then y I s t rue . If y I s true, then z is t r u e . x is true.

Suppose you have the fol lowing data: If w is true, then v is true.

If u is true, then w is true. If x is true, then u is t r ue - v is not true. What conclusions can you draw? Did you use indirect reason-

ing at any point? What conclusion Follows from the following data?

(1) Nobody is allowed to j o in the swimming club unless he

can play the piccolo.

(2) No t u r t l e can play the piccolo. ( 3 ) Nobody I s allowed t o wear striped trunks In the club

pool unless he is a member of the swimming club.

( 4 ) I always wear s tr iped trunks In the c lub pool , (Hint : This problem becomes easier if you convert it to i f - then form, as in several preceding problems. For

example, let A be "someone is a member of the swimming

club", let 3 be "someone can play the piccolo", e t c . )

If A is green, then B is red.

If A I s blue, then B is black.

If I3 is red, then Y is white.

a. A is green, so 3 is and Y is

b . B Is black . Is it possible to draw a conclusion concerning A? If so, what conclusion?

Prove that the bisector of any angle of a scalene t r i a n g l e

cannot be perpendicular to the opposite side.

L e t us now prove the o the r theorems of Chapter 3 . For

convenience, we first restate the postulates on which these proofs are based:

Postulate 1. Given any t w o different points, there is exactly one line which contains both of them.

[ sec. 6-21

1 Postulate 5. a. Every plane contains a t least t h ree non-

1 collinear points. b. Space contains a t l e a s t f o u r non- 1 coplanar po in t s .

Postulate 6. If two points l i e in a plane, then the line

containing these points lies in the same plane.

Pos tu la te 7. Any three points lie in a t l ea s t one plane, and any three non-colllnear p o i n t s l i e in exactly one plane, More brief ly , any three points are coplanar, and any three non-col l lnear points determine a plane.

Theorem s. If a line In te rsec t s a plane not containing

it, then the intersection is a single point.

Proof: By hypothesis, we have a line L and a plane E, and

(1) L intersects E In a t least one point P, and

(2) E does not contain L.

We are going to g ive an i n d i r e c t proof of the theorem and therefore we start by supposing t h a t the conclusion is false. Thus our supposition is that

( 3 ) L in tersects E In some o t h e r point Q,

To give an indirect proof, we need to show that our suppo- s i t i o n contradicts a known fact. And it does: If P and Q

lie i n E, it fol lows by Postulate 6 that the l i n e conta in ing

them lies in E. Therefore

( ' 1 ) L lies in E, This contradicts (2). Therefore t h e supposition (3 ) i s

Impossible. Therefore Theorem 3-2 is true.

Notice that the figures tha t we use to Illustrate Indirect proofs look peculiar. In the figure f o r Theorem 3-2, we have indicated a point Q, merely to remind ourselves of the notation

of the proof. The proof itself shows t h a t no such point Q can possibly e x i s t . In fac t , the figures for indirect proofs always look ridiculous, for a good reason: they are pictures of impos- s ib le situations. If we had drawn a figure for Theorem 3-1, it would have looked even worse, perhaps like this:

Th is I s a picture of an impossible situation in which two different lines intersect in two different points.

Theorem 3-5. Given a line and a point not on the l i n e , there is exactly one plane containing both of them.

Proof: By hypothesis we have a l i ne L and a point P

not on L. By the Ruler Postulate we know that every l i n e contains Inf ini te ly many points , and so L contains two poin ts Q and R. By Postulate 7 there exists a plane E which contains

P, Q, and R. Since by Postulate 6, E contains L, we have shown that there exists a plane E containing both L and P.

A t this p o i n t we actually have proved only half of the

theorem, since Theorem 3-3 says there is exactly - one such plane.

It remains to prove that no other plane containing L and P

exists. We do t h i s by Indirect proof Suppose t h a t there is another plane E' containing L and

P. Since by Postulate 1 L is the only line containing Q and

R, we know that Q and R, as w e l l as P, lie in E* This

con t r ad ic t s Postulate 7 which says t h a t exactly one plane contains

three non-collinear points Since E was established as a plane

containing P, Q and R, E1 can not exist, and E is the

only plane containing L and P, The t w o p a r t s of the proof of Theorem 5-3 bring up the

distinction between existence and uniqueness. The first hal f

o f t h e proof shows the existence of a plane E containing L and P. This leaves open the possibility t h a t there may be more than one such plane. The second half of the proof shows the

uniqueness of the plane. When we prove existence, we show that t he r e is a t l eas t one object of a certaln klnd. When we prove

uniqueness we show t h a t there is a t most one. If we prove both --- existence and uniqueness, th i s means that there is exactly E.

For example, f o r t he f l e a s on a stray dog, we can usually prove existence, but not uniqueness. (it I s a very lucky dog tha t has only one f lea . ) For t h e eldest daughters of a given woman, we can obviously prove uniqueness, but not necessarily existence; some women have no daughters a t a l l . For the points

common t o two different segments, we don't necessarily have e i t he r existence or uniqueness; the Intersection may contain many points, or exactly one point, or no points at all.

The phrase o n e and only o n e is of ten used instead oi tr exactly one' since I t emphasizes the double nature of t he

statement.

The following theorem breaks up into two parts in exactly t he same way:

Theorem E. Given two intersecting lines, there 2s exactly one plane containing them.

For variety we give the proof in double-column form. Note

the two par ts and the way we handle the Indirect proof In the

second part .

Proof: We have given the l i n e s L, and Ln, Intersecting in the point P.

Statements Reasons

L conta ins a point Q, different from P.

Q is not on %. There I s exactly one plane E, containing 5 and Q. E contains L-,.

1. By t he Ruler Postulate, every line contains in- f i n i t e l y many po in t s .

2. Theorem 3-1.

3 . Theorem 3-3.

4. By Postulate 6 , since E contains P and Q.

E I s the only plane containing Ll and L2.

Suppose tha t another plane F also contains L, and L2.

8. Step 7 contradicts Theorem 3-3.

P contains Q. E and F each contain L2 and Q.

Problem Set 6-2b --

6 . Q is on L,. 7. Steps 3 and 4, and 5 and 6.

Is a t r i ang le necessar i ly a plane figure? Explain,

Theorem 3-4 says, In effect, "Two intersecting l ines determine a plane". How many different planes are determined by pairs of intersecting l ines in this figure?

Assume that the three lines are not a l l in the same

plane. L i s t each plane by naming the two intersecting l i n e s t h a t determine i t .

5 . How many d i f fe ren t planes are determined by pairs of the < - Ã ˆ < Â ¥ Ã ˆ *

four d i f ferent l ines AQ, BQ, CQ, and DO., no three of whlch l i e in the same plane? L i s t the planes by naming f o r each the two intersecting lines that determine it.

4. If, in a plane 2, 3 1 line L and "pS -1 line *- Ã‡Ã

what conclusion can you draw regarding PQ and PT?

As indicated in this figure, A and B l i e In plane P. <->

Q lies above plane P. Does l i n e AB l i e entirely in P? Quote a postulate or theorem to support your conclusion. There is a second plane implicit In the situation. Name it by the three points which determine It. What is the intersection of these two planes? A t what point wi l l <-Ã

QB intersect plane F? 6. If A, B, C, D are four non-collinear points, list

a l l the planes determined by subsets of A, B, C, D.

6-3 . Theoxems about Perpendl culars . Some of the basic theorems about perpendicular lines are

good examples of existence, uniqueness, and i n d i r e c t proofs. Theorem 6-1. In a given plane, through a given point of

a given line of the plane, t h e r e passes one and only one l i n e perpendicular to t he given l i ne .

Given: E is a plane, L a l i n e in E, and P a point of L.

To prove: (1) There is a line M In E, such that M contains P and M 1 L;

(2) There I s at most one line in E, containing

P and perpendicular to L. [sec. 6-31

Le t H be

Proof of (1):

I M one o f t h e two ha l f -plan

J

Y '

- P

es in E t h a t ha

>

ti

- x - L

as an edge, and let X be a po in t of L, d i f f e r en t from

the Angle Construction Postulate, there is a point Y of H, such that L X P Y 18 a right angle. Let M be t h e l i n e % Then M -L L. Thus we have proved that there 1& at least one -- l i n e satisfying the conditions of the theorem.

Proof of (2 ) : We now need to prove t h a t there is a t mos t

one S L C ~ line. Suppose that there are two of them, MI and M2. L e t X be a point of L, different from P.

Then the l i n e s M, and Mp contain rays PY., and FY2 lying

in t he same half-plane H having L as its edge. By definition

of perpendicular lines, one of the angles determined by L and

M, I s a right angle, and by Theorem 4-8 all four of these angles are right angles. Thus r n f X P Y l = 90. Similarly, mLXPYp = 90. But this contradicts the Angle Construction * Postulate, which says that there I s only one ray PY, with Y in H , such t ha t m L X P Y = 90. This contradiction means that our assumption of two perpendiculars M, and M2 must be

false, which proves the second half of the theorem.

The condition 'ln a given p lane ' I s an important p a r t of the statement of t h i s theorem. If this condition were omitted the first (existence) part of t he theorem would s t i l l

be true but t he second (uniqueness) part would not. This is

easi ly seen by thinking of the relation between the spokes of a wheel and t h e axle. Thus l eav ing out t h i s condition would

give us an example of a geometric existence theorem with no corresponding uniqueness theorem. The opposite situation, a uniqueness theorem with no corresponding existence theorem,

has already been considered in this chapter. Can you i d e n t i f y i fc?

Defini t ion. The perpendicular bisector of a segment, In a plane, is the l i n e In t h e plane which is perpendicular to the segment and contains the mid-point .

Every segment has exactly one m i d - p o i n t , and through the

mid-point t he re i s exactly one perpendicular line in a given plane. Thus, f o r perpendicular b isectors in a given plane, we have both existence and uniqueness.

The following theorem gives a useful characterization of the points of a perpendicular bisector:

Theorem 6-2. The perpendicular bisector of a sewenfc, in a plane, i s the s e t of all points of the plane that are equi-

distant from the end-points of the segment. Restatement: Let L be the perpendicular b i sec to r of the

segment in a plane E and let C be the mid-point of m. Then

(1) If P I s on L, then PA = PB, and (2) If P is in E, and PA = PB, then P is on L. Notice that the restatement makes It plain that the proof

of the theorem w i l l consist of two parts . In the first par t we

prove t h a t every p o i n t of the perpendicular bisector satisfies

the characterization, that is, is equidistant from t h e end-po in t s

of the segment. But t h e theorem says that the perpendicular bisector I s the set of - all such points. To prove this, then, we must also show t h a t every such po ln t , character ized by belng

equidistant from the end-points of the segment, is on the

perpendicular b i s e c t o r . This last I s the second p a r t of t he

restatement.

Proof of (1): Given a point P of L. If P l i e s on <-> AB, then P = C, and this means that PA = PB by the def in i t ion of mid-point of a segment. If P is not on the line AS, then

PC = PC by i d e n t i t y , and, by hypothesis, CA = CB and

LPCA LPCB. Hence by the S.A.S. Postulate,

A PCA S APCB.

Therefore PA = PB, which was to be proved.

Proof of (2): Given tha t P lies in the plane E and

PA == PB. If P is on S, then P is the mid-point C of - m, and so P is on L. If P is no t on E, l e t L* be

<-> t h e l i n e PC:

[sec. 6-31

then PC = P C , CA =CB, and PA =PB. (Why?) By theS .S .S .

Theorem,

AFCA = A PCB.

Therefore L P C A Qt L PCB. Therefore, by definition, L! ( and so L' is the perpendicular bisector of m. Therefore, by Theorem 6-1, L1 = L, and P is on L, which was to be proved.

Next we prove the analog of Theorem 6-1 f o r t h e case in which the given point is not on the given line. Since the

proof is considerably more complicated than that of Theorem 6-1,

we w i l l state and prove the existence and the uniqueness parts as separate theorems. Because it I s the simpler, we s ta r t with

uniqueness.

Theorem 6-3. Through a given external point there is at most one line perpendicular to a given l i n e .

Proof: Like most uniqueness proofs, t h i s is an indirect

one. Suppose Li and Lg are distinct lines through point

P, each perpendicular - to L. . .a

L e t L., intersect L in A and Lg i n t e r s e c t L in 3.

Since the l ines are distinct and both go through P we must have A # B (Theorem 3-1). +

On the ray opposite to AP take AQ = AP (Point Plotting

he or em). Then AQ = AP, AB = AB, m L P A B = mLQAB = 90, and

so AQAB 2 A PAB by the S.A.S. Postulate. It follows t h a t

<-Ã

and so BQ J. there is only

containing L

L. This contradicts Theorem 6-1, which says tha t one perpendicular t o L a t 3 ly ing in the plane

and Li. Hence our supposition that there could be two perpendiculars t o L through P Is fa l s e .

Cp.rollary 6-51. A t most one angle of a triangle can bp a right angle.

For If In" A ABC, L A and L B were both right angles we would have two perpendiculars f r o m G to fS.

Definitions. A right t r i ang l e is a t r i ang le one of whose

angles is a right angle. The side opposite the right angle is

the hypotenuse; the s i d e s adjacent to the r igh t angle are the

legs.

Theorem 6-4. Through a given external point there is a t

least one line perpendicular to a given l i n e ,

Restatement: L e t L be a line, and l e t P be a point not

on L. Then there is a line perpendicular to L and containing

F i r s t we will explain how the perpendicular can a c t u a l l y be

constructed, on paper, using a r u l e r and a protractor. From the

method of construction, it will be c l ea r how the theorem can be

proved from t he postulates . Step 1. k t Q and R be any two points of t he l i n e L.

Measure the angle L PQP.

St-ep 2-. Using the prot~actor, construct an angle L RQS, with the same measure as LPQR, taking S on the opposite side of t he line L from P.

+ Step 3 . Measure the distance QP, Take a point T on

QS, such t h a t QT = QP, <->

Step 4 . Now draw the line TF. This is the perpendicular t ha t we were looking f o r . For the reasons, see the proof below. F i r s t , however, you should try this construction with your ruler and prot rac tor , and try t o see for yourself why It works.

Let us now write down the proof In t h e double-column form. Each of t h e first few statements on t he l e f t corresponds t o one of t h e th ings that we were doing with our drawing instruments.

Statements

1. L contains two po in t s Q and R.

2. There is an angle LRQS, congruent to L RQP, with S and P on d i f f e r e n t sides of L.

3 . There is a oint T of the ray d such that

4. T and P are on opposite sides of L.

- 5 . TP i n t e r s e c t s L, in a point U,

6 . APQU s ATQU.

7 . LQUP 2 L QUT.

8. LQUP is a r ight a n g l e . <-Ã

9. PT 1 L.

Reasons

This proof somewhat resembles the

The Ruler Postulate.

The Angle Construction Postu la te .

The Point Plotting Theorem.

P and S are on opposite sides of L, and 3 and T are on the same side of

Definition of opposite sides.

Statement 2, statement 3 , and the S.A.S. Postulate. Definition of a congruence between t r iangles . D e f i n i t i o n of right angle. Definition of perpendicu- l a r i t y .

proof of the S.S.S. Theorem h he or em 5-61, Like t h i s e a r l i e r theorem it has seve- r a l cases, only one of which (that In which U and R lie on t he same side of Q) is completely covered by t h e above proof.

[aec. 6-31

The modifications necessary for* the other t w o cases (U = Q and Q I s between R and TJ) are l e f t as exercises for the student,

Problem Set 6-2 If BC = DC and t? J. Is?, prove without the use of congruent triangles that C E , m = ED.

shown in the figure, with

lengths o f segments as Indicated, find x, y and

3 , Given: PA = PB, M I s the

midpoint of m, and Q I s Â¥<-

on line PM as shown In

the figure. Prove: QA =QB. ( U s e paragraph proof. )

B 4 . Given: The l i n e m is the perpendicular bisector of the

segment m. P I s on the same side of m as Q. R is

the in tersect ion of m and "FT. T Prove : PT = PR + RQ.

m

[sec. 6-31

COPY t e x t

COPY

the figure below. Following the steps outlined in the construct perpendiculars from A, B and X to l i n e L.

' X A

B

t h e figure. Using ruler and p r o t r a c t o r construct

perpendiculars from A and F t o

Does Theorem 6-4 state the existence of a unique perpendicular to a line from a po in t off the line? If we confine our

t h i n k i n g to a plane, does Theorem 6-1 s t a t e the existence

of a unique perpendicular to a l i n e through a point on the line?

Given i sosce les t r iangle ABC with AC = 3C and b i sec to r s 2 and BE of LA and L B. AD and BE intersect a t point I?. Prove that C? is perpendicular to m. (It is no t necessary to use any congruent triangles in your p r o o f . ) One diagonal of a quadrilateral b i s e c t s two angles of the quadr i la tera l . Prove that It bisects the other diagonal. In this figure given:

RC = sc, Q I s midpoint of L RCA 3 SCA.

Prove: ^&1 RS".

6-4. Introducing You probably

Auxi liary Sets Into Proofs . -- noticed t h a t in proving some theorems, most

recently, Theorems 6-2 and 6-4, we Introduced certain points, rays and segments I n t o the figure in addition to those speci- fled i n the theorem. Possibly two questions concerned you:

1. How can we justify introducing such additional sets

i n t o proofs on the basis of our postulates? 2. How do we know which of these sets, If any, should be

introduced into the proof of a theorem?

The f i r s t question is easy to answer. In working with

theorems we usually are concerned with various relationships among certain points, lines, planes and subsets of these, and as a p rac t i c a l m a t t e r in proving theorems, we choose c e r t a i n planes or l i n e s and certain points on them. Frequently we do not concern ourselves with j u s t i f y i n g this procedure. For

example, if we are given a l i n e we may immediately name I t = When asked to give a reason, however, w e can re fe r t o the Ruler Postulate, which says that a line con ta in s infinitely many points, and thereby the two points P and Q exist. Similar ly,

<-Ã

given two points A and B we may talk about AB with complete confidence since It s t ands for a line whose existence and unique-

ness are guaranteed by Postulate 1. (See Sect ion 6-2 . ) The careful concern over Justifying existence and uniqueness

becomes especially Important when we int roduce into the proof certain points, lines, segments, and so on, not accounted for by

the theorem being proved. Certainly we can not have these sets in our proofs if they do n o t exist under t h e conditions of our geometry, except, of course, In an Indirect proof, where She object is to show they can't exist.

In the t a b l e below we list the postulates and theorems occurring so f a r which may be used, appropr ia te ly , to Introduce auxiliary sets into proofs .

Geometric Set Existence Uniqueness

Point. a. Midpoint.

Line.

Perpendicular at point on line, In a plane.

Perpendicular bisector, in a plane. Perpendicular from point not on l i ne ,

Plane.

Ray as used in angle measure.

a. Bisector of an angle.

Segment.

Postulates 3 and Theorem 2-5. Postulates 1 and

he or ems 2-5 and 6-1

Theorem 6-4.

Pos tu la t e 7 . Theorems 3-3 and 3-'4. Postulate 12.

Theorem 5-3.

Postulate 1 and Definition of segment

Theorems 2-14, 3-1, 3-2. Theorem 2-5. Postulates 1 and 8. Theorem 6-1.

Theorems 2-6 and 6-1.

Theorem 6-3 .

Postulate 7. Theorems 3-3 and 3-4. Postulate 12.

Theorem 5-3.

Postu la te 1 and Definition of segment.

From this table you may see tha t you already know a l o t about the nature of our three basic undefined terms.

The answer to the second question presents a problem quite different from the answer to the f i rs t . Getting to know when to introduce a u x i l i a r y s e t s into a proof is l a rge ly part of the process of learn ing t o reason logically- It requires consider- able practice. Let 's try an example to see how t h i s works.

Example 1.

Given: The plane figure with AD = AE and CD = CE.

To prove: L D '= LE. A

Since a l l of our postula tes and theorems concerning congruence have dea l t with triangles, I t seems reasonable that our

figure should show some triangles. We can accomplish this easily by in t roducing e i t h e r 17 or HE. -

Suppo8e.we introduce DE so that our figure looks l i k e t h i s :

This allows us t o complete the proof, since mLADE = mLAED and m L C D E = mLCED g ives us m L A D C = m L A E C by the Angle

Addition Postulate.

Had we Introduced 'KQ Instead of BE, our proof, in two- column form this time, would have looked l i k e this:

Proof:

Statements Reasons

1. Introduce TO?.

2. AC = AC. 5 . AD = AE and CD = CE. 4. A ADC &AEC.

5. L - D s LE.

1. Postulate 1 and Def in i t i on of segment.

2, Identity. 3 . Given. 4. S.S.S. Theorem.

5. Definition of congruent t r iangles .

Each of the solutions to Example 1 is correct. The choice of which one you use is up to you. But it i s worth no t ing that i n many problems where a choice exists, t he choice you make w i l l

determine the degree of d i f f i c u l t y of the proof . It is he lp fu l t o think through each solution before writing one down formally.

An important aspect of learning what t o introduce in a proof can be illustrated if we remove from the hypothesis of Example 1 the condition that the figure is a plane f igure . If

D i s not coplanar with A, E, and C, at least one of the solutions does not hold. Does ei ther solution hold? I f one

does, which one?

One final word of warning before you begin t o Introduce auxiliary s e t s Into your proofs. In answering Question 1

we were careful to say that each such step must be justifiable,

that I s , that every poin t , l i n e , plane, and so on must exist

under our postulates. Students often make the mistake of not recognizing this. For example, you might t h i n k you could prove the statement " ~ l l angles are congruentt by the following arm- ment .

Example 2,

Given any A ABC , prove tha t LB 3 LC.

Proof: In AABC Introduce m, bisecting L A and perpendi-

cular t o EC'. 0 D C Then L BAD S L C A D by defi-

nition of the bisector of an angle, AD = AD by Identity, and LBDA a LCDA by t he d e f i n i t i o n

of perpendicular and the f ac t that all r igh t angles are congruent.

Therefore A BAD A CAD by A . S . A . , making i B 3 Z. C. It does not take long t o see the serious error o f t h i s so-

c a l l e d proof. The segment m, as angle bisector and the per- - pendlcular to the base, does no t exist -under our postulates.

Moreover, the figure makes A ABC appear to be Isosceles and - thus makes AD appear as introduced above. Were the f igure like this,

you c e r t a i n l y would no t consider using 755 as it is use! d. This leads us once more to say t ha t the figure 5 s merely 5 convenience

to aid you in th inking through your reasoning in logical and care fu l ly chosen words.

Problem Set 6-4 -- 1. Given: A, B, C and D are 0

coplanar. AD = CD.

m L A = - m L C .

Prove; A B = C B .

Does the proof work if A, B, C, D are no t coplanar?

Q c B

Y

2. Given: XY = AB, AY = XB. Prove : A XOY A AOB.

3. Given: E, A, S and Y are coplanar. L E S LA, - YE S ^S. Prove: LY L S .

4. Devise a second so lu t ion t o Problem 5 above by introducing aux i l i a ry segments different from the ones you used in the s o l u t i o n of Problem 3 .

5. If AC = AB and CD = BD In the plane figure, show L A C D S L. ABD.

Devise a proof that works if the figure does not lie in the plane.

c A B

Betweenness - and Separation. Critical students may have discovered two places in

Chapter 5 where the given proofs are not qui te complete. These

d e f e c t s occur in Theorems 5-3 and 5-6, and are similar in the

two places , consisting of a f a i l u r e t o show why a cer ta in p o i n t l ies i n the i n t e r i o r of a certain angle. I n Theorem 5-3 we

must know t h a t D is In the i n t e r i o r of LBAG before we can + conclude that AT> bisects t h i s angle. And I n s t e p s 9 and 10

of Theorem 5-6 we must know t h a t H is I n the interior of L ABC and of L AE^C before we can apply the Angle Addition Postulate.

I n these places It Is not enough to observe t h a t i n the figure the po in t s l i e i n the proper places. Remember f i r s t t h a t a drawing is only an approximation t o t h e true geometrical situation, and secondly that this Is only one figure and the

theorem i s supposed to be proved f o r a l l cases.

You probably wonder why an Incomplete proof should be presented In a text-book. The reason Is that the proofs of

of such separation properties as t h i s one a re o f t e n long, comp- l icated, and uninteresting, and that they contribute little or nothing to the essential Idea of the proof. If you understand

t he proof of these theorems as given but d i d no t notice the incompleteness of these particular steps, you need not worry about your competence in geometry. For many centur ies learned

men disputed whether s teps like these needed any justification.

However, mathematicians now agree that even such "obvious" steps require a logical proof, and so we present here two theorems and some problems t o f i l l t h e gaps i n these (and l a t e r ) proofs .

Theorem 6-5. If M Is between A and C on a line L,

then M and A are on the same s ide of any other 15ne t ha t

contains C.

Proof: The proof will be Ind i r ec t , If M and A are on opposite sides of L* (in the plane t h a t contains L and L')

then some point D of Lr lies on the segment 7S?I. Therefore

D is between A and M, by definition o f a segment. But D lies on both L and L*. Therefore D = C. Therefore C Is

between A and M. This is impossible, because M is between

A and C. (see Theorem 2-3. ) We can now prove a theorem which completes the proof of

Theorems 5-3 and 5-6: Theorem 6-6. If M is between A and C, and B is any

M point not on the l ine AC, then M I s in the i n t e r i o r of LABC.

Proof: By the preceding theorem, we know that M and A

are on the same s ide of %?. By another application of the

[sec. 6-51

preceding theorem (interchanging A and C) we know that M and C are on the same side of St. By definition of the Interior of an angle, these two statements t e l l us that M is in the

interior of L ABC, which was to be proved.

Problem S e t - Note: On this problem s e t no Information I s to be read from a

figure .

Given A ABC with F between A and C, X between A

and B and Q In the i n t e r io r of A A3C. Complete the

following statements, and give reasons to justify your

answers. a. F l i e s in the interior of /

b. X lies In the In t e r io r of / c. Q lies In the In t e r io r of /

f L f

2. The following faulty argument tha t an obtuse angle is congruent to a r ight angle emphasizes the importance of knowing the side of a l i n e on which a l i e s .

c '

Suppose t h a t ABCD is a rectangle as shown and t ha t the side B5" is swung outward so that EC1 = BC and ,LABCt I s

obtuse. Let the perpendicular bisector of intersect the perpendicular bisector of BET1 at X, If X is below

as shown, we have AAXD S A EXC1 by the S.S.S. Theorem, and hence mLDAX = m L C I B X . Also, A EAX S A EBX by S . S . S . , and so m L E A X = mLEBX. It follows by sub t rac t ion that mLDAE = rnLCIBE.

<-Ã In case X lies above AB, as in the figure below,

A E 0 we g e t , exactly as before, m L D A X a mLC'BX, rnLEAX = rnLEBX,

and t h e desired equa l i ty , m LDAE = m LC^S fol lows by addition.

What I s wrong w i t h the above argument?

c *5. Suppose ABC is a t r i ang le

and & is a point between B and C. Show that if L is

a line In the plane of A ABC <- > L

which i n t e r s e c t s a t D, then L i n t e r sec t s "SS" or m. H~

(~int: If L contains B, A 3

then L i n t e r s e c t s m. I f L

does not contain 3, then let Hi, Hp be the two half-planes into which L separates the

the plane of AAEC, H i . being the one that contains B. Since A belongs to e i ther L, Hi, o r Hz, there are three cases to consider. )

-4. A theorem whose truth appears obvious Is of ten di f f i cu l t to prove. The following such theorem is assumed In the

proof of Theorem 7-1 of the next chapter. Suppose ABC is a t r i ang le , D is a po in t between +

A and C and E I s a p o i n t of BC beyond C . Then each + point F of BD beyond D is in the i n t e r i o r of LACE.

The t h i n g to be proved is that F is on the same side -<Ã‘ -<-Ã

of BC as A and t h a t F is on the same side o f AC as E. a. How do we know that A and A

D are on the same side of Â¥*Ã BC? What theorem Implies

t h a t D and F are on t h i s same side?

b. Prove t h a t if Hp Kg are t h e t w o half planes I n t o

<-> which AC separates t he

plane of the figure and B c

belongs to H,, then each

of E, F belong to This shows t h a t E and F "2.- are on the same s ide of A C ,

*5. Another theorem whose truth is f requent ly accepted without

proof is the following: If D is a point In the interior of L ABC, then 3 intersects E.

We suggest below a t r i c k y t proof I n which we consider A EAC, where E I s a point of 8 beyond B. This

enables us t o apply the r e s u l t s of Problem 2, Parts a and

b below are used to show tha t %8 does not Intersect ES'. a . Suppose H,, Hp are the

two half-planes i n t o which M BC d iv ides the plane of A EAG w i t h A In Hi. Why is D I n HA? What

theorem implies t h a t each + point of BD other than B is In Hi? Why I s E

i n Hp? What theorem implies t h a t each point of E5" other than - C is in Hp? Why does f a l l to intersect BD? +

b. Why does S? f a i l to intersect t h e ray opposi te BD? + c . Why does BD I n t e r s e c t 15'? + d . Why does the ray opposite BD f a i l to intersect A(??

* 6 . The following theorem may be used Instead of Parts a and b of Problem 5 to show that A and C l i e on different

Â¥<Ã sides of BD.

Theorem: If point D is in the I n t e r i o r of L A B C ,

then A is not in the i n t e r i o r of L DBC nor is 0

in the i n t e r i o r of L ABD.

Prove t h i s theorem.

'7. There are studies or geometry that use other systems of postulates than the ones we have adopted. A postulate taken from one such system is the following:

If A, B, C, D, E are points such that A, B and C are non-colllnear and B is between A and E and D I s

between B and C, then there is a point X such t h a t X is between A and C while D Is between E and X .

This statement can be proved in our system of postu-

la tes . C

a . Why are A, B, C, D, E coplanar?

b. Show from the Plane Separat ion P o s t u l a t e * that ED intersects - AC a t a point x be-

tween A and C. A 0

c . It can be shown t h a t D is between E and X by showing that E and X are on opposite sides of some l i n e . What l i n e ?

8. Given points P and Q on ^P

opposite s i d e s of plane E I with Intersecting E in M. Identify the following

statements as t r ue or false. L o -

a. If L is a line in E perpendicular to PQ, then P

and Q are on opposite sides of L in the plane determined by P and L.

b. If L is a l i n e In E through M, then P and Q are

on opposite sides of L in the plane determined by P and L,

c. If L is a line In E, then P and Q are on opposite s i d e s of L in the plane determined by P and L.

d. P and Q are on opposite s ides of every plane through - M no t containing PQ.

Chapter 7

GEOMETRIC INEQUALITIES

7-1. Making Reasonable ConJectures. Up to now, in our study of the geometry of the triangle, we

have been dealing only with conditions under which we can say that

two segments are of equal length, err t w o angles are of equal measure. We wil l now proceed to study conditions under which we can say that one segment I s longer than another, (that I s , has a greater length), or one angle Is larger than another, (that I s , has

a greater measure). We shall not start, however, by proving theorems. Let us

start, rather, by making some reasonable conjectureB about the sort of statements that ought to be true. ( ~ h e s e statements should not be called theorems unless and until they are proved.)

An example: Given a triangle with two aides of unequal length, what can we aay about the angles opposite these sides?

Notice that t h i s problem is naturally suggested by Theorem 5-2, which says that if two sides of a triangle have the same length, then the angles opposite them. have the same measure.

You c a n Investigate t h i s situation by sketching a triangle with two a i d e s of obviously unequal lengths, like t h i s :

Here BC is greater than AB, and m L A I s greater than m/- C. After sketching a f e w more triangles, you will become pretty well convinced that the following statement ought to be true:

If two aides of a triangle aFe of unequal length, then the ----- angles opposlte t h e m are unequal measure, and the larger angle -- LEI oppo8ite the longer stde. - -

Now try the same sort of procedure with the following problems.

Problem Set - Here are some experiments for you to try. 1. Consider triangles w i t h two angles of unequal measure. Write

a statement which you think may be true concerning the sides

opposite those angles. 2. Consider several tr iangles ABC. How does AB + X compare

with AC? EC + AC compare with a? A B + AC compare with EC? These responses suggest a general conclusion. If you think t h i s conclusion is true for a l l triangles, write it as a proposition.

3. Consider a quadrilateral RSTQ. How does RS + ST + TQ com-

pare w i t h RQ? Write a proposition suggested by your answer. 4. Draw several triangles In which the measure of one angle I s

successively greater but the adjacent s i d e s remain unchanged in length. What happens to the length of the third s ide?

5. Draw A DEF and A XY2 such that DE = XY, FS = ZY, and m L DEF > m L XYZ. Compare DF and XZ.

6. Regarding A PDQ and A JUN such that m <t- FIQ = m Â£ JUN, PD > JU, and QD = NU, a hasty person might conclude that

PQ > JN. Draw a figure showing that the conclusion is not just if i e d .

7. A is a point In plane E, TS' is a ray not lying in E, and * AC is a ray lying in E. Con- sidering different positions + of AC, describe as accurate! -3 as you can t h e position of AC

L* which makes L BAG as small as possible; as large as possible. .No proof I s expected but you are asked to guess the answer on the basis of your knowledge of space.

8. On the basis of drawings decide whether or not an angle can be trisected by the following procedure:

Let A ABC be an isosceles triangle - with congruent sides A 3 and (T.

Trisect s i d e with poin ts D, E so that BD = DE = EC.

Is f BAD Z L DAE 2 i EAC?

B 0 E C

7-2. Algebra Inequalities.

Before considering geometric Inequalities we review some of the facts concerning inequa l i t i es between real numbers. Note first that a < b and b > a are merely two ways of writing the same th ing; we u s e whichever is more convenient, e.g. 3 < 5 or 5 > 3.

Definitions. A real number is positive If It is greater than

zero; It is negative If It I s less t h a n zero. We now restate t h e order postulates, giving examples of t h e i r

use. 0-1. (uniqueness of order.) For every x and y, one and

only one of the following re la t ions holds: x < y, x = y, x > y.

0-2. (~ransitivity of o rde r . ) If x < y and y < z , then x < 2.

Example 1. 3 < 5 and 5 < 9, hence, 3 < 9. Example 2. If we know t ha t a < 3 and b > 3, we can

conclude that a < b. Proof: If a < 3 and 3 < b, then a < b.

Example 3. Any positive number is greater than any negative number.

Given; p I s positive, n is negative. To prove: p > n.

Proof : 1. p is positive.

2. p > 0. 1. Given. 2. Definition of posltlve.

3 . Relation between < and >. 4. Def l n i ti on of negative . 5. Postulate 0-2. 6 . Relation between < and >.

Example 7 . If a + b < c then a < c - b. Proof left to the student.

Example 8. If a < b, then c - a > c - b for every c.

Proof l e f t to the student.

0-3. (Addition for inequalit ies . ) If x < y, then x + z < y + 2, for every z.

Example 4. Since 3 < 5 I t follows t h a t 3 + 2 < 5 + 2, or 5 < 7; that 3 + ( - 3 ) < 5 + ( - 3 ) , or 0 < 2; t h a t 3 + (-8) < 5 + ( - 8 ) , or -5 < -3 .

Example 5. If a < b then -b < -a. Proof: a + (-a-b)< b + ( -a-b) , or -b < - a .

Example 6. If a + b = c and b is positive, then

a < c .

Proof:

0-4. (~ultiplication for ~nequa l i t i e s . ) If x < y and

z > 0, then xz < yz.

1. b I s positive.

2. b > 0. 3 . O < b .

4. a < a + b . 5. a < c .

Example 9. From 3 < 6 we can conclude that 3000 < 6000; also, that -g- 1 1 - 3 < & - 6 , o r 5 < 7 .

1. Why?

2. Why? 3 . Why?

4 . Why?

5. MY?

Example 10, If x < y and z < 0 , then xz > yz, Proof l e f t to the student.

0-5. (Addition of ~nequalities.) If a < b and x < y, then a + x < b + y .

This Is not a postulate but a theorem; I t s proof is given in Section 2-2. However, I t is convenient to list it, for

reference, along with the postulates.

7-3. -- The Basic Inequality Theorems. In the figure below, the angle L E D is cal led an exterior

angle of A A X . More precisely:

D e f i n i t i o n . If C is between A and D, t h e n L BCD Is an

exterior angle of A ABC.

Every triangle has double -headed arrows in

s ix exter ior angles, as Indicated by the

the figure below:

These six angles form three pairs of congruent angles, because they form three pairs of vertical angles.

Definition. L A and LB of the triangles are called the

remote Interior an~les of the exterior angles L E D and L A C E .

Similarly, L A and L C of A A N are the remote In te r ior angles of the exterior angles L ABF and LC%.

Theorem 7-1. h he Exterior Angle Theorem. ) An exterior angle of a triangle is larger than e i t h e r remote i n t e r i o r angle.

Restatement: Note first that the two exterior angles at vertex C, above, have equal measures (ve r t i ca l angles), and so it

doesn't matter which of them we compare with L A andL3. It turns out to be easiest to compare A L BCD with mL B and m L ACE w i t h mL A . Since t h e proofs of these two cases are exactly similar we need prove only one.

Given triangle ABC. If C is between A and D, then

m L B C D > m L B.

Proof: Statements

1. L e t E be the mid-point of E.

2. Let F be a o nt of the ray opposite to @ such t h a t EF = EA.

6. m L B C D = m L E O F + m L FCD.

Reasons

1. Ety Theorem 2-5 there is such a mid-point.

2. By Theorem 2-4, there is such a po in t .

3. Vertical angles are eon- gruent .

4. Statements 1, 2, 3 and the S.A.S. Postulate,

5. Corresponding parts of congruent triangles.

6 . Postulate 13 (The Angle Addit ion Postulate. )

7. Statements 5 and 6. 8. By algebra from s tep 7.

(Since rn L FCD is a p o s i t i v e number, Example 6 of Section 7-2 applies.)

Problem Set 7-3a

1. a. Name the remote I n t e r i o r angles of the ex t e r i o r angle L ABE in t he f igure .

b . L A X and L BAC are the

remote i n t e r io r angles of E which exterior angle?

2. a. In the figure, which angles are exterior angles of the

t r iangle?

b. What I s the relationship of m L DBC fco m L A? Why?

c . What is the relationship of

m L DBC t o m L C? Why?

d . What Is the re la t ionsh ip of m L DBC to m L CBA? Why?

3. U s i n g the f igure, complete the following: a. If x = 40 and y = 30 , then m L BCE > b . If x = 72 and y = 73, then

m L BCE

c . If y = 54 and z = 68, then m L BCE

d . If m f BCE = 112, then x

e . If m L BCE = 150, then z

f. If x = 25 and 2 = 90, then m L BCE a

g If x = 90 and y = 90, then A C E

196

4. The accompanying figure I s an illustration o f t h i s statement:

An exterior angle of a quadrl-

lateral is greater than each of the remote In te r ior angles. Is

t h i s a true statement? Explain. A B E

*5. Prove the following theorem: \ The sum of the measures of any two angles of a triangle i s less than 180. Given: A ABC with angle

measures as in the

figure . Prove: a + b < 180.

b + c < 180. a + c < 180.

* 6 . Prove the following theorem: The base angles of an i sosce les triangle are acute. (~int; Base your proof on t h e statement of the previous problem. )

Theorem 7-1, while perhaps not very exciting in Itself, is

extremely useful in proving other theorems. ( A theorem of this type is sometimes cal led a lemma.) For* example, the following is a u s e f u l corollary.

Corollary 7-1-1. If a triangle has a right angle, then the other two angles are acute.

Proof: If m L A = 90, then m . BCD > 90, and theref o r e , m L EEA < 90. In a similar way we can prove m LAX < 90.

We next use Theorem 7-1 t o prove two more congruence theorems. Theorem 7-2. (The S . A . A . heo or ern) Given a correspondence

between two triangles. If two angles and a side opposite one of them in one triangle are congruent to the corresponding parts of

and

then

t he second t r i ang le , then t h e correspondence is a congruence. Restatement; L e t ABC-DEF be a correspondence between two

t r iangles . If

L A ~ L D, L B ~ L E , 15" Z'BF,

Aril A ABC = A DEF.

Proof :

2. A A X C ~ A DM. E . S.A.S. ~ostulate.

Statements Reasons

3 . m L A X C = m L DEF. 3 . D e f init Ion of congruence.

+ 1. On A 3 take X so that

AX = DE.

4. m L A X C = m L A X . 4 . Step 3 and given.

1. Point P l o t t i n g Theorem.

Now suppose t h a t X is not the same point as B.

6. In either case one of LAXC and L ABC is an e x t e r i o r angle of ABXC and t h e o ther I s a remote I n t e r i o r angle.

7. m L A X C + m L ABC.

8. X = B . 9 . A A B G Z A DEF.

5. Either X Is between A and B or B is between A and X.

6. Definition of exterior angle and remote I n t e r i o r angle.

5. Step 1 and definition of ray.

7 . Step 6 and Theorem 7-1. 8. Step 7 contradicts Step 4. 9. Steps 2 and 8.

Although I t was pointed out in connection w i t h t h e S.A.S.

Postulate t h a t an S.S.A. theorem cannot I n general be proved, there is one special case; namely, the case in which the angle is

a right angle, that follows from Theorem 7-2.

Theorem 7-3. h he Hypo tenuse-Leg he or em. ) Given a corres-

pondence between two right triangles. If the hypotenuse and one leg of one triangle are congruent to the corresponding parts of the second triangle, then t h e correspondence Is a congruence.

Restatement: In AABC andA DEF let m L A = m L D = 90.

Let ABC-DEF' be a correspondence such that

Then A ABC 2 A DEF.

- Proof; On the ray opposite to DF take Q such that

DQ = AC. Then & DEQ ZA ABC by the S.A.S. Postulate, and so EQ = BC. A EQF I s thus an Isosceles triangle, and SOL EQD 2 EFD. In A DB& and A DEF we thus have

Hence, by the S .A. A. Theorem, A DEF 2 A DEQ . Since we have already establ ished A DEQ 2 A ABC we conclude that A DEF 2 A A X , which

I s what we wanted.

Problem Set 7-3b

1. If in this figure AQ = BQ and

2. Given that m l m , AB = HF, AK = HQ.

Prove that KF = QB. , ,

3 . If AX = FH In this figure, prove that FB = AB.

it.

5 .

6 .

If two altitudes of a triangle are congruent, the triangle is isosceles.

In this figure: L C 2 L a .

AQ = AF.

Prove: OS = FK.

In this figure i f L a 2 L c, B

-- ABJAH and Elm, F prove that AH = FH.

Theorem 7-4. If two s ides of a triangle are not congruent, then the angles opposite these two s i d e s are not congruent, and the larger angle is opposite the longer s i d e .

Restatement: Given A ABC . If AB > AC , then m L C > mL B.

B * -- --- \ C \ \ --- \ \ -- - \ -\

\ -- \ --lD

Proof: Let D be a point of 2, such that AD = AB.

( ~ y the Poin t Plotting Theorem, there is such a point. ) Since the

base angles of an Isosceles tr iangle are congruent, we have

(1 1 rn f ABD = mL D. Now AD > AC, slnce AD = AB and A 3 > AC, and so C I s between A and D by Theorem 2-1. Theorem 6-6, C is in the i n t e r i o r of ABD, and so

(2) m f ABD = m L A E 4- rnL CBD by t he Angle Additlon Postulate. S1nc-e m L CBD > 0 it follows

that

(31 mL ABD > m L A=, Therefore

( 4 1 rnL D > m L AX, from (I) and (3). Slnce ACB i8 an e x t e r i o r angle of A XD, we have

(5) m L ACB > rn L D. w ( 41 and (5L

mf. ACB > th L A=, that I s ,

m L c > m L B, which was t o be proved.

T h e o ~ e m 7-5. If two angles of a triangle are not congruent, then the sides oppo3ite them are no t congruent, and t h e longer s i d e Is opposite the lapger angle.

Restatement: In any triangle A A X , Tf m L C > m L B, then A 3 3 AC.

Proof: We want t o prove t h a t AB > AC. S m c e AB and AC

are numbers, there are only three possibilities: (1) AB = AC,

(2) AB < AC and ( 3 ) A3 > AC. The method of the proof is to show

that the first two of these ~ ~ p o s s l ~ i l ~ t l e ~ ' l are in fact impossible.

The only remaining possl .bi l i ty will be (31, and t h l s will mean

tha t the theorem is t rue. (1) If AB = AC, then by Theorem 5-2 it follows that

L B 2 L C; and this is false. Therefope, I t is imposa3ble that

A 3 = AC. ( 2 ) If AB < AC, then by Theopem 7-4 I t follows thaC

m L C < m L B; and t h i s is false. Therefore, It is impossible

that A% < AC.

The only rernainlng pussibLllty is that A 3 > AC, whlch was to be proved.

The proof of Theorem 7-5, as we have given it, I s rne~ely a handy way of stating an l n d i ~ e c t proof. It could have been written mope formallyy llke this;

"Suppose that Theorem 7-5 is false. !hen e i ther AB = AC or

AB < AC. It is impossible that A 3 = AC, because . . . . And I t

I s irnpossTble that AB < AC, because . . . . Therefore, 7-5 I s

not false ." The proof is probably easier t o read, however, the way we

gave it the first t i m e . We will be using t he same sort of scheme again. That is, we will list the possibillt~es, in a given aitu-

ation, and then show that all but one of these ''posslbi1lt~es~~ are in fac t ImpossEible; it will then fol low t h a t the last remaining poss ib l l ity must represent what actually happens.

11 That process starts upon the supposition that when you have eliminated all which is Impossible, then whatever remains, how eve^

imp~obable must be t he t r u t h . " (sherlock Holmes in "The Adventurn

of the Blanched ~ o l d l e ~ " .

Theorems 7-4 and 7-5 are related in a s p c i a l way; they are called converses of one another. To g e t one f ~ o m the other, we

interchange the hypothesis and the conclus5on. Me can exhib3t

thls fac t by restating t he theorems t h i s way: Theomm 7-4. Given A A=. If A 3 > AC, then m L C > m L B. Theorem 7-5. Given A A=. If m L C > m L B, then AB > AC.

We have seen lots of palm of theorems tha t are related t h i s

way. For example, we showed t h a t if a trlangle is isosceles, then

I t s base angles are congment; and l a t e r we showed that If t he

base angles of a t~iangle are congruent, then the triangle is

isosceles. Each o f these theorems is the converse of t h e other.

We showed that every equilateral triangle 1s equiangular; and l a t e r we proved the cmverse , which s t a t e s that every equiangular

t~iangle 2 s equilateral.

It is v w y important to remember that the converse of a true theorem is not necessarily t r u e at all. For example, the theorem

1'vert5cal angles are congruent" is always true, but the converse, "congruent angles are vertlcalt ' 1s cer ta inly not true In all cases.

If two triangles are congruent, then tney have the same area, but if t w o triangles have the same area, it does not follow that they

2 2 are congruent. If x = y, then it follows that x = y ; but I f 2 2 x = y , it does not fol low that x = y. h he other p o s s i b i l i t y 1s that x = -y.) It is true t h a t every physicist 1s a scientfs t , but it is n o t t rue t h a t every scientist is a physicist.

If a theorem and its converse are both true, they can be con-

venlentl$ combined into a single statenent by w i n g the phrase "if

and only i f 1 ' , Thus, if we say:

Two angles of a t r i ang le are congruent if and only if t h e

oppo s ft e s ides are congruent ; we are including 3n one statement both theorems on isosceles tr3- angles. The first half of t h i s double statement:

TWO angles of a t r i a n g l e are congment If the oppos i te sides are congruent;

is Theorem 5-2; and t h e second h a l f :

Two angles of a triangle ape congruent only if the opposlte

s i d e s are congruent; is a restatement of Theorem 5-5*

P ~ ~ b l m 7 - 3 ~

1. In AGHK, GH = 5, HK = 14, KG = 11. Name the largest angle. Name the smallest angle.

2. In A AEC, m L A = 36, m L B = 74, and m L C = 70. Name the

largest side. Nme t h e shortest s i d e .

3 . Given the figure with HA = HB,

m L HEX = 140, and m L Am = 100,

fill in the blanks below:

a. m L A = ---. b. m L RHB = - -- c. I s t h e longest s i d e B K

A - - - of A ABH.

4 , What conclusion can you reach a b u t the length of in

A KLM if:

a. m L K > m L M? b. m L K < m L L ? c . r n L M > m L K > m L L? d. r n L M > m L L ?

e . m L K > m L M and m L K > m f L?

f. m L K > m L L and m L M < m L L?

5. If t he f lgure were c o r r e c t l y dram which segment would be the longest?

c .

6 . Name the sides of the figure

In order of increasing lewh. c

- 7 . If in the figure AF I s the

shortest s i d e and is the

longest s ide , prove t h a t

m L F >mL B. (Hint! w e - diagonal FB.)

4

*8. If the base of an isosceles triangle is extended, a segment which joins the vertex of the trtangle w i t h any point In this extenszon is greater than one of the congruent sides of the triangle.

F

9. Write the converse of each statement. Try to decide whether

each statement, and each converse, is true or false. a. If a team has some spirit, it can win some games. b. If two angles are right angles, they are congruent. c. Any two congruent angles are supplementary. d . The i n t e r i o r of an angle is the i n t e r s e c t i o n of two

half-planes.

e . If Joe has scarlet f e v e ~ , he is ser ious ly ill. f. If a man lives in Cleveland, Ohio, he lives in Ohlo.

g. lf the tmee angles of one triangle are congruent to the

corresponding angles of another triangle, the triangles are congruent.

h. If two angles are complementary, the sum of their measures I s go.

10. When asked to give the converse of t h i s statement, "If I hold

a lighted match too long, 1 will be burnedn, John said, ''1 w311 be burned if 1 hold a IZghted match t o o long." Was

Johnts sentence the converse of t h e orlgznal statement? D ~ S C U S S .

11. a. Is a converse of a true statement always t rue? Which

parts of Problem 9 illustrate your answer? b. May a converse of a false statement be t r u e ? Which parts

of Problem 9 Illustrate your answer?

[see. 7-31

Theorem 7-6. The shortest segment Joining a point to a line is the perpendicular segment.

Restatement: L e t Q be t he foot of the perpendicular to the line L through the point P, and l e t R be any other point on L. Then PQ < PR.

Proof: Let S be a point of L, such that Q is between S and R . Then L PQS I s an exter ior angle o f A PQR. Therefore, m L PQS > m L PRQ. But mL PQS = m L PQR = 90, and so m L PQR > m L PRQ. By Theorem 7-5 it follows that PQ < PR, which was to be proved.

&finition, The distance between a line and a po ln t not on --- it is the length of the perpendicular segment f r o m t h e point t o the

l i n e . The distance between a l i n e and a point on the line l a de-

f ined to be zero.

Theorem 7-7. (The Triangle Inequality.) The sun of the lengths of any two sides of a triangle Is greater t han t h e length

of the third s i d e . Restatement: In any triangle A A X , we have AB + BC > AC.

+ Proof: Let D be a point of the ray opposite to EC such

that Dl3 = AB. Since B is between C and D

Then Also

D C = D B + B C . (1) DC = AB+ BC.'

(2) m L DAB < rn L DAC,

because B is in the i n t e r io r of L DAC.

Since A DAB is i sosce les , with AD = DB, it follows that

( 3 ) m L AD3 = m /. DAB.

By (2 ) and (3) we have

m L ADB < m L DAC.

Applying Theorem 7-5 to A ADC, we see- t h a t

( 4 ) DC > AC.

By (1) and ( 4 ) it follows that

AB + BC > AC, which was t o be proved.

Problem Set 7'-3d

1. Here A H < - and AH < . m < - and BT < . State the theorem invo 1 ved . I A

2. With angle measures as shown in the f igure, Inse r t HA, HF, HB below In correct order.

State theorems t o support

your conclusion.

3. Suppose that you wish to draw a triangle with 5 as t h e length

of one side and 8 as the length of a second side. Your third

side must have a length greater than -' a n d l e s s t h a n .

4. Suppose that you wish to draw a triangle w i t h j as t h e

length of one s i d e and k as the length of a second side. It is known t h a t j < k. Indicate, as e f f i c i e n t l y as you can, the restrictions on the length, x, of the t h i r d s i d e .

Prove t h a t the sum of the lengths

of the diagonals of this quadrl- lateral I s less than the sum of the lengths of its sides.

Given: Quadrilateral ABCD.

To prove: DB + CA < A 3 + BC + CD +

* 6 . L e t A , B, C, be points , not necessarily different. Prove that AB + BC > AC and that A 3 + BC = AC if and only if - B I s on t he segment T^.

*7. Prove that the shortest polygonal path f r o m one point to another Is the segment Joining them.

Given: n points Al, An, ......, *n Prove: A A + AgAg+ ..... 1 2 + 2 AIAn-

*8. Given two segments and Intersecting at P.

Prove that If X is any point in the plane of ABCD o t h e r than P, then XA + XB + XC + XD > PA + PB + PC + PD. Will this result be true if X is n o t in the plane of ABCD?

*9. Given a line m and two points P, Q on the same s ide of m.

Find the point R on m f o r which PR + RQ I s as small as

possible.

We w i l l now prove a theorem which is a l i t t l e like Theorem 7-5, except t h a t it deals w i t h - two triangles instead of one.

Theorem 7-8. If two sides of one triangle are congruent respectively t o two sides of a second triangle, and t h e included

angle of t h e f i rs t triangle is larger than the Included angle of the second, then the oppos i te side of the f i rs t triangle is longer

than t h e opposite side of the second. Restatement: Given A ABC and& DEF. If AB = DE, AC = DP

and r n f A > rn L D, then BC > EF, B

Proof: Step 1. We construct A AKC, with K in the i n t e r i o r

of L BAG, such that A AKC 2 A DEF, like t h i s :

To do t h i s , we use t h e Angle Construction Postu la te , t o get a ray Â¥<-Ã s, w i t h Q on the same side of AC as B such that L QAC 2 L D.

On AQ we take a po in t K such that AK = DE. 'By t h e S.A.S.

Postula te , we now have A AKC 2 A DEF, which is what we wanted.

Step 2. Now we b i s e c t LBAK, and let 14 be the po in t where t h e b isec tor crosses E, like this:

The marks on the f igure indicate that AK = AB, and t h i s is true,

because AK = DE and DE = AB.

We are now almost done. Ey the S.A.S. Postulate, we have

A ABM 2 A AKM. Therefore, MB = MK. By Theorem 7-7, we know t h a t

Theref ore, CK < CM + MK.

CK < OH + MB,

because MB = MK. Since CK = EF and CM + MB = BC, we get

EF < BC, which is what we wanted.

The converse of thie theorem 1s also true, Theorem 7-9. If two sides of one triangle are congruent

respectively to two s ides of a second triangle, and t he third side of t h e first triangle is longer than the t h i r d side of the second, then the included angle of the first triangle is larger than the

included angle o f the second.

The proof Is similar to t h a t of Theorem 7-5, use being made of Theorem 7-6 and the S . S . S . Theorem t o eliminate the t w o unwanted

cases. The student should fill In the details.

Problem - Set 7-3e

1. S t a t e the combination of Theorems 7-8 and 7-9 In the "If and only if" form.

8 2, In this figure AC = BC,

and BD < AD. Prove: mL. x > m L y.

3. In i sosce les triangle RAP with - RA = RF and B a point on AF

such that m L ARB < m L BRF.

Prove: A 3 < BF.

4. Given A ABF with median

and m L ARB = 80. Prove: m L A > m L F.

- 5. In A A B C , BC > AC and Q is the midpoint of AB. Is /. CQA

obtuse or acu te? Explain.

[ s e c . 7-31

6. In t h i s f igure FH = AQ.

AH > F'Q. Prove: AB > FB.

A non-equllateral quadrilateral has two pairs of congruent adjacent s i d e s . Prove t h a t the measure of the angle included

between the smaller s i d e s Is greater than the measure of the angle between t h e larger sides.

Prove the following theorem: If a median of a triangle is not perpendicular* to the s i d e to which it is drawn, then t he

lengths of the other two sides of t he triangle are unequal. A

8 F

9 Given AB 3 AC and FC = DB c

in t h i s figure. Prove that FB > CD.

21 4

7-4. Altitudes.

Defini, t ion. A n altitude of a triangle is the perpendicular segment Joining a vertex of the triangle to the line that contains the opposite s i d e .

e In the figure I s called the altitude - from - B - to - AC, or simply the altitude- from (~ottce that we say - the a l t i t u d e

from B instead o f - an altitude from B, because Theorem 6-3 te l ls us that there I s only one. )

Notice that the f o o t of the perpendicular does not necessarily

lie on the s ide AC of the triangle. The figure may look l i k e

this:

Notice also that every triangle has three altitudes, one from each of t h e three ver t ices , like t h i s :

- Here T^F I s t h e altitude f r o m A , BD I s t h e a l t i t u d e from B and is t h e al t i tude from C.

It is customary to use the same word "alt i tude" f o r t w o

other d i f ferent , but related, concepts.

(l) The number which is the length of t h e perpendicular segment Is called altitude; thus one may say h he alt itude from B

is 6 " , meaning that BD = 6. (2) The line containing the perpendicular segment is also

ca l l ed a l t i t u d e ; a property of the above f i g u r e can be expressed

by saying that the three altitudes of the triangle Intersect In

one point. (This property is true f o r a l l triangles and will be proved in Chapter 14. )

This t r ip le use of the one word could cause t rouble but generally does not, since it is usually easy to tell In any particular case which usage is being made.

Problem Set 7-4 -- 1. Define: a. Altitude of a triangle.

b. Median of a triangle.

2. Draw an obtuse triangle (a triangle having an obtuse angle) and its three altitudes.

[sec. 7-41

In an equilateral triangle a median and an altitude are drawn !

to the same s i d e , Compare the lengths of these two segments. I

Prove that the perimeter of a t ~ l m g l e is greater than the 1 I sum of the three a l t i tudes .

Prove the following theorem: The altitudes of an equilateral triangle are congruent,

Review Problems

Three guy wires of equal length are being used to support a

newly planted tree on level ground. If they are a l l fastened t o the t r e e at the same height on the tree, will they be pegged t o the ground at equal d l s t ~ c e s frorn %he foot of the

tree? Why?

If t h i s figure were drawn correctly, which segment in A

the figure would be $he ~ h o r t e s t ? Explain your reasoning.

c Prove the following theorem:

If two oblique ( no t perpendicular) line segments are d~awn t o a l i n e from a po ln t on a perpendicular to t h a t line,

the m e containing the polnt more remote frorn the foot of the perpendicular is t he longer.

In this planar figme, LK AK = HQ, AF = HB, WlE. Prove f Q g L !A. - B e s bisect BF?

5. In & A E , AC > AB. Prove that my l i n e segment from A to a - - point on EC between B and C 1 s s h o r t e r than AC.

6 , Segments drawn from a point in the Interior of a t r t a n g l e t o the three v e r t i c e s have lengths r* s, t. Prove that r +- s + t I s g ~ e a t e r than half' the pertmeter of the triangle.

7, In th i s planar f igwe is H

the shor%est side and i s t h e longest side. Prove m L F > m L A .

8. Prove t h e following theorem: The length of the longest side

of the t r i ang le I s less than half its perimeter,

*g. Given Isosceles A Am with FA = l?B, AB < AF, and H on H AF, so t h a t F is between A

and H, Prove no two s ides of A ABH are equal in length.

A

*lo. On the basis of the a s ~ m p t l o n s we have accepted and the

theorems we have proved in this course we are not able at present to prove that t h e sum of the measures of t h e three

angles of a t r i a n g l e is 180 (an idea w l t h which you have been familiar for some t ime). But* we can easily construct a trlangle and prove that the sum of the measures of the angles of this triangle is less than 181. - Let L FCG have measure 1 { Angle - - c A Construction P o s t u l a t e ) . On CF I* + and CG take points A and 3 8 G

so t h a t CA = CB ( P o ~ n t Plotting heo or em),

Why i s the sum of the measures of the angles of this tr iangle less than 181?

*ll, Tine sum of t h e measures of the three angles of a triangle is less than 270.

*12. In this f lgure : L C 1s a right angle.

m L 3 = 2 m L A .

Prove: AB = 2 CE.

(Hlnt : Introduce auxlliaxy segments, )

*l3. Prove this theorem: The sum of the distances from a point w l th in a triangle to the ends of one ~ i d e is less thm t h e

sum of t he lengths of t h e o t h e r two sides,

w * *14, Su.ppose AC i n t e r s ec t s 3D at a poin t B between A and C ,

w Pe~pendiculars are dropped from A and C to BD BtPlkIng

it at P and Q ~ e s p e c t i v e l y . S~IOW that P and Q are not on the same side of 33.

Chapter 8

PERPENDICULAR LINES AND PLANES IN SPACE

8-1. The BBIC Deflnitlon, In t h l s chapter we shall be specifically concerned with prop-

erties of figures t h a t do not lie in a single plane. The funda-

mental properties of such figures are s ta ted in Postulates 5b, 6, 7, 8 and 10, and inTheorems 3-2, 3-3 and 3-4. It would be worth your wh5le to review these.

Definition. A l i n e and a plane are perpendicular If they Intersect and if every llne lylng in the plane and passlng through

the polnt of intersection is perpendimlap to t h e given line.

If l b e L and plane E are perpendicular we write L ~ E or ELL. We have indicated, i n the figure, three lines in E passing

through P. Notice that in a perspecttve drawtng, perpendicular llnes dontt n~cessar~ly look perpendicula~~ Notice also tha t if we merely required t h a t E contain - one l i n e through P perpendicula~

t o L, t h i s would m e a n very l i t t l e ; you can fairly eas i ly convlnce you~self that every plane through P contalns such a line.

Problem -- Set 8-1

1. The f l g u ~ e at t h e r i g h t

represents plane E.

Do any points outside t h e

qua& i lateral shown / belong t o plane E ? u Is plane E Intended to include

every polnt outside the quadri-

l a t e ra l ?

Sketch a plane perpendicular t o a vertical line. (See

Appendix V, 1 Sketch a plane perpendicular t o a horizontal llne. Does each of your sketches represent a llne perpendicular

to a plane?

Repeat the sketch of Problem 2b. Add to t h e sketch th ree

l h e s In the plane which pass through t h e point of i n t e r - section. What I s the relatLonship between each of the three lines and the orfginal line?

4. Reread the deflnitlon of perpendicularity between a llne and a plane and decide whether the following statement is t rue if' that definition i s accepted:

T I If a l i n e is perpendicular to a plane, then it is perpendicu la r t o every line lylng In the plane and passing through

the point of intersection. "

5* Given that B, R, S and T are - I n plane E, and t h a t AB 1 E, whlch of the fol~owlng angles must be rlght angles:

E \ !--jj/ T

L Am, Am, L RBT, L TBA, I

L SBR? 4

6. If L PQH is a r igh t angle and Q and H are jn E, should you infer

from the definition of a line and e

a plane perpendicular that PQ 1 E? Why o r why not?

7. In the figure plane E contains points R, S, and P, but not T. a, Do points R, S and T determine

a plane? + b, If SP is perpendtcular to the

plane of R, S and T, which angles In the figure must be r i g h t angles? M

8. a. If a poin t 5s equldlstant from each of two o t h w p o h t s , are the three polnts coplanar?

b. If two points are each equidistant from each of two other

p o h t s , are the four points coplamar?

*g, a. Given: P Q Collinear points A, B and X as in the figure; I3 equidistant from P and Q; and A equldlstant f r o m T and Q. Prove: X is equidistant from F and Q.

b. b e s the proof require t ha t

Q be in the plane o f A, B, X and P?

10. Look ahead t o Theo~em 8-1 and make a model for it from s t i cks , wire coat hangem, or straws.

8-2. The Easic Theorem. The bask theorem on perpendicularity In space says tha t if a

plane E con ta in s two llnes, each perpendicular to a line L at the same point o f L, then L 1 E. The proof of t h i s Is easler If we prove two preliminary theorems (lemmas} .

Theorem 8-1. Xf each of two points of a 13ne I s equldlstant

from t w o given polnts, then every point of the line is equidistant

Restatement: If P and Q are two p o i n t s and L is a l i n e such t h a t two po in t s A, B of L are each equidistant from P and Q, then every point X of L is equidistant from P and Q. he above figure

8hows three poss ib l e positions f o r X. Of course , X might be a t

A or B,) Proof: First we consider t h e case whem X I s on the sane

side of A as B. X mLght be a t X B, or X but for convenience 1 ' 2

in the flgure we show it beyond 3 at XI. In this case LPAB =LPAX

and L QAB = L Q h X , We t reat t h i s case in 3 steps.

1. Since AP = AQ (given) , BP = (given), and AB = AB (identity),

AABP Z f i Am (S.S.S.) . Hence, L PA3 Z L QA3.

2. L PAX ,L QAX. This is because L PA3 2 L QAB by Step 1. (we are considering the case where L PAX = L PAB and

LQAX = L WB.1

3. Using Step 2 and t h e f a c t s that AP = AQ (glven), and AX = AX

( i d e n t i t y ) we flnd that A PAX 2 A W (s ,A.s) . Hence PX = QX.

+ The case where X ltes on the ray opposite AB 1s proved in a

similar fashion.

1. A piece of paper AXE& as

pictured here, 1s fo lded along m* Imagine A and B as both

being in the foreground of the

picture and in the background. Under these conditions will a polnt K of be equtdistant

from A and B? State a theorem to support your answer. If AF = 6, 3F = *

2. Here mine plane AXE obscur- ing part of plane A m . It I s

given that XA = X% and YA = YB, 6

T, W and 2 ape three other - p o i n t s of XY. b e 8 TA = TB? b e 8 WA = WB? Lbes ZA = ZB? State a theorem that supports each conclusion. . \

Theorem 8-2. If each of three non-collinea~ po in t s of a plane is equid is tan t from two po ln t s , then every po ln t of' the plane 2s equidistant from these t w o points,

0 P Given: Thee non-colllnear polnts A, B and C each equidistant from P and Q,

Prove: Zvery point of t he

plane determined by A, B ahd C Is equldlsta,nt from P and Q

Proof: The proof is given in t h ~ e e steps. .P

1. Since A and B are each glven equidistant from P and Q, each

M poin t of AB is equidistant from P and Q. This follows f ~ o m - ///

/ . .

Theorem 8-1. Similarly each / * p o l n t of IX is equidistant from Q

P and Q.

2. Let X be any other polnt of the H

plane. If X I s on ei ther AB or a P ++ CB, X i s equidistant from P and Q by S t e p l . If X is on one - side of W , choose Y, some potnt

w of AB ori the other side of ?%. The Plane Separation Postulate

/AT Q

assures us that there Is such a - poin t Y and tha t XY will I n t e r -

@ P s e c t C B in some polnt 2.

w 3 * Slnce Z is on CB it I s equ3-

distant from P a Q by Step 1. since Y i s m % t t i s e q u i -

d i s t a n t from P and Q by Step 1 -

Therefore by Theorem 8-1 every @ Q polnt of YZ is equld3stant from P and Q. X is one of' these points .

[ sec. 8-21

Since we have shown that each point X of the plane determined

by A, B, C is equidistant from P and Q, Theorem 8-2 I s established.

We are now ready to prove the basic theorem.

Theorem 8-3. If a l i n e i s perpendicular t o each of two In ter - secting l i n e s a t their point o f intersection, then it i s perpen- d icular to t he plane of these l i n e s .

Restatement; Let L, and Lp be l i n e s in plane E intersecting at A and let L be a l i ne through A perpendicular to L and L2. Then any line Lg In E through A is perpendicular t o L.

Proof: Statement Reason

1. Let P be a point on L, 1 1. By the Ruler Postulate, each

on Ln, none of these points

coinciding with A.

3, a point on L-,, Bg a point on Lp, and Bo a point

of these lines has an i n f i n i t e number of points.

2. Let Q be the oint on the ray

AQ -; AP. 1 oppos i te to such that

2. Point Plotting Theorem.

3. In t h e plane containing L and Li, L, is the perpendicular

b i s e c t o r of T^.

3 . Defini t ion of perpendicular bisector (section 6-3) .

4. B is equidistant from P and 6.

4. Theorem 6-2.

[sec. 8-21

Bn is equidistant from P and

a. A is equ id i s t an t from P and Q,

B3 is equidistant from P

and Q.

In t h e plane containing L and L3, L3 is the perpen-

dicular b i sec to r of m.

5. Similar t o 3 and 4.

7. Steps 4 , 5 and 6, and Theorem 8-2.

8. Theorem 6-2.

9. Definition of perpendicular bisector.

.O. Definition of perpendlcular- ity of line and plane, since L3 I s line in E through A.

Problem Set 8-2b --

Suppose A, B and C are each ,lq P

equidistant from P and Q. ex Explain in terms of a definition or theorem why each

A I

B point X of plane ABC is equi-

4 1 Q distant from P and Q.

Ã

Explain the relationship between the line of Intersection L of two walls of your classroom and the plane of the floor. How many lines perpendicular to L could be drawn on the f loor? Is L perpendicular t o every line that could be drawn on the f l oo r?

Figure FRHB I s a square. _[ E. A I s not In plane FRHB. a. How many planes are determined by pairs of segments in

t h e figure? Name them. b. At least one of the segments in this figure is perpendicular*

t o one of the planes asked f o r In Part (a). Which segment?

Which plane? A systematic approach t o such a problem is

t o write down every pair of perpendicular segments you see

in the figure. Then you can observe whether you have one line perpendicular to two intersecting lines.

4 .

A ABF is isosceles w i t h B as vertex. AH = FH. 1 m. R is not in plane AFB. a. How many different planes are determined by the segments

in the figure? Name them. b. Do you find a segment tha t is perpendicular t o a plane?

If so, tell what segment and what plane and prove your st at ernent .

- 5. In t h i s figure, FB 1 plane P, and in A RAB, which lies in

plane P, BR = EA. Prove A ABF ^ A RBF and L FAR Z L FRA.

*6. Given the cube shown, with BR = BL. Does KR = KL? Prove that your answer is correct.

(Since we have not yet given a precise definition of a cube we state here, for use in your proof, the essential proper t ies

of the edges of a cube:

The edges of a cube consist o f twelve congruent segments, related as shown In the p ic ture , such that any two intersecting segments are perpendicular. )

<Ã‘

7. In the accompanying figure WX is a line in plane E. Plane F 1 % at Q. In plane F, % 1 ti?. IS la the intersection of E

*-> and F. Prove RQ 1 E.

For all we know up to now t h e conditions specif ied in t h e

definition of a line and a plane perpendicular might be impossible

to achieve. To reassure us, we need an existence theorem. The next theorem enables us to see that we are not talking about things

that cannot exist In speaking of perpendicularity between lines and planes.

Theorem 8-4. Through a given poin t on a given l i n e there passes a plane perpendicular to the line.

Proof: Let P be a point on a line L. We show In six steps that there is a plane E through P perpendicular to L.

1. Let R be a point n o t on L. That there is such a point

follows from Pos tu la te 5a. 2. Let M be the plane determined by L and R. Theorem 3-3

tells us there is such a plane. 3. Let Q be a po in t not on M . Postu la te 5b assures us that

there is subh a po in t . 4. L e t N be the plane determined by L and Q.

5. In plane M there is a line L, perpendicular to L at P h he or em 6 - l ) , and in plane N there is a line Lp perpendicular to L at P.

6 . By Theorem 8-3, the plane I3 determined by L1 and Lp I s

perpendicular to L at P.

If E 1 L at P then every line In E and through P I s perpendie - ular to L, by definition. May there be some l i n e s not in E but still perpendicular to L at P? The next theorem says, "?Jon.

Theorem 8-5. If a l i n e and a plane are perpendicular, then

the plane contains every line perpendicular to the given line at its point of intersection with the given plane.

Restatement: If line L is perpendicular to plane E a t p o i n t P, and if M is a line perpendicular t o L at P, then M lies Â¥1 E.

Proof: Statements

t

Reasons

1. L and M determine a plane F. 2. Planes F and E intersect in

a l i n e N. 3. N I L .

4. M l L *

5* M = N. his means M and N are the same line.)

6 . M lies in E.

1. Theorem 3-4. 2. Postulate 8.

3. Defin i t ion of perpendicularity of l i n e and p l a n e .

4. Given.

5. M and N both l i e in plane I? by Steps 1 and 2, are both 1 L by Steps 3 and 4, but Theorem 6-1 says there is only one such perpendicular.

6 , M = N by Step 5 and N lies in E by S t e p 2.

This theorem enables us to prove the uniqueness theorem that goes w i t h Theorem 8-4.

Theorem 8-6. Through a given point on a given line there is at most one plane perpendicular to the line..

Proof: 3ince a perpendicular plane contains perpendicular l i n e s through the point , and since two di f fe ren t planes have only one l i n e In common (Theorem 3-41, there cannot be two such planes. -

Just as in a plane where the characterization Theorem 6-2

followedthe existence and uniqueness Theorem 6-1, so now we can prove a similar characterization theorem f o r space.

Theorem 8-7. The perpendicular bisecting plane of a segment is t h e s e t o f all po in t s e q u i d i s t a n t from the end-points of t h e

segment. Note that this theorem, l i k e Theorem 6-2, has two parts. Restatement: Let E be the perpendicular bisecting plane of -

AB. Let C be the mid-point of m. Then

(1) If P is in E, then PA = PB, and (2) If PA = PB, then P is in E.

The proof is lef t t o the student.

Problem S e t 8 -2c -- a. A t a point on a line how many lines are perpendicular t o

the line? b. At a point on a line how many planes are perpendicular to

the line?

Planes E and F i n t e r s ec t in <->Â KQ, as shown in this f igure .

Â¥<-

E. BR lies in plane E. '<Ã‘Ã

Plane ABR intersects F i n BC.

w * 1f G]E at P and QP 1 PR,

<-Ã why does PR lie in E?

, Assuming here t h a t

AX = EX,

AY = BY,

AW = EW, A 2 = BZ,

why are W, X, Y and Z coplanar?

P l a n e E is the perpendicular bisecting plane of E, as shown In the figure.

a . m 2 E2 a2

m L A F W =

AAKF 2

b. Does FW = FK = FR? Explain. v 0

* 6 . Prove t h i s theorem: If L is a line intersecting plane E at point M, there is at least one l i ne Lf in E such that L1 1 L.

The next theorem is a lemma which is useful In proving later theorems.

Theorem 8-8. Two lines perpendicular to the same plane are coplanar .

Proof: L e t l ines Li and L2 be perpendicular to plane E at

t h e points A and B respectively. Let M be the mid-point of m, l e t L be the line In E which is the perpendicular bisector of m, and l e t P and Q be two po in t s on L such that PM = QM* L e t C be

a point on L, dis t inc t from A .

By the S.A.S. Postulate , A AMP 2 A AMQ, and so AP = AQ.

Since L _[ E, L CAP and L CAQ are right angles, and the S.A.S.

Postulate gives A CAP 2 A CAQ, so t h a t CP = CQ.

From AP = AQ and CP = CQ it follows, by Theorem 8-7, that C and A both l i e in the bisecting plane El of R. Hence, L, lies In Ef . In exactly the same way we prove that Lp lies in E l . Hence, L, and L2 are coplanar.

8-3. Existence - and Uniqueness Theorems. The following theorems cover a l l possible relations between

a point , a l i ne and a perpendicular plane. They are s ta ted here f o r completeness and for convenience i n reference.

Theorem 8-9. Through a given point there passes one and only one plane pemendicular t o a given l i n e .

Theorem 8-10. Through a given point there passes one and only one 1 i n e perpendicular t o a given plane .

The proof of each of these theorems has two cases, depending

on whether or not t he given poin t lies on the given line or plane, and each case has two parts, one for proving existence and one for proving uniqueness. T h i s makes a to ta l of eight proofs required. Theorems 8-4 and 8-6 are two of these eight; the remaining six, some of which are hard and some easy, are given In Appendix VI.

Theorem 8-10 assures us of the existence of a unique perpendicular t o a given plane from an external point. Hence, we are justif ied In giving the following definition, analogous t o the one following Theorem 7-6,

Definition. The distance t o a plane f r o m an external point is the length of the perpendicular segment from the po in t t o the plane.

Theorem 8-11. The s h o r t e s t segment t o a plane from an ex-

ternal point is t he perpendicular segment. The proof is similar to tha t of Theorem 7-6.

Review Problems

1. Use a drawing if necessary to help you decide whether each

statement is true or false. a. The Intersection of two planes may be a segment. b. If a l i n e i n t e r s ec t s a plane in only one point , there are

at least two lines in the plane perpendicular to the line. c . For any four points, there I s a plane containing them all.

d. If three lines intersect in pairs, but no point belongs to all three, the lines are coplanar.

e. It is poss ib l e f o r three lines to Intersect in a point , so that each is perpendicular to the other t w o ,

f . Only one l i n e can be drawn perpendicular to a given l i n e at a given point.

g. A t a point in a plane there is only one line perpendicular

to the plane. h. The greatest number of regions into which three planes can

separate space is eight. -

2. F r o m a po in t R outside plane E, RB 1 E and I B Intersects t h e

plane in B. 'SK is any other segment from R, intersecting E in A . Compare the lengths arid RE. Compare the measures

of L A and L B.

3 . If the goal posts at one end of a f o o t b a l l f i e l d are perpendicular to the ground, then they are coplanar even without a brace between them. Which theorem supports this conclusion? Can the goal posts still be coplanar even if they are not perpendicular to t he ground? Could they fail to be coplanar

even with a brace between them?

4 , Do there always exist a. two lines perpendicular to a given line at a given point

on the line?

b. two planes perpendicular to a given line at a given point

on t h e line? c. two lines perpendicular to a given plane at a given point

on the plane? d. two planes perpendicular to a given line? e. two intersecting lines each perpendicular to a given plane?

5. The assumption t ha t two l i n e s L, and Lp are perpendicular t o plane E and L, and L2 intersect in point P not in plane E can be shown t o be false by

proving that the assumption leads to a cont radic t ion of a

theorem about figures In a plane. Which theorem?

<-Ã

6 . Given MQ 1 plane E, and 31 to plane E. How many di f fe ren t planes are determined by w[, <-Ãˆ MW, WF and ?I Explain.

AABF is isosceles with vertex at 3. HF = HA. m l m . R is not In the plane AFB. a. How many di f fe ren t planes are determined by the segments

in the f igure? Explain.

b. Locate and describe a line t h a t is perpendicular t o a

plane.

8. Given: P is in plane E which contains A, B, C; P I s equl-

distant f r o m A, B, C; line L 1 E at P. Prove: Every point, X, in L is equidistant from A, B, C .

9. Given: Line L J plane A X a t Q; point P of L is equidistant

from A , B, C .

Prove: Every point of L I s

e q u i d i s t a n t f r o m A, B, C. (~int: Consider any point x # Q on L and show XA = XB = XC.)

- 10. Given: AP l P ^ and D I E ; A

11? a t Q. Prove; Ad 1 S?.

<- (~int: Take R on QC so that

QB = O R , Drawm, m.) c

11. Prove the following theorem: If from a poin t A outside a plane, a perpendicular and oblique (non-perpendicular ) segments and AH are drawn, meeting the plane at unequal

distances from 3, the segment which meets the plane a t t h e

greater distance from 3 has the greater length.

- Given: A 3 1 plane E. F and

H are points of E such that BF > BH. Prove: AF > AH.

- + + + 12. Prove that each of fou r rays AB, AC, AD and AE cannot be

perpendicular to the other three. <-9Ã <-Ã <-*Â

1 3 . Given: XB and YB are two l ines in plane E; m is a plane 1 XI3 <-Ã <-Ã

at B; n is a plane 1 Y 3 at B; AB is the intersection of m and

n. <-Ã

Prove: AB j E.

Chapter 9

PARALLEL LIMES IN A PLANE

9-1. Conditions Which Guarantee Parallelism. Thus far in our geometry we have been mainly concerned w i t h

what happens when l ines and planes In te r sec t in cer ta in ways. We are now going to see what happens when they do in te r see t . It will turn out t h a t many more Interes t ing things can be proved.

We first consider the case of two l ines . Theorem 3-3 gives us some information right away, since it says tha t if t w o lines intersect they l i e in a plane. Hence, if two lines are not coplanar they cannot in te rsec t .

Definition: Two l ines which are not coplanar are said t o be

skew. - You can easily f ind examples-' of skew lines in your classroom. This still leaves open the question as to whether two

coplanar lines must always in te r sec t . In Theorem 9-2 we shall prove t h e existence of coplanar lines that do not i n t e r s e c t , but

are parallel, like t h i s :

L e t us first make a precise d e f i n i t i o n . Defini t ion: Two l i n e s are parallel if they are coplanar and

do not; in tersect . Note that f o r two l ines t o be parallel two conditions must be

sa t i s f i ed : they m u s t no t intersect; they must both lie i n the same plane.

Theorem 9-1. Two parallel lines lie in exactly one plane.

Proof: If L., and Lp are parallel lines it follows from the above definition that there is a plane E containing L-, and

L p . If P is any point of % It follows from Theorem 3-3 that

there is only one plane containing Li and P. Hence, E la the

only plane containing L, and Lp.

We will use the abbreviation Ll 11 Ln to mean that the lines Li and LA are parallel. As a matter of convenience

we will say that two segments are parallel If the lines that contain them are parallel. We will speak similarly of a line and

a segment, or a line and ray, and so on. For example, suppose we have given that L-, \\ Lp, in the figure below;

+@ 4 + Then we can a l so write It?%, AB 1 1 Lg, Ll 1 1 CD, BA 11 z, and so on. Each of these statements I s equivalent to the s ta te-

ment that L, 11 Ln. It does not seem easy to t e l l from the def ini t ion whether

two lines which seem to be parallel really are parallel. Every

line stretches out i n f i n i t e l y far in two directions, and to tell whether two lines do no t in tersect , we would have to look at - all

of each of the two l i n e s . There i s a simple condition, however, which is su f f i c i en t t o guarantee t ha t two lines are parallel . It goes like this:

Theorem - 9-2. Two lines In a plane are para l le l if they are both perpendicular to the sme line.

Proof: Suppose that Li and L2 are two lines In plane E, each perpendicular to a line L, at points P and Q.

There are now two possibilities: (1) L, and L2 Intersect in a point R.

(2) L-, and Lp do not Intersect.

In Case (1) we would have two lines, L, and Lp, each perpendicular to L and each passing through R. This i s impossible by Theorem 6-1 i f R l i e s on L, and by Theorem 6-3 if R I s not on L. Hence, Case (2) i s the only possible one, and so, by definition, L, 11 Lp

Theorem 9-2 enables us to prove the following important existence theorem.

Theorem 9-3, Let L be a line, and let P be a point not on L. Then there is at least one line through P, parallel to

L

Proof; Let Li be a line through P, perpendicular to L.

(By Theorem 6-1, there is such a line. ) Lst L, be a l i n e through P , perpendicular to Li in the plane of L and P. By Theorem 9-2, L2 11 L.

I t might seem natural, at this point , to try t o prove that

the parallel given by Theorem 9-3 is unique; that is, we might try to show t ha t in a plane through a given po in t n o t on a given l i n e there is only one paral le l to the given line. Astonishing as it

may seem, t h i s cannot - be proved on the basis of the postulates ----- that we have s ta ted so far; it must be taken as a new postu la te . -7- -- We will discuss t h i s in more de ta i l in Sect ion 9 - 3 , In the mean-

time, before we get to work on the basis of this new postulate we shall prove some additional theorems which, like Theorem 9-2, tell us when two lines are parallel.

We f i r s t give some definitions.

Definition: A transversal of two coplanar l ines is a line which intersects them in t w o d i f ferent points.

We say the two lines are "cut" by the transversal.

Definition: Let L be a transversal of L-, and L2, intersecting them in P and Q. L e t A be a point of Ll and B a point of L2 such that A and B are on opposite sides of L. Then PW and WA are alternate Lnte~ io r awlea formed by the transversal to the two lines.

Notice that In the definition of a transversal, the two lines that we star t with may or may not be parallel. But if they in te r - s e c t , then the transversal is not allowed to intersect them at their common point. The situation in the figure below is not allowed:

That is, in thi3 figure L is a transversal to the lines Ll and Lp.

Notice also that a common perpendicular to two lines in a

plane, as In Theorem 9-2, is always a t ransversal .

Theorem s, If two l h e s are cut by a trmsver8a1, and if one pair of alternate I n t e r i o r angles are congruent, then the other pair of alternate In t e r io r angles are also congruent.

That I s , if / a = / a 4 , then / b E / b l . And if / b ~ / b t ,

then / a ^ / a! . The proof is left to the student.

The following theorem I s a generalization of Theorem 9-2, t h a t is, It Includes Theorem 9-2 as a special case:

Theorem s. Xf two lines are c u t by a transversal, and if a pair of alternate in te r io r angles are congruent, then the lines

are parallel.

Proof: Let L be a transversal to L-, and Lp, i n t e r - secting them in P and Q. Suppose t h a t a pair of alternate in t e r io r angles are congruent. There are now two possibilities:

(1) L-, and Lp intersect in a point R.

(2) L, I I L,* In Case (1) the figure looks l ike this:

L e t S be a point of Ll on the opposite side of L from R .

Then / SPQ is an exterior angle of A P Q R , and / PQR is one of t h e remote i n t e r io r angles. By Theorem 7-1, t h i s means that

m / SPQ 3 m / PQR.

But we know by hypothesis t ha t - one p a i r of alternate interior angles are congruent. By the preceding theorem, both pairs of

alternate In te r io r angles are congruent. Therefore,

m / SPQ = m / PQR.

Since Statement (1) leads to a contradiction of our hypothesis, Statement (1) is false. Therefore Statement (2 ) is true.

Problem - Set

1. a . Does the definition of parallel lines s t a t e that the

lines must remain the same distance apart?

b. If two given lines do not l i e in one plane, can the lines be paral le l?

2. Two lines in a plane are parallel if , or if 9

or if

3 . If two lines in a plane are Intersected by a transversal, are the alternate i n t e r io r angles always congruent?

4. In space, if two lines are perpendicular t o a third line, are the two lines parallel?

5. a. If the 80' angles were correctly drawn, would 4 so* L, be parallel to Lp

* L,

according to Theorem 9-5? Explain.

4 B04

>Ln b. How many different

measures of angles

would occur in the drawing? What measures?

6 . In the figure, If the angles At-,

were of the s i z e indicated, M1< SO-

which lines would be parallel? M2<

^v v v 491- +

1-2

89' 90,

3

>Â

91.

7 . Given a line L and a point F no t on L, show how pro- t r a c t o r and ruler can be used to draw a para l le l to L

through P.

8. Suppose the following two definitions are agreed upon:

A vertical - line is one containing the center of t h e

earth.

A horizontal - line l a one which I s perpendicular to some

vertical line.

Could two horizontal l i ne s be paral le l?

Could two vertical l i ne s be parallel?

Could two hor izonta l l ines be perpendicular?

Could two vertical l ines be perpendicular?

Would every vertical line also be horizontal?

Would every horizontal l ine also be vertical?

Could a hor izonta l line be p a r a l l e l to a vert ical line?

Would every l i n e be hor izonta l?

It possible t o f ind two lines in space which are nei ther parallel nor in tersect ing?

D C 10. Given; DAB =I I I /CBA = 90,

and A D = C B .

prove: rnL ADC = m /BCD. C a n you also prove m / ADO

Â¥s

'\ 0 #'

\ 0

X' , ^

'*Ã

,/' '., = m / BCD = 90? A 0

11. Given the figure with AR = RC = PQ,

AP = PB = RQ,

BQ = QC = PR. Prove : m / ~ + r n / ~ + m / ~ = l 8 0 . B (~int: Prove m / a = m / A, m / b = m / B , m L c = m / C . )

12. Given: A 3 = AC, AP = AQ. <Â¥Ã Ã‡-

Prove: PQ 11 BC . ( ~ i n t : L e t the bisector of - / A Intersect PQ a t R - and BC at D.)

13. Given: The figure with

/ A = / B , AD = BC,

AT = TB, SD = SC. Prove :

9-2. Correspondin& Angles. In the figure below, the angles marked a and at are

called corresponding angles:

S imi l a r ly , b and b 1 are corresponding angles; and the pairs

c, c! and d, d t are also corresponding angles.

Definition: If two lines are cut by a transversal, if / x and / y are alternate In t e r io r angles, and If / y and / z

are vertical angles, then / x and / z are corresponding angles.

You should prove the following theorem.

Theorem - 9-6. 1f two lines are cut by a transversal, and if one pair of corresponding angles are congruent, then the other three pairs of corresponding angles have the same property.

The proof is only a little longer than that of Theorem 9-4,

Theorem u. If two lines are cut by a tmmversal, and if a pair of corresponding angles are congruent, then the lines are parallel. The proof is l e f t to the student.

It looks as though the converses of Theorem 9-5 and Theorem 9-7 ought to be t rue . The converse of Theorem 9-5 would say that if two paral lel ' l ines are c u t by a transversal, then the alternate i n t e r i o r angles are congruent. The converse of Theorem 9-7 would say that If t w o parallel lines are cut by a transversal, then corresponding angles are congruent. These theorems, however, cannot be proved on the basis of the postulates tha t we have

stated so far*. To prove them, we shall need to use the Parallel Postulate, which w i l l be stated i n the next section.

The Parallel Postulate i s essential to the proofs of many

other theorems of our geometry as well. Some of these you are

already familiar with from your work in other grades. For example, you have known f o r some t i m e that the sum of the measures of the

angles of any triangle is 180. Yet, without the Parallel

Postulate It is impossible to prove this very Important theorem. L e t us go on, then, to the Para l le l Pos tu l a t e .

9-3. - The Parallel Postulate .

I Postulate - 16. he Parallel Postulate . ) Through 1 1 a given external point there is at most one line I 1 parallel to a given line. 1

Notice that we dont t need to say, i n the postulate, that there is a t least one such parallel, because we already know t h i s by

Theorem 9-3.

[see. 9-31

It might seem natural to suppose that we already have enough pos tu l a t e s to be able to prove anything tha t l a "reasonable"; and

since the Parallel Postu la te is reasonable, we m i g h t try to prove it instead of ca l l i ng it a postulate. A t any rate , some very clever people felt this way about the postulate, over a period of a good many centuries. None of them, however, was able to find a proof. Finally, In the last century, I t was discovered t h a t no such proof is possible. The po in t is that there are some mathe-

matical system that are almost like the geometry that we are studying, but not quite. In these mathematical systems, nearly all of the postulates of ordinary geometry are satisfied, but the

Parallel Postulate is not. These on-~uclidean ~eometries " may seem strange, and In f ac t they are. (FOP example, in these "geometries" there is no such thin& as a aquare. 1 Not only do ---- -- they lead to interesting mathematical theories, but they also have

important applications t o physics. Now that we have the Parallel Postulate we can go on to prove

numerous important theorems we could not prove without it. We

star t by proving the converse of Theorem 9-5.

Theorem - 9-8. If two parallel lines are cut by a t ransversal , then alternate i n t e r io r angles are congruent.

Proof: We have given parallel lines L, and 5, and a transversal Ln, intersecting them in P and Q.

Suppose that / a and / b are - not congruent. L e t L be a line through P for which alternate interior angles - are congruent.

(BY the Angle Construction Postulate, there Is such a line.)

Then L # Li, because / b and / c are not congruent.

Now l e t us see what we have. By hypothesis, L, I [ L2. And

by Theorem 9-5, we know t ha t L 1 f Lp. Therefore there are two

lines through P, parallel to Lp. This is impossible, because It

contradicts the Parallel Postulate. Therefore / a a / b , whloh was to be proved.

The proofs of the following theorems are short, and you should write them for yourself:

Theorem u. If two parallel lines are c u t by a transvernal, each pair of corresponding angles are congruent.

Theorem 9-10. If two parallel lines are c u t by a transversal, In ter ior angles on the same side of the transversal are supplementary.

Restatement: Given Ll 1 1 Lg and T intersects Ll and

Lp. Prove tha t / b I s supplementary to / d and /_ a I s supplementary to / e .

Theorem 9-11. In a plane, two lines parallel to the same

l ine are parallel to each other.

Theoremy-12. In a plane, if a line is perpendlcula~ to one two parallel lines it I s perpendicular to the other.

Problem - S e t

Given : m L ~ = r n L B = r n L C = 9 0 . Prove: m / D = 90.

Prove that a l ine parallel to the base of an isosceles triangle and intersecting the other two sides of the triangle forms another Isosceles triangle.

Given: In the figure, RT = HS, % [I '8.

Prove: PQ = PT.

Q T

4. Review indirect proof as illustrated by the proof of Theorem 9-8. Give an Indirect proof of each of the following statements, showing a contradiction of the Parallel Pos tu l a t e .

a . In a plane, if a third line M intersects one of two parallel l ines Li a t P, It also M intersects the other

La - -

b. In a plane, if a l i n e R intersects only one of t w o other lines Ll

and Lp, then the l ines L, and L2 intersect .

- La

G i v e n ; R i n t e r s ec t s L, a t P.

R does not in te rsec t L2.

Prove: Li Intersects L2

5. a. Prove: Two angles In a plane which have their s ides respectively parallel and extending both in the same

( o r both in opposite) direct ions are congruent.

G i v e n : BA 1 1 ?? , BE [ I 3 ,

b. Prove: Two angles in a plane which have t h e i r s ides respectively parallel but have only one p a i r extending in the same d i r e c t i o n are supplementary.

--+ G i v e n : BA 1 [ 5 ,

B? 11s. Prove: In (a ) LABC 2 / X Y Z .

In (b) ~LABC + rn&CYz = 3.80.

(~ote: Only cer ta in cases are illustrated and proved here. All other cases can also be proved easily. The

term "direction" I s undefined but should be understood.)

6 . Make drawings of various pairs of angles ABC and DEF such

tha t B? 1 and 5E 1 3. Sta te a theorem that you think may be true about the measures of such angles.

*7 . If Theorem 9-8 I s assumed as a postulate, then the Parallel

Postulate can be proved as a theorem. { m a t is. it must be shown that there cannot be a second parallel to a l i n e through a point not on it.)

Given: Li and Lg are

two l i n e s containing P,

and Ll 1 1 M.

Prove: Lp not parallel

t o M. M

*8 . Show that if Theorem 9-12 (1f a t ransversa l is perpendicular to one of two parallel lines, it I s perpendicular to the

other.) I s assumed as a postulate, the Pa ra l l e l Postulate can be proved as a theorem.

Glven: Ll 1 1 M and Ll and

Ln contain P. (L, # L,.) Prove: LA not parallel t o

9-4. Triangles.

Theorem 3-13, The sum of the measures of the angles of a triangle is 180.

Proof : Given A ABC , l e t L be t h e line through B, - parallel to AC. L e t / x , / x*, / y, / yt and / be as in the figure.

<-> Let D be a point of L on the same side of AB a s C.

<-Ã Since AC 11 =, A is on the same side of BD as C. Therefore C is in the i n t e r i o r of / ABD (definition of i n t e r i o r of an angle), and so, by the Angle Addition Postulate, we have

m / A B D = m l z + m / y l .

By the Supplement Postulate, m / X I + m / ABD = 180.

Therefore m / x l + + m / z + m / y 1 = 180.

But we know by Theorem 9-8 that m / x = m / xt and m / y =

m / y f , because these are alternate Inter ior angles. By sub- sti-bufcion we get

m / x + m L z + m / y = l 8 0 , which was t o be proved.

From th i s we get a number of Important corollaries:

Corollary 9-13-1. Given a correspondence between t w o

triangles. If two pairs of corresponding angles are congruent,

then the th i rd pair of corresponding angles are a l so congruent.

The corollary says that if / A A ' and / B S / B', then / C S / C 1 . As the figure suggests, the corollary applies to cases where the correspondence given is no t a congruence, as

well as to cases where A ABC 9t A A t B I C t .

Corol la ry 3-13-2. The acute angles of a right triangle are complementary.

Corollary 9-l3-3. FOP any triangle, the measure of am exterior angle is the sum of the measures of the two remote

Interior angles.


1. If the measures of two angles of a triangle are as follows, what I s the measure of the third angle?

a . 37 and 58. d. r and a .

b. 149and30. e. 45 + a and 45 - a. c . n and n. f. 90 and At.

2. To find the distance from a point A to a distant point P, a surveyor may

A measure a small distance AB and also measure / A

and / B. From this in-

formation he can compute the measure of / P and pp 0 by appropriate formulas then compute AP. If

m / A = 87.5 and m / 3 = 88.3, compute rn / P.

3. Why is the Parallel Postulate essential t o the proof of Theorem 9-13?

4. On a drawing like the one on the

values

right fill in

of a l l of the the

angles,

5. Given: / A s / X and / 3 a/ Y, can you

correctly conclude that:

a. LC S L Z ? -

b. E i - XY? A A.

6 . Given: BD bisects / EBC , and % 11 Prove: AB = BC.

7. The b i sec to r of an exterior angle at the vertex of an Isosceles t r iangle is parallel to the base. Prove t h i s .

8. Given: The figure.

Prove: s + r = t + u. ( ~ i n t : Draw DB.)

*9. Given: In the figure, / BAG is a right angle

and QB = QA.

Prove: QB = QC.

*lo. Given: In A m , / C

is a r i gh t angle, AS = AT and BR = BT.

c Prove: m / STR = 45.

( ~ i n t : Suppose m / A = a.

Write formulas in turn for

the measures of other angles i n t h e figure In terms of a.) A

9-5. Quadrilaterals in Plane. A quadrilateral is a plane figure with four sides, like one

of the following:

The t w o figures on the bottom illustrate what we might call

the most general case, In which no two sides are congruent, no two sides are parallel, and no two angles are congruent.

We can state the definition of a quadrilateral more precisely, in the following way.

Definition: L e t A, B, C and D be four points lying in

the same plane, such tha t no three of them are collinear, and such - - - t h a t the segments AB, BC, CD and DA intersect only In t h e i r

end-points. Then t he union of these four segments is a quadri-

lateral.

For short, we will denote this figure by ABCD. Notice tha t

in each of the examples above, with the exception of the last one, the quadrilateral plus its i n t e r i o r forms a convex se t , in the sense which was defined in Chapter 3. This I s not true of the figure at the lower right, but this figure is s t i l l a quadrilater-

a l under our definition. Notice, however, that under our

def in i t ion of a quadrilateral, figures like the following one are ruled out.

[set- 9-51

Here the figure is no t a quadrilateral, because the segments

and DA intersect in a point which la not an end-point of e i ther of them. Notice also, however, that a quadrilateral can be formed, using these same four points as vertices, like t h i s :

Here ABDC Is a quadrilateral.

Defini t iom : Opposite s i d e s of a quadrilateral are two side3 that do not i n t e r sec t . Two or Its pngles a r e opposLte if they do

not contain a common side. Two s ides are called consecutive if they have a common v e r t e x . Similarly, two angles are called consecutive if they contain a common side. A diagonal is a

segment joining two non-consecutive vertices.

In a quadrilateral ABCD, and CD are opposite sides, - as are BC and AD. AD and CD or AB and are con- - aecutive sides. AC and are the diagonals of ABCD.

Which angles are opposite? Which consecutive?

Definition: A trapezoid l a a quadrilateral in which two, and only two, opposite sides are parallel.

Definition: A ~parallelopyam is a quadrilateral in which both pairs of opposite sides are parallel.

You should not have much trouble in proving the bas ic theorems on trapezoids and parallelograms:

Theorem $l-14. Either diagonal separates a parallelogram

i n t o two congruent triangles. That is, if ABCD is a parallel-

ogram, then A ABC s A CDA . Theorem 3-15. In a parallelogram, any two opposite sfdes

are congruent.

Corollaq 9-15-1. If Ll 11 L, and If P and Q are any two points on Li, then the distances of P and Q from Lp are equal.

This proper ty of parallel lines is sometimes abbreviated by saying tha t "parallel l ines are everywhere equidistant" .

Definition: The distance between - two parallel lines is the

distance from any point of one l i n e to the other line.

Theorem 9-16. In a parallelogram, any two opposite angles are congruent.

Theorem - 9-17. In a parallelogram, any two consecutive angles are supplementary.

Theorem 9-18. The diagonals of a parallelogram bisect each other.

In Theorems 9-14 through 9-18 we are concerned with several p rope r t i e s of a parallelogram; tha t is, if we h o w that a quadrilateral is a parallelogram we can conclude certain facts about It. In t h e following three theorems we provide for the converse relationship; that is, if we know certain facts about a quadri- la teral we can conclude that I t is a parallelogram.

Theorem - 9-19. Given a quadrilateral in which both pairs of opposite sides are congruent. Then the quadrilateral is a

parallelogram.

Theorern 9-20. If two sides of a quadrilateral are papallel and congruent, then the quadrilateral is a parallelogram.

Theorem 9-21. If the diagonals of a quadrilateral bisect

each other, then the quadrilateral is a parallelogram.

The fol lowing theorem s t a t e s two useful f a c t s . The proof of this theorem is given In full.

Theorem 9-22. The segment between the mid-points of two

s ides of a triangle is paral le l to the third s ide and half as

long as the t h i r d side,

Res taternent : Given ^AN. L e t D and E be the mid- 1 points of and E. Then BE 11 AC , and DE = -y AC .

Proof: Using the Point Plotting Theorem, let F be the Ã‘Ã

po in t of the ray opposi te to ED such tha t EF = DE. We give

the rest of the proof in the two-column form. The notation f o r angles i s that of the figure,

Statements 1. EF = ED.

1 10. DE = 7 AC.

Reasons F was chosen so as to make this t rue. E is the mid-point of e. Vertical angles are congruent.

The S . A . S . Postulate. Corresponding parts of congruent triangles . Theorem 9-5. AD = DB, by hypothesis, and DB = PC, by statement 4. Theorem 9-20.

Definition of a parallelogram*

1 DE = -y DF, by statement 1,

and DF = AC, by Theorem 9-15.

9-6. Rhomb*a, Rectangle - and Square.

Definitions: A rhombus is a parallelogram a l l of whose sides are congruent.

A rectangle i s a parallelogram a l l of whose angles are flight angles.

Finally, a square is a rectangle all of whose sides are cohgruent,

As before, we leave the proofs of the following theorems f o r

the student.

Theorem - 9-23. If a parallelogram has one right angle, then it has four right angles, and the parallelogram I s a rectangle.

Theorem - 9-24. In a rhombus, the diagonals are perpendicular to one another.

Theorem - 9 - 2 5 , If t h e diagonals of a quadrilateral b i s e c t each other and are perpendicular, then the quadrilateral is a

rhombus.

Problem Set 9-6 -- 1. For which of the quadr i l a te ra l s -- rectangle, square,

rhombus, parallelogram -- can each of the following properties be proved?

a . Both pairs of opposite angles are congruent.

b. Both pairs of opposite sides are congruent.

c . Each diagonal bisects two angles.

d. The diagonals b i s e c t each other.

e. The diagonals are perpendicular.

f. Each pa i r o f consecutive angles I s supplementary.

g. Each pair of consecutive sides Is congruent.

h. The figure is a parallelogram.

i. Each pair of consecutive angles Is congruent.

j. The diagonals are congruent.

2. With the measures of t h e angles as given in parallelogram ABFH, give the degree measure of each angle.

3. In th i s figure ABHQ and AFRM are parallelograms. What is the relationship of

/ M to /H? of / R to

/H? Prove your answer.

A B F

4. Would the following information about a quadrilateral be

sufficient to prove it a parallelogram? a rectangle? a

rhombus? a square? Consider each item of information separately.

Both pairs of Its opposite sides are parallel.

Both pairs of I t s opposite s ides are congruent. 3

Three of its angles are right angles. A

Its diagonals bisect each other.

Its diagonals are congruent.

Its diagonals are perpendicular and congruent.

Its diagonals are perpendicular b i sec to r s of each other.

It is equi la teral .

It I s equiangular.

It is equilateral and equiangular.

Both pairs of Its opposite angles are congruent.

Each pair of its consecutive angles I s supplementary.

5. Given: ABCD is a parallelo- - gram with diagonal AC. 0 r. AP = RC.

Prove: DPBR I s a parallel-

ogram. A B

6 . Given: Parallelograms AFED and PBCE, a8 shown in t h i s plane figure.

Prove : ABCD is a parallelogram.

7. If lines are drawn parallel to t h e legs of an Isosceles t r iangle through a point in the base of the triangle, then a parallelogram is formed and Its perimeter is equal to the sum,

of the lengths of the legs.

Given: In the figure - - RS a RT, 11 E, PY 11 E. Prove: a. PXRY is a

Parallelogram.

b. PX + XR + RY + YP S

/\ P T

= RS + RT.

8. In this figure, if ABCD l a a parallelogram w i t h diagonals AC and %6 intersecting in Q and is drawn through Q, prove tha t is bisected by Q.

Q A F 8

9 . Given the I s o s c e l e s trapezoid

ABCD In which AD = GB and - CD 11 E. Prove / A S / B.

10. The median of a trapezoid is the segment Joining the mid- points of its non-parallel sides.

a. Prove the following theorem: The median of a trapezoid

is parallel t o the bases and equal in length to half

the sum of the lengths of the bases.

Given: Trapezoid ABCD

with CD 11 E, P the midpoint of AD and Q the midpoint of BC.

Prove: PQ 1 1 AB (A3 + cD). PQ = - 2

<-Ã ( ~ i n t : Draw DQ meeting - AB a t K . )

b. If AB = 91n. and DC = Tin, then A B PQ =

1 c . If DC = % and A 3 = 7 , then PQ =

11. A convex quadrilateral with vertices labeled consecutively ABCD is called a - k i t e if AB = BC and CD = DA. Sketch some k i t e s . S t a t e as many theorems about a k i t e as you can and prove at least one of them.

Given: Quadri la tera l ABCD

with P, Q, R, S the

midpoints of the s ides .

Prove: RSPQ is a parallel- - ogram, and PR and SQ bisect each other.

13+ Given: In the figure - AD < BC, DA 1 E,

Prove: m / C < m / D.

*l4. Prove that t he sum of the lengths of the perpendiculars

drawn from any point in the base of an Isosceles triangle to the legs is equal t o the

length of the altitude upon

either of the legs.

(~int: Draw % 1 E. Then - the figure suggests that PX

and Q3' are congruent, and t h a t and % are congruent. )

*l5. Prove that the sum of the lengths of the perpendiculars drawn from any point in the i n t e r i o r of an equilateral triangle to the three sides is equal t o the length of an altitude.

(Hint: Draw a segment; perpendicular to the altitude used, f r o m the interior p o i n t .)

16. Given a hexagon as In the F A figure w i t h AB 11 OC, - - BC [ O D , CD 11 E , - - DE 11 OF, EF 1 c.

- E Prove: FA I[ m.

D C

- - 17. a . Given sl, BBt , CC1

are paral le l and - - - 11 AfB ' , BC I ~ B ' C I

A as in figure.

prove: AC J J A ' G ~ . B

b. Is the figure necessarily

a plane f igure . Will

your proof apply If it

18. Given A3CD is a square and the poin ts K, L, M

H divide the sides as shown, a and b being

lengths of the indicated

segments.

Prove : KLMN is a square.

*19. Show that if ABCD is a parallelogram then D is in the i n t e r i o r of / A3C.

*20. Show that the diagonals of a parallelogram in tersect each

other .

9-7. Transversals - To Many Parallel Llnes. - Deflni t ions: If a transversal intersects two lines Ll, L2

in points A and 3, then we say that Ll and L2 intercept - the segment AB on t he transversal.

Suppose that we have given three Hnes Ll, L2, L3 and a t ransversal intersecttng them in poln ts A , B and C. If AB = BC,

then we say that the three line3 3mtercept congruent segments on the transversal.

We shall prove the following:

Theorem - 9-26. If three parallel l lnes intercept congruent segments on one tramversal, then they Intercept congruent seg-

ments on any o t h e r t ransvernal .

Proof : L e t Ll, L2 and L3 be pa ra l l e l lines, cut by a t r a n s v e r s a l TI in points A , B and C. Let T2 be another t ~ a n s v e r s a l , c u t t l n g these lines in D, E , and F. We have given tha t

AB = BC;

and w e need t o prove that DE = EF.

We w i l l first prove the theorem for the case in which Tl and T2 are - n o t p a r a l l e l , and A # D, as In-the figure:

Let T3 be the line through A, para l le l t o T2, htw- secting L2 and L3 in G and H; and l e t T4 be the l ine

through 3, paral le l to T2, intersecting L3 in I. L e t x, y, w and 1 z be as i n d k a t e d in the figure.

Statements Reasons

L x S L z . = BC.

9 11 T ~ - L w z L y .

ABQ BCI.

AG = BI.

AGED and BIFE are papal lelograms .

AG = DE a d BI = El?.

DE = EF.

1. Theorem 9-9. 2. Hypothes I s ,

3 . Theorem 9-11. 4. Theorem 9-9. 5. A . S . A .

6. Def in i t ion of congruent triangles.

7. Definition of parallelogram3.

8. Upposite s i d e s of a parallelogram are congruent.

9. Steps 6 and 8.

This proves the theorem for the case Ln which the two trans-

versals are not parallel, and intersect Ll In two different p o i n t s . The other cases are easy.

(1) If the two tmnsversals are paral le l , llke T2 and T3 in the f igwe, then the theorem holds, because opposite siaes of a paral le logrm are congruent. ( ~ h u s , if AG = I%, It follows that DE = EF.)

( 2 ) If the two transversals Lntersect at A , 1lke !TI and

T3 Ln the figure, then the theorem holds; In fact, we have already proved that If AB = BC, then AG = GH.

he fol~~wing coro l l a ry generalizes meorem 9-26.

Corollary 9-26-1. If three or more parallel line8 intercept congruent segments on one t r ansver~a l , then they in tercept congruent segments on any other transver~al.

That I s , given that

BIB2 = B2B3 = B3B4 = . . . , and so on, This follows by repeated applications of the theorem t h a t we have just proved.

DefInTtion: Two or more sets are concurrent if there l a a

po in t which belongs to a l l of the sets.

In particular, three or mom llms are concurrent if they a l l pass through one p o i n t .

The following theorem is an interesting application of

Co~ollary 9-26-1.

Theorem - 9-27, The medians of' a tr langle are concurrent in a point two-thirds the way from any vertex to the mid-point of the opposite side.

Given: In AABc, D, E and F are the mld-point3 - c of iz, zi and AB re-

spectively.

To Prove: There I s a p o i n t P - -

which lies on AD, BE and - 2 CF; and AF = VJ AD, A F B

Sketch of proof:

(1)

e L e t Ll, L2, L3, LQ and L with L3 = AD be five 5 '

parallel lines d i v t d b g i n t o four congruent segments, Then

(b) Ll, L2, L3, Lh divide BE into three congruent sewents, and so if P 1s the point of in te r sec t ion of - 2 AD and E, then BP = 3 BE.

In t he same way, w i t h l ines parallel to OF, we f ind that If PI 1s the i n t e r s e c t i o n of BE and &, then BPI =$BE.

( 3 ) From (1) and (2) and Theorem 2-4 it follows t h a t Pv = P, and therefore the three medians are concurrent.

( 4 ) Since we now know that passes through P we can easily 2

g e t CP = -7 CF from the figure in (1), and similarly ge t

AP = ' AD from the figure in (2 ) . 7 Definition: The centroid of a tr iangle I s the point of con-

currency of the medians.

Problem Set Q-7

a . Prove ZY = YX. <-Ã <-> <Ã‘

b. Do AC, TR and ZX have to be coplanar* to

carry out the proof?

The procedure at the right

can be used to rule a sheet

of paper, 3, I n t o columns of equal width. If A is

an ordinary sheet of ruled

paper and B is a second

sheet placed over I t as shown, explain why

3. Divide a given segment i n t o f ive congruent parts by t h e

following method: + -

(1) Draw ray AR (no t co l l inear with AB. ) - (2) Use your ruler t o mark o f f congruent segments AN,, - - - -

N1N2, N g M 3 3 N3N4 and N N of any convenient length.

4 5 - (3) Draw NcB.

( 4 ) Measure AN? and use your p r o t r a c t o r t o draw corre-

sponding angles congruent to / AN B with vertices a t 5

Explain why AB is divided into congruent parts.

4. The medians of AABC meet at Q, as shown in this

f igure.

If BF = 18, AQ = 10,

CM = 9, then BQ = J

GH = , CQ =

it-

In equilateral AABC if one median is 15 Inches long,

what I s the distance f r o m the centroid to A? To the mid- p o i n t of AB? TO side AC?

- Given: CM bisects at M. % bisects ?% a t P.

Prove: Q is a trisection point of E; t h a t is, AQ = 2QC.

( H i n t : On the ray opposite 4

to CB take po in t E such

t ha t CE = CB and show c

tha t % is contained in a median of &BE, )

What is the smallest number of congruent segments in to which AC can be divided by A

4 \ bR 8ome set of equally spaced parallels which will include the parallels IS, ?% and e8 CT if:

4 B a. A B = 2 and BC = I?

< \c @S

1 b. A3= +arid BC = I ? * T c. AB = 21 and BC = 67

e. A3 = y " and BC = I?

*8. Prove t h a t t he l i n e s through opposite vertices of a

parallelogram and the m i d p o i n t s of t h e opposite s i d e s

trisect a diagonal.

( ~ i n t : Through an extremity of the diagonal, consider a parallel to one of the lines .)

GLven: ABCD I s a parallelogram. X and Y are midpoints.

Prove: AT = TQ = QC.

Review Problems

1. Indica te whether each of the following statements I s true in ALL cases, true in SOME cases and fa l se in others, o r true in NO case, using the l e t t e r A , S o r N:

a Line segments In the same plane which have no p o i n t in common are parallel.

b. If two s ides of a quadrilateral ABCD are parallel,

then A3CD is a trapezoid.

c . Two angles in a plane which have their s i d e s respectively perpendicular are congruent.

d . If two parallel lines are c u t toy a transversal, then a pair of alternate exterior angles are congruent.

If two lines are c u t by a transversal, then the rays bisecting a pair of alternate interior angles are parallel.

In a plane, if a line is parallel to one of two parallel

l i n e s , it is parallel t o the other.

In a plane two lines are e i ther p a r a l l e l or they intersect . In a parallelogram the opposite angles are supplementary.

The diagonals of a rhombus bisect each other.

All three exterior angles of a triangle are acute.

A quadrilateral having two opposite angles which are right angles I s a rectangle.

The diagonals of a rhombus are congruent.

If a quadrilateral is equilateral, then a l l of its angles are congruent.

If two opposite sides of a quadrilateral are congruent and the o t h e r two sides are parallel, the quadrilateral I s a parallelogram.

The diagonals of a rhombus bisect the angles of the

rhombus.

If the diagonals of a parallelogram are perpendicular, the parallelogram is a square.

If a median to one side of a triangle is not an altitude, the other two sides are unequal in length.

E i t h e r diagonal of a parallelogram makes fcwo congruent

tr iangles with the s ides .

If a diagonal of a quadrilateral divides it Into two congruent triangles, the quadrilateral Is a parallelogram

If t w o lines are intersected by a transversal, the alternate interior angles are congruent.

u . A l l four sides of a rec tangle are congruent.

v . A l l four angles of a rhombus are congruent.

w . A square is a rhombus . x. A square is a rectangle.

2. Would the following information about a quadrilateral be sufficient t o prove it a parallelogram? A square? A

rhombus? A rectangle? Consider each item of Information separately.

I ts diagonals bisect each other . Its diagonals a re congruent . It is equilateral.

It is equi la te ra l and equiangular.

A diagonal bisec t s two angles.

Every two opposi te sides are congruent.

Some two consecutive sides are congruent and perpendicular.

The diagonals are perpendicular.

Every two opposite angles are congruent.

Each diagonal b isects two angles.

Every two consecutive angles are supplementary.

Every two consecutive sides are congruent.

and / B have the i r sides respectively parallel.

If only one pair of corresponding sides extend in the same direction the angles are

If corresponding sides extend in opposite directions, then the angles are .

In Problems 4, 5 and 6 below select the one word or phrase

that makes the statement true.

4. The bisectors of the opposite angles of a non-equilateral

parallelogram (a) coincide, (b) are perpendicular, ( c ) intersect but are not perpendicular, (d) are parallel.

5. The figure formed by Joining the consecutive mid-points of the sides of a rhombus I s (a) a rhombus, (b) a rectangle,

(c) a square, (d) none of these answers.

* 6 . The figure formed by joining the consecutive mid-points of the sides of quadrilateral ABCD is a square (a) if, and only if, the diagonals of ABCD are congruent and perpen-

dicular, (b) if, and only if, the diagonals o f ABCD are congruent, ( c ) I f , and only If, ABCD is a square, (d) if, and only if, the diagonals of ABCD are perpen-

dicu la r .

7 . In the left-hand column below, certain conditions are speci- f l ed . In the right-hand column, some deducible conclusions are l e f t for you t o complete.

diagona 1s of MKWR .

Conditions : Conclusions:

a . MKWR is a parallelogram, m/d = and ~ / R W K = m/a = 30. and

b. MKWR is a rectangle and m/d = and m/b = m/a = 30.

c . MKWR Is a rhombus, m/b = and BK = m/a = 30 and MK = 6 .

A

8. Given: In the figure A E = E B , G F = 8 ,

<-*Â <-Ã

CF = FB, DE 11 CB.

Find: DO. C F 8

9 . If the perimeter (sum of lengths of s i d e s ) of a triangle I s

18 inches, what is the perimeter of the t r iangle formed by Joining the mid-points of sides of the first tr iangle?

10. a. If m / A = 3 0 and C

m/ C = 25, w h a t is

t h e measure of / CBD?

b. If r n / A = a and a what is 0 / . c s 7 , >

m/ CBD? m/ ABC? A D

A 11. Show that the measure of

/ E, formed by the

bisector of / ABC and E the b i sec to r of exterior / ACD of A ABC, I s

1 equal to -y 4 A.

B D

< - ^ < - > + 12. In the figure AB 11 CD, EG 9 >B

bisects / BEP, m/ G = 90. 6 If the measure of / GEF = 25,

what is the measure of / GPD? D

4 Y

- - \F

13. Given: AB and CD which

bisect each o t h e r at 0. - - Prove: AC 1 BD.

C B

14. Given: ABCD i s a

parallelogram with - diagonals AC and - 1 DB. AP = RC < -y AC

Prove: DPBR is a

parallelogram.

15. Prove or disprove:

If a quadrilateral has one pair of parallel sides and one

pair of congruent sides, then the quadrilateral is a parallelogram. - -

*16. In AABC, median AM I s congruent t o MC. Prove that AABC I s a r igh t t r i angle .

17. Prove: If t h e b i sec to r s of two consecutive angles of a

parallelogram in te r sec t , they are perpendicular to each other .

Given: ABODE I s a <Ã *Ã

as shown. AE 11 CD. P is mid-point of

K is mid-point of

- Prove: KE bisects

A E

pentagon AE = CD. 7

AB . - BC . - PM .

19. When a beam of light is reflected from a smooth surface, the

angle between the incoming beam and the surface is congruent to the angle between the reflected beam and the surface.

In the accompanying figure, m/ ABC = 90, m / BCD = 75, and

the beam of light makes an angle of 35' with E. Copy

the figure and complete the path of the light beam as it

ref lects from E, from E, from E, and from AB again. A t what angle does the beam r e f l ec t from AB the second

time?

20. Given t~iangle ABC wlth and medians . If is

extended its own length to D, and CS is extended its o m length to F, prove that 3?, B and ll as shorn are

collinear.

Chapter 10

10-1. Parallel Planes.

Defini t ion: Two planes, or a plane aria a line, are paral lel if they do not i n t e r s e c t ,

If planes El arid E2 are parallel we w r i t e El 11 E2; if

line L and plane E are parallel we w r l t e L 1 1 E or I3 1 1 L. As we will soon see, parallels in space behave in somewhat the same way as parallel lines Ln a plane. To study them we do not

need any new postulates. However, in sp i t e of the similarities I t is neces saq , 3.n

studying theorem and t he i r proofs in t h i s chapter, to distinguish

carefully between parallel llnes and parallel planes. Two

parallel plane3 such as E and F in the f i r s t Figure below contain lines such as Ll and L2 which a m - n o t parallel And

the second Figure shows parallel l i ne s MI and M2 ly ing in

intersecting planes G and H.

The following theorem d e s c r l b e ~ a common situation in whlch parallel planes m d parallel lines occur in the same Figure .

Theorem - 10-1. If a plane Intersects two parallel planes,

then tt intersects them in two parallel lines .

Proof: Glven a plane E, in temect ing two parallel planes El and E ~ . By Postulate 8, the btersect~om are lines Ll and L2. These lines are In the same plane E; and they have no point In cornon because El and E2 have no poLnt in comon. There-

fore, they are parallel by the definition of parallel l ines.

Theorem - 10-2. If a line is perpendicular to one of two parallel planes, it is perpendicular to the other.

Proof: L e t planes El and E2 be parallel and let llne L be perpendicula~ to El. In E2 take a p i n t A no t on L, and

l e t l3 be the plane determined by L and A . By the preceding

theorem E in te rsec t s El and E2 in parallel lines Ll and

L2. L 1 Ll since L 1 Bl, and so by Theorem 9-12 (look it up)

L L2. Now take a point A1 In E2 not on L2 and repeat the process. We thus obtatn two llnes in E2 each perpendZcular to L, and so L 1 E2, by !?heorem 8-3.

Theorem - 10-3. Two planes perpendicular to the same line aye

parallel.

Proof: The figure on the l e f t

El 1 L at P and E2 1 L at Q: shows w h a t happens when

we wish t o show El 1 1 E2. If El and E2 are not parallel, they intersect . Let R be a - 4+ e common point. Consider the lines PR and a. Then L 1 PR a d

L 1 % because L i8 perpendlcula~ to every line in El through P and every line in E2 through Q. ThZs glves two perpendicul-

ars to a line from an eiternal point , which is impossible, by

Corol lary 10-3-1, If two planes are each parallel to a th ird

plane, they are parallel to each other.

Proof: L e t El 1 1 E3, E2 1 1 E3. Let L be a line perpendicular to E3. By Theorem 10-2 L 1 El and L 1 E2. Thus El

and E2 ape each perpendicular t o L and El 1 1 E2 by the

Theorem 10-3.

Theorem - 10-4. Two l ines perpendicular t o the same plane are parallel.

proof: By meorem 8-8 two such lines are coplanar* Since they are perpendicular to the given plane, say at points A and

B, they are perpendicula~ to s. Hence by Theorem 9-2 they aye

parallel.

Corol lary 10-4-1. A plane perpendicular to one of two parallel lines I s perpendicular to the other.

Proof: L e t Ll 1 1 L2, Ll

cular to E through any poln t 8-9. Then by Theorem 10-4 Ll

1 E m L e t L3 be a 1Lne perpendi- A of L2. L~ exists by Theorem

11 L3. Hence, by the Parallel

Postulate L3 = L2, and 80 L2 E.

Corollary 10-4-2. If two ltnes are each parallel to a t h i r d

they are parallel to each other.

Proof: Let Ll 1 1 L2, Ll [I La. Let E be a plane per-

pendicular t o Ll. By the above corolla^ E 1 L2 and E 1 L3, and so by the above theorem 11 L3.

Theorem 10-5. Two parallel planes are everywhere equidistant.

That is, a l l segments perpendicular to the two planes and having t he i r end points in the planes have the same length.

Proof; Let and be perpendicular segments between

t he p a r a l l e l planes E, and Ep. By Theorem 10-2, each of the segments Is perpendicular to each of the planes. By Theorem 10-4, Â¥fr PQ 11 %; and this means, In particular, tha t ?S and % lie

# *->Â i n the same plane ET By Theorem 10-1, QR 11 PS. Therefore,

PQRS is a parallelogram. Opposite sides of a parallelogram are congruent. Therefore, PQ = RS, which was t o be proved.

(obviously PQRS is a rectangle, but this f a c t does not need to be mentioned in the proof .)

Problem Set 10-1

1. D r a w a small sketch to illustrate the hypothesis of each of the following statements. Below each sketch indicate whether the statement is true or false.

If a line is perpendicular t o one of two parallel planes

it is perpendicular to the other.

Two l i n e s parallel to the same plane may be perpendicular to each other.

Two planes perpendicular to the same line may intersect.

If a plane intersects two i n t e r s e c t i n g planes, the l i n e s of intersection may be parallel.

If two planes a r e both perpendicular to each of two

parallel lines, the segments of the two lines in te rcepted between the planes are congruent.

If two planes, perpendicular to the same line, are in tersected by a th i rd plane, the lines of Intersection are parallel.

If a line lies in a plane, a perpendicular t o the line is perpendicular to the plane.

If a line lies in a plane, a perpendicular to the plane

at some point of the line is perpendicular to the line.

If two lines are parallel, every plane containing only one of them Is paral le l t o the other line. . . If two lines are parallel, every line Intersecting one of them in te rsec t s the other.

If two planes are parallel. any line in one of them is parallel to t h e other.

If two planes are parallel, any l ine I n one of them i s parallel to any line in the other.

2. Given lines L, and Ly intersecting parallel planes m, n, and p

at points A, B, C,

and X, Y, 2, w i t h

B the mid-point of z. Prove : XY = YZ .

3 . Given: plane s 11 plane r, - A3 1 r. OX = CY in

plane s . Prove : AX = AY.

4. Given: A, C in m;

Given: In the figure m 11 n,

Prove: AD = CB.

Planes E and F are <->

perpendicular to A , . <->Â <->

Lines BK and BH, in plane F, determine

<-Ã

with AB two planes which Intersect E in <-> Â¥(-

AD and A C . Certain lengths are given, as in the figure.

A r e BKDA and BACH parallelograms? Can you give a further description of them? Is ABHK Â \CAD? C a n you give the - length of CD?

In the figure half planes n and m have a common

<-Ã edge AB and i n t e r sec t

<

parallel planes s and t < Â ¥ Ã ˆ Â ¥ < - >

in l ines AD, AE, BG,

and BP as shown,

Prove that / DAE S / GBF.

Show how to determine a plane containing one of two skew lines and parallel to the other. Prove your construction.

Given: PL and PM lie - in plane E . RL 1 SM 1 MP. RL 11 SM.

- - Prove: RL 1 E, SM 1 E. (flint: A t P draw

10-2. Dihedral Angles, Perpendicular Planes.

We have considered perpendicularity between two l ines, and between a line and a plane. We have y e t to define perpendicular-

i t y between two planes. This can be done in various ways, and we choose the one tha t has the closest analogy with the definition of perpendicular lines.

Definitions: A dihedral angle is the union of a line and two non-coplanar half-planes having t h i s l i n e as t he i r common edge.

(Compare wlth the definitLon of angle in Chapter 4,) The line is called the e d ~ e of the dihedral angle. The union of the edge and either half-plane I s called a -3 face or - s i d e , of the dihedral angle.

<->Â If PQ is the edge, and A and B points on different

sides, we denote the dihedral angle by /A-PQ-B.

Analogous to the discussion on page 88 we see tha t two

intersecting planes determine four dihedral angles.

Terms such as v e r t i c a l , interior, exter ior , e tc . can be applied to dihedral angles. Definitions of these terms can be considered an exercise f o r the student.

To de f ine r i gh t dihedral angles, however, we need to ta lk

.-about the measure of a dihedral angle. One might a t first think

that we must introduce four new postu la tes , analogous to those In

Section 4-3. However, t h i s is not necessary, for we can r e l a t e each dihedral angle with an ordinary angle, as follows:

Definition: Through any point o n t h e edge of the dihedral

angle pass a plane perpendicular to the edge, intersecting each of the aides i n a ray. The angle formed by these rays is called a plane angle of the d ihedra l angle.

The sides of the plane angle are perpendicular t o the edge

o f the dihedral angle, so another way of defining the plane angle would be the angle formed by two rays, one in each s i d e o f the

dihedral angle, and perpendicular t o its edge at the same point.

It is natural a t t h i s p o i n t to use the measure of the plane

angle as a measure of the dihedra l angle, but before we do this we must prove tha t any two plane angles of a dihedral angle have

the same measure.

Theorem 10-6. Any two plane angles of a given dihedral angle - are congruent,

Figure A . Figure B .

Proof: L e t V and S be the vertices of two plane angles of /A-PQ-B. (~igure A . ) On the s ides of / V take points U

and W distinct f r o m V. On the sides of / S take poin ts R - and T such that SR = W, ST = W. (Figure B. ) VU and - are coplanar and perpendicular to PQ; hence they are parallel by Theorem 9-2. Hence by Theorem 9-20 (look it up) WRS I s a

parallelogram and UR = VS and UR 11 E. Similarly, WT = VS and 11 m. Hence UR = WT and 11 m, the latter f ac t following from Corollary 10-4-2. URTW is thus a parallelogram, and UW = RT. It follows f r o m the S .S .S . Theorem t h a t

A U W at A RST, and so rn/ UW = m/ RST.

Thus we can make the following definitions.

Defini t ions: The measure of a dihedral angle is the real number which is the measure of any of I t s plane angles. A dihedral angle is a right dihedral angle If its plane angles are ~ i g h t

angles. Two planes are perpendicular if they determine right dihedral angles.

The following are some immediate consequences of these def in i t ions . Their proofs are left as exercises.

Corollary 10-6-1. If a line is perpendicular to a plane, then any plane containing this line is perpendicular to the given plane.

Given: E, F contains <-> AB.

Prove: F 1 E. <-Ã <->

(~int: Take BC 1 PQ in

E.)

Corollary 10-6-2. If two planes are perpendicular, then any line in one of them perpendicular to t h e i r line of intersection,

is perpendicular to the o the r plane.

(~intt In the above figure; given F 1 E, AS 1 z; prove w El E. Take BC as before.)

Problem S e t 10-2 --

1 . ~ ; m e the 8 ix dihedral angles in t h i s three dimensional f igure .

2. Eachof G , ?k and

is perpendicular to the

other two . a = b = c = 45. What I s the measure of / C-PA-B? of / CAB?

3 . Draw a small sketch to i l lus t ra te the hypothesis of each of the following statements. Then indicate whether each is

True (1) or False (0).

If a plane and a line not in it are both perpendicular to

the same l i n e , they are parallel to each other.

If a plane and a line not in it are both parallel to the

same line they are parallel to each other.

If parallel planes E and F are cut by plane Q, the lines of in te r sec t ion are perpendicular.

I f two planes are paral le l t o the same line they are parallel to each other.

Two lines parallel to the same plane are para l le l to each other.

Segments of parallel lines intercepted between two parallel planes are congruent.

<-Ã If planes E and F are perpendicular to AB, then

they intersect In line %$. Two planes perpendicular to the same plane are paral lel

to each other.

Two l ines perpendicular to t he same line at the same

point are perpendicular t o each o t h e r .

A plane perpendicular t o one of two intersecting planes m u s t in te rsec t the other.

[sec. 10-21

k. If two intersecting planes are each perpendicular to a

th i rd plane, their line of intersection is perpendicular

t o the third plane.

Prove: If two intersect-

ing planes are each perpendicular to a third

plane, t he i r i n t e r s e c t i o n is perpendicular to that third plane. / x \

Y Given: Planes r and s

intersect in PQ (P being chosen f o r convenience on plane E) . r 1 E and

s 1 E. - Prove: QP 1 E. (I-iint: In plane E, draw 1 DC and -

fl 1 E, and use Corollary 10-6-2.) <-Ã <->

5. CD and FK are perpendlcular to plane E. Other t given information is as shown in the figure.

Which two segments have the same length?

' 6 . Prove the following theorem: If three planes El, Ey and E, intersect in pairs and determine three lines Li2, Ln

and Lgn, then ei ther the three lines are concurrent or each pair of the lines are parallel.

(~int; The figure shows E, and Eg meeting in Lip.

If En 11 LIP will the three

lines L *, L13 and Lgg, be concurrent or parallel? @ ,' Give proof. If En in te r - sects Lip In some point P will the three lines be con- Lie

current or parallel? Give

proof. )

-7. Desarppesr Theorem. If two triangles lying in non-parallel planes are such that the lines joining corresponding vertices are concurrent, then if corresponding side-lines intersect , their poin ts of in te r sec t ion are collinear.

Restatement. Given the triangles AABC and AA'B' C' in non-parallel planes such t h a t z* , st and CC! in te r sec t

<-> at U. L e t the lines ^S and i'B meet at X, CA and - <Ã‘ - C I A 1 meet at Y, and AB and AtB1 meet a t 2. Prove tha t the po in t s X, Y, 2 l i e on a line.

10-3. Projections . You are familiar with a slide pro jec to r which projects each

point of a slide onto a screen. Each figure in the s l ide is pro- jected as an enlarged figure on the screen. In t h i s sec t ion you

will no t i ce ce r ta in differences and cer ta in similarities between this familiar kind of projection and the kind of geometric pro- j ec t ion which is presented.

Definition: The projection -- of a point i n t o a plane is the

foot of the perpendicular f r o m the po in t t o the plane. (BY Theorem 8-10 t h i s perpendicular exists and is unique. )

In the f igure, Q is the projection of P into E.

Definition: The projec t ion --- of a line into a plane is the set of points which are p ro j ec t ions into the plane of the poin ts of the l i n e .

In the figure, P* is the projection of P , Qt is the projection of Q, and so on. It looks as if the pro jection of the

line I s a line; and in fact this is what always happens, except when the l i ne and the plane are perpendicular.

Theorem 10-7. The projection of a line into a plane is a - line, unless the l i n e and the plane are perpendicular.

Proof: L e t L be a line not perpendicular t o plane E .

Case 1. L lies in E. Then eachpointof L lies In E and is its own projection, (That is, a line through such a po in t P, perpendicular to E, In te r sec t s E in P. ) Thus, the pro-

jection of L is just L itself, and so is certainly a line.

Case 2. L does not lie in E. L e t P be a point of L

that i s not in E, l e t PI be the projecttion of P into E, and l e t F be the plane determined by the intersecting lines L and

EL I? and E have point P I In common, ane so, they In te r sec t in a line which we c a l l L1 . (pos tu la t e 8 . ) We want to show that L1 i s t h e pro jec t ion of L.

To do t h i s we must show two things:

(1) If R is a point of L, then i t s projection is a point of L' . This will show that the projection of L lies

on L t , but It will not assure us that the projection

of L constitutes all of Lt . To show the l a t t e r we - must prove

(2) If S t is any p o i n t of L* there is a point S of

L whose projection is S t .

We can prove these two p a r t s of Case 2 as fo l lows :

Proof of (1): If R = P, then R1 = Pt and so R1 l ies on Lt . So suppose R is different from P. Then EP*' and

^n?a are coplanar, by Theorem 8-8. Since F I s the only

plane containing P, R and PI ( (~ostulate 7.1, R r I s

in F . R* is also in E. Therefore R1 is on L', since '

L t , being the intersection of E and F, contains all - points common to E and P.

Proof of (2): If S Ã is any point of Lt, then the line - M through S t perpendicular to E I s coplanar with P P t

( o r coincides with it if S t = PI) and so lies in F . Therefore M in te rsec t s L (why?) at some po in t S .

S t is the projection of S. This completes the proof of Theorem 10-7.

If a line is perpendicular to a plane its projection i n t o

the plane is a single point.

The idea of projection can be defined more generally, f o r any set of po in t s . If A is any s e t of points, then the pro-

jection of A Into the plane E is simply the set of all pro- . jections of points of A . For example, the projection of a triangle is usually a triangle, although In certain exceptional cases

it may be a segment. Q

On the l e f t , the projection of APQR Is ASW. On the right,

the plane that contains APQR is perpendicular to E, so that - the projection of APQR is simply the segment ST.

Problem Set 10-3 --

1. Using the kind of projection explained in Sec t ion 10-3 answer the following:

Is the projection of a p o i n t always a point?

Is the projection of a segment always a segment?

Can the projection of an angle be a ray? A line?

An angle?

Can the projection of an acu te angle be an obtuse angle?

Is the projection of a right angle always a right angle?

Can the length of the projection of a segment be greater

than the length o f the segment?

If two segments are congruent will their projections be congruent?

If two lines Intersect can t he i r projections be two

parallel lines?

If two lines do not intersect can the i r projections be

two in tersect ing lines?

If two segments are paral le l and congruent, will t h e i r

projections be congruent?

3 . Given the figure with - not in plane m, XY the

projection of i n t o plane rn, M the mid- point of E, and N the

p r o j e c t ion of M, prove

N is the mid-point of E.

Front

2 Top View

In mechanical drawing the top view or "plan" of a solid may be considered the projection of the various segments of the s o l i d i n t o a horizontal plane m, as shown in perspective a t the left.

The t o p view as it would actual ly be drawn is shown at the right. (NO attempt is made here to give dimensions to the segments. )

a. Sketch a f r o n t view of the s o l i d shown above - tha t is, sketch the resu l t of projecting the segments of the s o l i d Into any plane parallel to its - front face.

b. Sketch the right side view of the solid.

5. The projection of a tetrahedron (triangular pyramid) into the

plane of i t s base may look like the figure at the right. How else may it appear?

- 6 , Given: BD is t he projection

U e s ~ p l a n e m ~ A B C /7/ -1 is a right angle .

Prove: / ABD l a a r igh t

angle.

( ~ i n t : L e t BE be perpendlcu- L___________/ lar to plane m. )

+ *7. Given: AQ Has projection + *

AR in plane m. AP is

any other ray from A in plane m. (~ofce: / QAR 1s called the m ~ l e that A$ makes with plane m.)

Prove: m/ QAR < m/ QAP ,

(~int: Let Q1 be the projection of Q i n t o m, On AP choose X so t h a t

AX = A Q 1 . - -

Draw QQ1 , Q f X and O X . ) '

*8. If t h e diagonal of a cube is perpendicular to a given plane, sketch the p r o j e c t i o n into the plane of all the edges of the cube. (NO proof required.)

Review Problems

Suppose / R-AB-S is an acu te dihedral angle w i t h

P a point on its edge. * 4 Can rays PX and PY be

chosen in the two faces so that

a. / XPY is acute?

b. / XPY is obtuse?

c . / XPY is r i g h t ?

2. Planes r and s Intersect Ã‡-

in TQ. 3 is a point between T and Q. i s i n r. - m/ TBA = 40. FB is in s .

m/ FBQ = 90. Is / ABF a

plane angle of dihedral

/ TQ? Can you determine m/ ABF? If SO, s ta te a theorem to support your conclusion.

3 . Planes x and r intersect In %. B is a point between + K and Q. BA i s ~ n r. "B? is in x. m/ ABK = 90.

m/ QBF = 90. Is / FBA a plane angle of dihedral

/ QK? If your answer is "yes", s t a t e a theorem or def in i t ion to support your conclusion. If m/ ABF = 80, is r l x ? If r l x , what

is m/ ABF?

4. Indicate whether each of the following statements is t rue in all cases ( A ) , tme in some cases md false in others (31, - or true in g caae (N).

a. Two lines parallel to the same plane are perpendicular to each other.

b. If a plane in te r sec t s each of t ~ o In tersect ing planes, the lines of in tersect ion are parallel.

c . If a line lies in a plane, a perpendicular to the line I s perpendicular to the plane.

If two planes are parallel, any line in one of them is parallel to the other.

If two planes are parallel t o the same line they are parallel to each other .

Two lines perpendicular t o the same line at the same point are perpendicular to each other.

If two intersecting planes are each perpendicular to a third plane, their line of in tersect ion is perpendicular to the th i rd plane.

The projection of a segment is a segment.

The projection of a right angle is a right angle.

Congruent segments have congruent projections.

Two lines are parallel if they are both perpendicular to the same line.

If a plane is perpendicular to each of two lines, the

two lines are coplanar.

If a plane Intersects two other planes In parallel l i n e s , then the two planes are parallel.

If a plane in te rsec t s the faces of a dihedral angle, the i n t e r s e c t i o n is c a l l e d a plane angle of t he dihedral

angle.

5. Given: F i s the projection of po in t A into plane E. - BH lies in plane E. / FBH is a right angle.

Prove: / ABH is a right

angle.

6. Given: Planes X, Y and

Z are parallel as shown, with in 2, and A

in X. AC cuts Y in B and cuts Y In D. A 3 = BC. AC = CE.

Prove: BD = BA.

7 . Given: R , 2, Y, X are t he mid-points of the re- A - - spec t ive s i d e s CB, BA, - - AD, DC of the non-planar

quadr i la tera l CBAD . Prove: RZYX is a paral le lo- gram.

*8. In the following incomplete statement it is possible to fill

in the solid blanks with "line" or "plane" and t he dotted

blanks w i t h I [ or 1 in eight ways so as to make the complet-

ed statement true: Give f i v e of these ways.

Given: ABCD is a paral lelo- - - gram. Each of E, BF, XY, - DH and are perpendicular

t o L, L is in the plane of parallelogram ABCD.

Prove: AE + CG = BF + DH*

Appendix I

A CONVENIENT SHORTHAND

There was a t i m e when algebra was all written out in words.

In words, you might state an algebra ic problem in the following way:

11 If you square a cer ta in number, add f ive times the number,

and then subtract s ix , the result is zero. What are possibilities for t h i s number?"

This problem can be more br ie f ly s ta ted in the following form:

' ~ i n d the r o o t s of the equation 2 + 5x - 6 = 0."

The n o t a t i o n of algebra I s a very convenient shorthand. A

similar shorthand has been Invented for talking about s e t s . It saves a lot of time and space, once you get used to It, and it is

a l l right t o use it In your written work, unless your teacher '

objec ts . Let us start with a picture, and say various things about i t

first in words and then in shorthand. L

2

Here we see a line L, which separates the plane E i n t o two half-planes H and H,,. Now l e t us say some things in t w o 1 ways.

In Words In Shorthand -

1. The segment PQ lie8 InHl.

2. The i n t e r s e c t i o n of and

L IS T.

1. m C H . 2. 1 0 L = T.

- The shorthand expression PQ C H, is pronounced in exactly

the same way as the expression on the left of It. In general, when we write A C B, t h i s means that the set A lies in the

set B. ; An expression of the type A n B denotes the intersection of

the sets A and B. The symbol " 0 '' 1 i s pronounced "cap," be- 1 -

cause I t looks a little like a cap. Notice that the sets PQ and 1 - RS do not intersect. If we agree to write 0 f o r the empty set, 1

I then we can express this fact by writing -

F Q ~ R S = 0 .

Similarly,

p^n L = O

and

mfl H, = 0.

Of course, a is a set which lies In H,. But the point P

above I s a member of H,. We write t h i s In shorthand l i k e t h i s

P C : H,. Thi s I s pronounced "P belongs to Hi."

The union of t w o sets A and B is written as A U B. This

is pronounced " A cup B." In the same way, we write A U 3 U C f o r the union of three s e t s . For example, in the figure above, the

plane E I s the union of H,, Hp and L. We can therefore write

E = H , U HpU L. Notice t h a t here (as everywhere else), a formula involving

the sign "=" means that the things on the left and r i gh t of "="

are the same thing. The sign "=I1 i s simply an abbreviation of the

word "is", as in the expression 2 + 2 = 4, which says that two

plus t w o is four.

Problem -- Set I

Consider the sets, A, B, C , and so on, def ined in the foilow-

Ing way:

A I s the set of all doctors. B I s the s e t of a l l lawyers. C is t he set of a l l tall people.

D is t h e s e t of all people who can play the v i o l i n . E is the set of all people who make a l o t of money. F is the set of a l l basketball players. Write shorthand expressions for t h e following statements:

1. All basketball players are tall. 2. No doctor is a lawyer.

3. No violinist makes a l o t of money, unless he is tall.

4. No basketba l l player I s a violinist, 5. Everyone who is both a doctor and a lawyer can also play

the violin.

6. Every basketball player who can play the violin makes a

lot of money. 7. The man X is a tall violinist.

8. The man Y is a prosperous lawyer. 9. The man Z is a tall basketball player.

Appendix I1 POSTULATES OF ADDITION AND MULTIPLICATION

The methods of manipulating real numbers by means of the oper- ation of addition and mu1 tiplication, and the related opera t ions of subtract ion and d iv i s ion , are a l l determined by the following eleven pos tu l a t e s . In t h e statement of these postulates and t h e

proofs of the following theorems It is to be understood that all

the l e t t e r s are real numbers. A - 1 . (closure under Addition.) x + y is always a real

number. A - 2 . (~ssociative Law for ~ddition.) x + (y + z ) = ( x + y)+z.

A - 3 . (cornmutat i ve Law f o r Addition. ) x + y = y + x. A - 4 . (Existence of 0. ) There is a unique number 0 such

that x + 0 = x f o r every x .

A-5. (Existence of ~ e g a t i v e s . ) For each x there is a unique number -x such tha t x + ( -x) Â¥ 0.

M - 1 . (Closure under ~ulti~lication.) xy is always a real

number.

M-2. ( ~ s s o c i a t i v e Law f o r ~ultiplication. ) x(yz) = (xy)z.

M-3. (commutative Law for Mult ip l ica t ion . ) xy = yx.

M-4. (~xistence of 1. ) There is a unique number 1 such

t h a t xal == x for every x. M-5. (~xistence of Reciprocals. ) For each number x other

1 l such that x-- = 1. than 0 there is a unique number 2 x D. (Dis tr ibut ive Law. ) x(y + z) = xy + xz.

The following basic theorems will illustrate how these pos-

tu la tes are used in simple cases .

Theorem 11-1. If b = -a,then -b = a. Proof: By A - 5 , b = -a means the same as a + b = 0. By

A-3 t h i s is the same as b + a = 0. By A-5 , this is the same as a = - b .

Another way of stating this theorem is that -(-a) = a.

Theorem 11-2. For any a , a-0 = 0.

Proof : a = a g l (M-41

= a ( l + 0) ( A - 4 ) = a - 1 + a.0

= a + a-0 04 -4 ) Hence by (A-41, a - 0 = 0.

Theorem 11-3. a(-b) = -(ab).

Proof:

ab + a(-b) = a [ b + (-b)] ( D l = a.0 (A-5) = 0 ( ~ h . 11-2)

Hence by A-4, a ( - b ) = -(ab). A s a special case of this theorem we have a(-1) = - a .

Def in i t i on . x - y shall mean x -I- (-y). Note that by this

definition a - a = 0.

Theorem 11-4. if a + b = c , then a = c - b. Proof: If a + b = c , t h e n

(a + b) + (-b) = c + (-b) ( a + b) + (-b) = a + [b + (-b)] (A-2)

a + 0 - (A-51 = a (A-4)

Hence a = c I- (-b) = c - b by definition.

Theorem 11-5. If ab = 0, then e i t h e r a Â¥Â 0 or b = 0.

Proof: To prove the theorem It will be enough to show that 1 If a # 0 then b = 0. So suppose a # 0. Then g e x i s t s , by

M-5. Therefore, 1 1 - (ah) = -*o = o a a ( ~ h . A-11-2)

also, 1 1 - (ab) = (Ã‘a b a (M-2)

= 1 - b 04-51

Therefore b = 0.

Theorem 11-6. (Cancellation Law.) If ab = ac and a # 0 then b = c .

Proof: If ab = ac then ab - ac = 0, By Theorem A-11-3

this i s the same as ab + a ( - c ) = 0, or, by D, as a ( b - c ) = 0.

Since a # 0 we get , by applying Theorem 11-5, that b - c = 0.

Hence b = c .

These are j u s t a few examples of t h e use of the postu-lates in proving basic algebraic theorems. Ordinarily we don't use the

postulates d i r ec t l y but make use of such properties as those stated In Theorems 11-4 and 11-6 in our algebraic work.

Problem Set I1 -7

1. Prove each of the following theorems. a . (-a) (-b) = ab.

b. a(b-c) = ab - ac.

c . If a - b = c , then a = b + c .

d. (a + b) (c + d) = ac + ad + be + bd. ( l i int: As a first

step apply D, regarding (a + b) as a single number. ) 2. Given the d e f i n i t i o n s :

2 x = X'X ,

prove that 3 ( a + b12 = a2 + 2ab + b .

2 2 3. Prove: (a + b) (a-b) = a - b .

-1 4. Defini t ion: = ab . Prove each of the following:

-1 a , (ab) = a -y1

a c ac b. - . -=- b d bd- a ac c . x==.

-1 a. ( - a ) 1 = - ( a ) . a -a a e. -5 = = -5 .

a + c f. b to b a c ad + be

g. 5 + g = bd

Appendix 111

NATIONAL AND IRRATIONAL NUMBERS

111-1. ----- How t o Show That a Number & Rational. By d e f i n i t i o n a number is rational if it i s the ra t io of two

integers. Therefore, it we want to prove that a number x is

rational, we have to produce two integers p and q, such that P = q

x. Here are some examples: (1) The number* x = - + Is rational, because

Therefore x = Â¥ where p = 13 amd q = 14. Q

(2) The number x = 1.23 is rational, because 123 1.23 = - 100'

which I s the ratio o f t he two integers 123 and 100.

(3) If t h e number x I s rational, then so is the number 2x. h hat is, twice a rational number is always rational.) For if

where p and q are Integers, then

-9, 2x - q

where the numerator 2p and t h e denominator q are both integers. ( 4 ) If the number x I s rational, then so is the number

2 x + 3. For if

then

where the numerator and denominator are both Integers. .. 5 ) If x is a rational number, then so I s xz + x. For if

P x = - , q

then

where the numerator and denominator are integers.


1. Show that .Â£35 Is a r a t i o n a l number. 2. Show that -7 + - 5 is ~ational. 7 3 . Show that if x is a rational number, then so I s x - 5. 4. Show that If x is rational, then so is 2x - 7.

is rational. 5. Show that 7 + - 17 6. Show t h a t the sum of two rational numbers is a

rational number. 17 23 7. Show that (-) ( ) I s rational. T7-

8. Show that t h e product of two ratlonal numbers is a

rational number.

9. show that - 23 1s ~at~onal. 17 +-? 10. Show t h a t the qu&ient of two rational numbers is a

rational number, as long as t h e d i v i s o r i s not zero. 11. Given t h a t i/^T is irrational, show t h a t t 1s also

I r r a t i ona l . (Hin t : This problem is a lot easier, now

that you understand about Indirect proofs . ) 12. Given that T is irrational, show that Is also

5 irrational.

13. Show that t h e reciprocal of every rational number d i f - ferent from zero is rational.

1 4 . Show that the reciprocal of every irrational number d i f - ferent from zero Is irrational.

15. Is it true t h a t the sum of a rational number and an ir-

rational number is always irrational? Why or why n o t ? 16. Is it true tha t the sum of two Irrational numbers is

always Irrational? Why or why not? 17. How about the product o f a rational number and an ir-

rational number?

111-2. Some Ewnples & Irrational N m b e ~ s .

I n the previous section, we proved that under certain condi- t i o n s a number must be rational. In some of the problems, you showed t h a t starting -- with an Irrational number we could get more irrational numbers in various ways. In all this we l e f t one very important question unsettled: are there any Irrational numbers?

We shall settle t h i s question by showing that a part icular number, namely-^/^ cannot be expressed as the ra t io of any two in tegers .

To prove this, we f irs t need to establish some of the facts about squares of odd and even Integers. Every integer I s e i ther

even or odd. If n is even, then n is twice some integer k, and we can write

n = 2k.

If n is odd, then when we div ide by 2 we get a quotient k and a remainder 1, so that

Therefore, we can write n = 2k + 1.

These are t he typical formulas for even numbers and odd numbers ,

respectively. For example,

6 = 2 - 3 n = 6 , k = 3 7 = 2-3 + 1 n = 7 , k = 3 8 = 2.4 n a 8 , k = 4

and so on. The fol lowing theorem is easy t o prove:

Theorem 111-1. The square of every odd number Is odd.

Proof: If n I s oda, then we can write n = 2k + 1,

where k is an Integer. Squaring both sides, we get 2 n = ( ~ k ) ~ + 2'21t + 1

2 = 4k + 4k 4- 1.

The right-hand side must be odd, because it I s written I n the form 2'[2k2 + 2k] + 1;

that l a , It Is twice an integer, plus 1. Therefore, n I s odd, I which was to be proved, I

From Theorem 111-1 we can quickly get another theorem:

Theorem 111-2. If n2 I s even, then n is even. Proof: If n were odd, then n2 would be odd, which is

fa lse . Therefore n I s even.

Notice that t h i s is an ind i rec t proof. We are now ready to begin the proof of Theorem 111 -3. f i is i r r a t i o n a l . Proof: The proof will be indirect. We begin by making the

assumption that i/^T I s rational. We w i l l show that this leads to a contradiction.

Step 1. Supposing that ./2" is rational, it follows that 2

can be expressed as

Â¥wher the fraction ^ is In lowest terms. q

The reason is that if i/T can be expressed aa a fraction a t

all, then we can reduce the fraction to lowest terms by div id ing

out any common fac tors of the numerator and denominator. We therefore have

in lowest terms. T h i s g ives P* 2 = - 0 )

which in turn gives

Step 2. p2 is even. Because p2 is twice an Integer. Step 3 . p I s even.

By Theorem 111-2. We therefore set p = 2k. Substituting In the formula at the

end of Step 1, we get 2 2

(2k) = 2q , which means t h a t

4k2 = za2 . Theref ore

2 q 2 = 2 k .

s tep . IS even.

Because Is twice an Integer.

Step 5. q is even.

By Theorem 111-2. We started by assuming t h a t */^ was ra t ional . From this we

got #/T= 2 in lowest terms. From t h i s we have proved t h a t p q '

and q were both even. Therefore was not in lowest terms, q

after a l l . This contradiction shows that our i n i t i a l assumption

must have been wrong, that Is, i/^T must not be rational.

Problem Set 111-2 -- These problems are harder than most of the problems in the

t e x t .

1. Adapt the proof that 6 is irrational, so as to ge t a

proof that 1/3" is Irrational. (!lint : Start with t h e fac t t h a t

every integer has one of the forms n = 3k

n = 3 k + 1 n = 3k + 2,

and then prove a theorem corresponding to Theorem 111-2.) 2, Obviously nobody can prove that f i is irrational, be-

cause 6 a: 2. If you try to "prove" this by adapting the proof

f o r /?, at what po in t does the "proof" break down? 3 . Show that 'fi I s Irrational.

Actually, t h e square r o o t of an in teger is e i t h e r another

integer or an irrational number; t h a t is, */n' e i ther "comes out very e v e n or "comes o u t very uneven.'' The proof of this fac t ,

however, requires more mathematical technique than we now have at our disposal. Problems like this are solved In a branch of mathe n i a t l c s called the Theory of Numbers.

Appendix IV SQUARES AND SQUARE ROOTS

Everybody knows what I t means to square a number: you multiply the number by itself. The facts about square roots, however, are considerably trickier, and the language in which most people talk about them is very confusing. Here we will try to state the f a c t s and poin t out t he p i t - f a l l s .

To say that x is a square root of a means t h a t -- 2 x = a.

For example, 2 Is a square root of 4,

3 is a square root of 9, -2 is a square root of 4, -3 i s a square roo t of 9,

and so on. You may wonder why we d i d not abbreviate these statements by using radical signs. The reason (as w e s h a l l soon see)

is that radical signs mean something s l i g h t l y different.

The following l a a fundamental fact about the real number system:

I

Every posi t ive number has exactly one p o s i t i v e square r o o t .

For example, 22 = 4, and no other positive number I s a root

of the equation x2 = 4. 4 = 16, and no other pos i t ive number I s a root of the equation x2 = 16. And so on.

Of course, if x I s a square root of a, then so is -x, 2 2 because ( - x ) = x . Therefore every positive number has exactly

two square roots, one positive and the o t h e r nega t ive . The meaning of the radical sign I s defined t h i s way:

If - a is pos i t ive , t h e n f i denotes the pos i t ive square root of - a.

We provide further that -/O = 0.

For example,

^ = 2,

8 = 3, ^ = 4, and so on. To i n d i c a t e the other square root -- t h a t is, the

negative one -- we simply put a minus sign In front of the radical sign. For example:

4 has two square roots , 2 and -2.

3 has two square roots, ,/3" and - a. 7 has two square roo t s , ./T and - Ã /T

The following two statements look alike, but i n f a c t they are

different: (1) x is a square root of a. (2) x =/S-.

The first statement means

statement means not only that Therefore the second statement

the first.

2 merely tha t x = a. The second 2 x = a, but a l so that x - > 0.

I s not simply a short-hand form of

L e t us now investigate the expression-i/x, where x is not

equal to zero. There are t w o possibilities: 2 I. If x > 0, then x is the positive square root of x ,

and we can write

11. If x < 0, then and I t is -x that is the

f o r x < 0, we have

2 x is the negative square root of x , 2 pos i t ive square root of x . Therefore,

^s -X. looks so appealing that I t seems almost

l i k e a law of nature. In fac t , however, this equation holds true only half of the time: it I s always true when x 2 O? and I t is ---- never true when x < 0.

Fitting together cases I and 11, we see that f o r every x without exception we have

J?- 1x1- To see tills, you should check it against the definition of 1 x 1 , in

Sect ion 2-3.

Problem -- Set IV

Which of the following statements are true? Why or why no t? 1. ^9 = 3 .

2. = -3 . 4- 3 . = - 1.414. (~pproxirnately.)

4 . r 2 = 1.414. (~pproximat ely . ) 5. = 2 5. 6. y e 5 = 5.

For what values of the unknowns (if any) do the following equations hold true? Why?

7. .(--= a - 1.

9 10. ,/X= -1x - 11. 11. ./(X = (x 4- 312

2 12. J+ = --(x + 3) 13. ./(x = 3)" (x + 3 I 2 l 14. ,/(xm= - l(x + 3 1 ~ 1 .

Appendix V HOW TO DRAW FIGURES IN 3-SPACE

V-1 . Simple Drawae. A course in mechanical drawing is concerned w i t h precise rep-

resentat lon of physical objects seen from d i f f e r e n t positions in space. In geometry we are concerned with drawing only to the ex- t e n t t h a t we use sketches t o help us do mathematical thinking. There I s no one correct way to draw pictures I n geometry, but there are some t echn iques h e l p f u l enough t o be In rather general

use. Here, f o r example, I s a technically correct drawing of an ordinary pyramid,

f o r a person can argue t h a t he is looking

at t he pyramid from d i r e c t l y above. Bat ca re fu l ruler drawing is not as helpful

a s this very crude free-hand sketch. The

first drawing does not suggest 3-space;

â‚¬ the second one does.

/q) The f i rs t part of t h i s discussion offers suggestions f o r

simple ways to draw 3-space figures. The second part in t roduces the more elaborate technique o f drawing from perspective. The

difference between the two approaches is suggested by these two

drawings of a rectangular box.

In t he f i rs t drawing the base is shown by an easy-to-draw para l l -

elogram. In the second drawing, the f ron t base edge and t h e back

base edge are parallel, but the back base edge is drawn shor te r

under the belief t h a t the shorter length will suggest "more remote".

No matter how a rectangular box is drawn, some sacrif ices must be made. All angles of a rectangular s o l i d are right angles, but in each of the drawings shown above two-thirds of the angles do not come c lose t o Indica t ing ninety degrees when measured w i t h

a protractor. We are willing to give up the drawing of r i g h t

angles t h a t look like r i gh t angles in order t h a t we make the f igure as a whole more suggestive.

You already know t h a t a plane is generally pictured by a

parallelogram.

It seems reasonable to draw a horizontal plane In e i t h e r L 3 n of t h e ways shown, and to draw a ve r t i ca l plane like t h i s . 0 If we want t o indicate two parallel planes, however, we can not be ef fec t ive if we just draw any two "hor izonta l" planes. Notice how t h e drawing t o t h e right below Improves upon the one t o the l e f t . Perhaps you p r e f e r s t i l l another kind of drawing.

Various devices are used t o Indicate t h a t one part of a f ig -

ure passes behind another part . Sometimes a hidden part Is simply omitted, sometimes it is indicated by dotted l i n e s . Thus

a line piercing a plane may be drawn i n e i ther of the t w o ways:

Two intersecting planes are I l lus t ra ted by each of these drawings.

The second I s better than the first because the line of In tersection is shown and parts concealed from view are dotted. The

t h i r d and f o u r t h drawings are better y e t because t h e l i ne of intersection I s v i sua l ly t i e d i n with plane P as well as plane Q by the use of parallel l ines In the drawing. Here is a drawing which has the advantage

A

of simplicity and t he disadvantage of suggesting one plane and one half-plane.

In any case a line of intersection is a par t i cu la r ly important part of a figure.

Suppose that we wish to draw two Intersecting planes each

perpendicular t o a th ird plane. An effective procedure is shown

by this step-by-step development.

Notice how the l a s t two planes drawn are b u i l t on the l i n e of In t e r sec t ion . A complete drawing showing a l l the hidden lines I s

just too Involved t o handle pleasantly. The picture below is much more suggestive.

1

I 1

A dime, from d i f f e r en t angles, looks l i k e this:

0 Qa Neither the f i r s t nor the last is a good p i c t u ~ e of a c i r c l e fn

3-space. E i t h e r of the o t h e ~ s is satisfactory. The thinner oval is perhaps better to use to represent the base of a cone,

C e ~ t a i n l y nobody should expect us to i n t e r p r e t the f igure shown

A few addL tional drawings, w i t h verbal d e s c r i p t i o n s , ape

shown.

A line parallel t o a plane.

A cylinder cut by a plane

parallel to t he base,

A cy l inde r c u t by a plane

not parallel to t h e base.

A pyrmld cut by a plane parallel t o t h e base.

It is important to remember that a drawing is not an end in

I tself but simply an aid to our understanding of the geometrical

sttuation. We should choose the kind of p i c t u r e that wlll serve us best f o r this purpose, and one personts chofce may be d i f f e r e n t

from another.

V-2. Perspec t ive . The rays a? b? c, d, e , f In the left-hand f i g u ~ e below

suggest coplanar l i n e s intersecting at V; the corresponding ray6

in the right-hmd Xigure suggest parallel lines in a three-dimensional drawing. Thlnk of a railroad t rack and telephone poles as you look at the rig11-k-hand figure

The right -hand figure suggests certain principles which are useful in making perspective drawings.

1) A s e t of pa~allel line^ which recede from t h e viewer are d~awn as concurrent m y s ; f o ~ example, rays a, b, c, d, e , f. The point , on the drawing? where the rays met is h o r n as the

" vanishing point" . 2) Congruent segments are dram smaller when they are

farther from t he vLewer. (Find examples 5n the drawing. )

3 ) Parallel lines which are perpendicular t o t h e l i n e of s i g h t of the viewer are shown as parallel l i n e s in t h e dra&.ng.

(FLnd examples An t he drawing. A pepson does no t need much a~tistlc ability to make use of these t hme principles.

The steps t o follow in sketching a rectangular s o l i d are ~ h o w n below.

m Draw the f ron t face as a rectangle*

Select a vanishing point and draw segments *om Jt to t h e v e r t i c e s . Omit segments t h a t cannot be seen.

Draw edges parallel t o those of the front face. Finally erase l i nes of perspec t ive .

Under this technique a single horizontal plane can be dram

as the t o p face of the s o l i d shown above.

A single vertical plane can be pepresented by the f ~ o n t face or the right-hand face

of the solid.

After t h i s brLef account o f two approaches t o t he drawing of figures in 3-space we should once again recognize the fac t t h a t

there is no one correct way to picture geometr5c ideas. However, t h e more "rezl" we want our picture to appear, t h e more a t t e n t i o n we should pay to perspective. Such an artist as k o n a ~ d o & V l n c i paid great attention to perspective. Most of us find this done fop us when we use ordinary cameras.

See some books on drawlng or look up "perspect ivet t in an en- cyclopedia i f you are in teres ted T n a d e t a l l e d treatment.

Appendix VI

PROOFS OF THEOREMS ON PERFENDICUURITY

In Section 8-3 two themems are s t a t e d , which, between them, cover all cases of existence and uniqueness involved In the perpendicularity of a l i n e and a plane. As s t a t e d there , e igh t

separate i tems must be proved to establish the p ~ o o f 8 of these

two theorems. Here we will s t a t e these items and prove those which have not already been proved.

We first r e s t a t e the two theorems.

Theo~ern u. Through a glven point there is one and only

one plane perpendicular to a given - line.

Theorem - 8-10. Through a given po in t there is one and only one line perpendicular to a given plane. -

We now consider the eight proofs , in a systematic order .

Read the statements carefully, for there are only slight differ-

ences in t h e i r wording: the presence or absence of' a "not", the substitution of "mo8t1' fo r "leastt', o r the interchange of "line"

and l1p1ane'l . Theorem V I - 1 . - Through a given point on a given l i n e there is

D a t least one plane pwpendicular to the line.

t This is meorem 0-4, which i s proved i n the text . I

Theorem VI-2, Through a given p o i n t on a glven l i n e there is a t m o s t one plane perpendicular to the l ine .

This is Theorem 8-6, which I s proved in the text .

meorern - VI-3. Through a given po in t not on a given l i n e there is at least one plane perpendicular to the given line-

Given: Line L and point P n o t on L.

To prove: There is a plane E through P, w l t h I3 1 L,

There 1s a line M through F perpendicular t o L (meorem 6-4) . ~t M and L i n t e r s ec t a t Q, and l i e In the plane F [Theorem 3-41.

There i~ a point R (~igure 2 1 not in F ( postulate 5b) , Let G be the plane conhlning L and R (meorem 3-31.

In G there 1s a llne N perpendicular to L at Q

( n e o ~ r n 6-11.

L e t E be the plane containing M and N. Then E 1 I, by Theorem 0-3,

Theorem VI-4. - Through a glven point not on a given llne there

1s at most one plane perpendicular t o the given line.

Proof: Suppose that there are two planes El and E2, each

perpendicular t o line L and each containing P. 1f El

and E2 intemect L in the same p o i n t Q, we have two planes perpendicular to L at Q, m d thls contradicts Theorem VI-2.

On the other hand, if El and Ep intersect L In dis t inct <Ã‘ @

points A and B, then PA and PB are d i s t i n c t l ines through

P perpendicular to L, contradicting Theorem 6-4. E i t h e r way, we g e t a contradiction, and so we cannot have two planes through P perpendicular to L.

T h i s finishes the proof of Theorem 8-9. The next four

theorems, which read like the prev ious four w i t h "line" and

\plane" Interchanged, will prove Theorem 8-10.

Theorem VI-5. - Through a given point in a given plane there is at least one line perpendicular to the plane.

A

Proof: L e t P be a p o i n t in plane E. By Postulate 5a there is another point Q i n E. Let plane F be perpendicular

<-Ã to PQ at P (?Theorem VI-1) .

Since F intersects E ( a t P) t h e i r i n t e r s e c t i o n is a l i ne M, by Postulate 8. L e t L be a line in F, perpendicular

to M (Theorem 6-1) . Since F 1 z, and L lies in F and contains P, we

have, from the definition of a line perpendicular to a plane, that <6Ã‘

L 1 PQ. A l s o , from above, L _^ M. Hence L 1 E, by Theorem 8-4.

Theorem VI-6. Through a given poin t in a given plane there - is a t most one line perpendicular to the given plane.

Proof: Suppose L-, and L2 are distinct lines, each perpendicular to plane E at point P. Li and L2 determine a

plane F h he or em 3-4) which intersects E in a line L. In F, we then have two perpendiculars t o L a t the same po in t P i con-

tradicting Theorem 6-1.

Theorem VI-7. Through a glven pomt not a glven plane there is at least one line perpendicular t o the given plane.

Proof: L e t P be a point not in plane E. Let A be any point of E, and M a line through A perpendicular to E (!Theorem VI-5).

If M contains P I t is the desired perpendicular. If M does not contain P l e t F be the plane containing

M and P h he or ern 3-31, and N the line of Intersection of F

and E. In F l e t 3 be the foot of a perpendicular f r o m F to N heo or ern 6-4).

L e t line L be perpendicular to E at 3 (Theorem VI-5) . By Theorem 8-8, L and M are coplanar, and hence, L lies in

P since M and B detennlne F. In F, L 1 N, s ince L 1 E and N lies in E, Since by

Theorem 6-1 there is only one line in F perpendicular to N a t w

3, L and BP must coincide. That l a , L contains P and so

is the desired perpendicular.

Theorem VI-8. Through a given point not in a given plane there I s at most one line perpendicular to the given plane.

The proof is word f o r word the same as that of Theorem VI-6, except f o r the replacement of "at point P" by "from p o i n t P" and of h he or em 6-1" by "!Theorem 6-3".

The Meaning and Use of Symbols - General.

= A = B can be read as "A equals 3 , "A Is equal to Bmj

"A equal B" (as In "Let A = B") , and possibly other ways to fit the structure of the sentence In which the symbol appears. However, we should not use the symbol,

, I n such forms as "A and B are ="; its proper use is between - t w o expressions. If two expressions are connected

by n= I t , I t is to be understood that these two expressions

stand for the same mathematical ent i ty , In our case either a real number or a point set.

/. " ~ o t equal ton. A # B means that A and B do - n o t represent the same entity. The same variations and cautions apply to the use of # as to the use of =.

Algebraic.

+, -, -, +. These familiar algebraic symbols for operating with real numbers need no comment. The basic postulates about them are presented in Appendix 11.

<, >, <, 2. Like =, these can be read in various ways In

sentences, and A < B may stand for the underlined part of " ~ f A I s less than B", "Let A be less than B", "A less than B implies ", e t c . Similarly for the other three symbols, read "greater than", "less than or equal to", "greater than or equal to". These Inequalities apply

only t o real numbers. Their properties are mentioned brief ly in Section 2-2, and In more detail I n Section 7-2.

n 6, [ A [ . Square root of An and "absolute value of A".

Discussed in Sections 2-2 and 2-3 and Appendix IV.

Geometric.

P o i n t Sets . A single l e t t e r may stand f o r any suitably described p o i n t s e t . Thus we may speak of a p o i n t P, a l i n e m, a half-plane H, a circle C, an angle x , a segment b, e t c .

^Â¤ The line containing the two points A and B ( P . 30). The segment having A and 3 as end-points ( P . 45).

"51?. The ray with A as i t s e n d - p o i n t and conta in ing p o i n t B (P. 45).

/ABC. The angle having B as vertex and B? and 57 as

sides (P. 71). AABC. The t r i ang le having A, B and C as vertices (P. 72) . / A-BC-D. The dihedral angle having ^SS as edge and with sides

con t a in ing A and D (P. 299).

Real Numbers. 7

AB. The p o s i t i v e number which is the distance between the two

p o i n t s A and B, and also the length of the segment (P. 34) .

~[ABC. The real number between 0 and 180 which I s the

degree measure of /ABC (P. 80). Area R. The p o s i t i v e number which is the area of the polygonal

region R (P. 320).

Relations. f\* . Congruence. A $ B I s read "A Is congruent to B", but

w i t h the same possible variations and restr ict ions as A = B. In the t e x t A and 33 may be two (not necessarily

different) segments ( P. log), angles ( P . l og ) , or triangles (P. 111).

( Perpendicular . A _^ B is read "A I s perpendicular to B", with t h e same comment as for s. A and B may be either two

1 ines ( P . 861, two planes (P. 301 ) , or a line and a plane (P. 219).

1 1 . Parallel. A 1 1 B Is read "A is parallel t o B", with the same comment as f o r 2. A and 3 may be either two l i n e s

(P. 241), two planes (P. 291) or a l i n e and a plane

(P. 291).

L i s t of Postulates

Postulate 1. (P. 30) Given any t w o d i f f e r e n t points, there is exactly one l i n e which conta ins both o f them.

Postulate 2. ( P . 3b) h he Distance postulate .) To every

pair of different points there corresponds a unique p o s i t i v e

number. Postulate 3. (P. 36) (The Ruler postulate .) The p o i n t s

of a l i n e can be p l a c e d in correspondence w i t h the real numbers

in such a way that

(1) To every point of the l i n e there corresponds exactly one real number,

(2) To every real number there corresponds exactly one p o i n t of the l i n e , and

( 3 ) The distance between two points Is the absolute value of t h e difference of the corresponding numbers.

Postulate 4. (F. 40) h he Ruler Placement ~ostulate .) Given two p o i n t s P and Q of a l i n e , the coordinate system can be chosen in such a way t h a t the coordinate of P i s zero

and the coordinate of Q I s p o s i t i v e . Postulate 5. (P. 54) (a) Every p l a n e confcalns at least

three non-colllnear points.

(b) Space conta ins at least four non-coplanar points . Postulate 6. (P. 56) if two p o i n t s l i e In a plane, t hen

the line containing these points l i e s In the same p l a c e .

Postulate 7- (P. 57) Any three points l i e In at least one plane , and any three non-collinear points lie In e x a c t l y one plane. More brief ly , any three p o i n t s are coplanar, and any

three non-collinear p o i n t s determine a plane. Postulate 8. ( P. 58) If two d i f f e r e n t planes intersect,

then t h e i r intersect ion i s a l i n e . Postulate 9. (P. 64) h he Plane Separation ~ostulate.)

Given a l i n e and a plane containing It, the points of the plane

t h a t do not lie on the line fonn two sets such t h a t

(1) each of the sets is convex and (2) if P I s I n one se t and Q I s in the o t h e r then the -

segment PQ i n t e r s e c t s t he l i n e .

Postulate 10. (P. 66) � he Space Separation postula te . )

The p o i n t s of space t h a t do no t lie I n a g iven plane form two se ts such t h a t

(1) each of the s e t s is convex and (2) if P is in one s e t and Q is in t h e o the r , t hen -

the segrnen t PQ intersects the plane. Postulate 11. (P . 80) h he Angle Measurement Postulate.)

To every angle /BAG there corresponds a real number between 0 and 180.

Pos tu l a t e 12. ( P . 81) (The Angle Construction pos tu la te .) L e t 7 be a ray on the edge of the half-plane H. For every

number r between 0 and 180 t he re is exactly one ray 'SP, w i t h P I n H, such t h a t ~/PAB = r.

Postulate 13. (P. 81) h he Angle Addition Pos tu la te . ) If D is a point in the Interior of /BAG, then ~ D A C = AD + ~ /DAC.

Pos tu l a t e 14. (P. 82) he Supplement ~ostulate.) If two

angles form a l i n e a r pair, t h e n t h e y are supplementary.

Pos tu la te 15. ( P . 115) h he S.A.S. ~ o s t u l a t e . ) Given a correspondence between two t r i ang les (or between a triangle and itself). If two sides and t h e i n c l u d e d angle of t h e first

triangle are congruent t o the corresponding p a r t s of the second

triangle, then the correspondence is a congruence. P o s t u l a t e 16. (P . 252) h he Parallel Postulate . ) Through

a given ex te rna l p o i n t there is at most one l i n e parallel to a

g i ven line.

Postu la te 17. (P. 320) To every polygonal region there

corresponds a unique positive number. Postulate 18. (p. 320) 1C two t r iangles are congruent,

then the triangular regions have the same area. Postulate 19. (P. 320) Suppose that the region R is the

union of two regions Ri and R2 . Suppose t h a t Ri and R p

intersect at most in a finite number of segments and p o i n t s .

Then t he area of R is t h e sum of the areas of Ri and R p . Pos tu l a t e 20. (P. 322) The area of a rectangle is the

product of the length of i ts base and the l eng th of its a l t i t u d e .

Postulate 21 . (P . 546) The volume of a rectangular parallelepiped is the product of the altitude and the area of t he base.

Pos tu la te 22. (P . 548) (Cavalier1 1 s Principle. } Given two

s o l i d s and a plane. If for every plane which intersects the

sol ids and is parallel to the given plane the two intersections

have equal areas, then the two s o l i d s have the same volume.

L i s t of Theorems and Corollaries -- - Theorem 2-1. (P. 42) Let A, B, C be three points of a

line, with coordinates x, y, z . If x < y < z, then 3 is between A and C.

Theorem 2-2. same line, one is

Theorem 2-3.

line, only one is

Theorem 2-4. +?

AB be a ray., and

exactly one p o i n t

Theorem 2-5.

point .

Theorem 3-1. most one point.

Theorem 3-2.

(P. 43) Of any three different points on the between t he other two.

(P. 44) Of three di f ferent points on the same

between the other two.

(P. 46) (The Point Plotting he or ern.) Let let x be a positive number. Then there I s

Ã‘ P of AB such that AP = x.

(P. 47) Every segment has exactly one mid-

( P . 55) Two dif ferent lines Intersect i n at

(P. 56) fa l ine intersects a plane not containing It, then the Intersection is a single p o i n t .

Theorem 3-3. (P. 57) Given a line and a point not on the line, there is exactly one plane containing both of them.

Theorem 3-4. (P. 58) Given two Intersecting lines, there is exactly one plane containing them.

Theorem 4-1. (P . 87) If two angles are complementary, then

both of them are acute.

Theorem 4-2. (P. 87) Every angle is congruent t o i t s e l f .

Theorem 4-3. (P. 87) Any two right angles are congruent.

Theorem 4-4. (P. 87) If two angles are both congruent and supplementary, then each of them is a right angle.

Theorem 4-5. (P. 87) Supplements of congruent angles are congruent.

Theorem 4-6. ( P . 88) ~ohplements of congruent angles are congruent .

Theorem 4-7. ( P . 88) Vertical angles are congruent.

Theorem 4-8. (P. 89) If two Intersecting l ines form one r i gh t angle, then they f o m four right angles.

Theorem 5-1. (P. 109) very segment is congruent to i t se l f .

Theorem 5-2. (P. 127) If two sides of a triangle are congruent, then the angles opposite these sides are congruent.

Corollary 5-2-1. (P . 128) Every equilateral triangle is

equiangular.

Theorem 5-3. (P. 129) Every angle has exactly one bisec to r .

Theorem 5-4. ( P . 132) (me A .S .A. Theorem.) Given a correspondence between two triangles (or between a triangle and itself) . If t w o angles and the included side of the first triangle are congruent to the corresponding parts of the second triangle, then the correspondence is a congruence.

Theorem 5-5. (P. 133) If two angles of a triangle are congruent, the sides opposite these angles are congruent.

Corollary 5-5-1. (P. 133) An equiangular triangle is

equi la teral .

Theorem 5-6. (P. 137) h he S .S .S. Theorem.) Given a correspondence between two triangles (or between a t r iangle and itself.) If a l l three pairs of corresponding sides are congruent, then the correspondence is a congruence.

Theorem 6-1. (P. 167) In a given plane, through a given

point of a given line of the plane, there passes one and only one

line perpendicular to the given line.

Theorem 6-2. (P. 169) The perpendicular bisector of a segment, In a plane, is the s e t of a l l po in t s of the plane that . are equ id i s t an t from the end-points of the segment.

Theorem 6-3. (P. 171 ) Through a given external point there is at most one line perpendicular to a given line.

Corollary 6-3-1. (P. 172) A t most one angle of a triangle can be a right angle.

Theorem 6-4. (P. 172) Through a given external point there is at l e a a t one l ine perpendicular to a given l ine . 1

Theorem 6-5. (P. 183) If M Is between A and C on a line L, then M and A are on the same s i d e of any other line that contains C .

Theorem 6 -6 . (P . 183,) If M is between A and C , and <Ã‘

B is any po in t not on line AC, then M I s in the i n t e r io r of / ABC .

Theorem 7-1. (P. 193) he Exterior Angle Theorem.) An

exterior angle of a triangle I s larger than either remote In ter ior

angle.

Corollary 7-1-1. (P-196) If a triangle has a right angle,

then the other two angles are acute.

Theorem 7-2. (P. 1 9 7 ) (The S . A . A . he or em.) Given a

correspondence between two triangles. If two angles and a side opposite one of them in one triangle are congruent to the corresponding parts of the second triangle, then the correspondence is a congruence.

Theorem 7-3. (P. 198 ) h he Hypotenuse - Leg Theorem.) Given a correspondence between two right triangles. If the hypotenuse and one leg of one triangle are congruent to the corresponding parts of the second triangle, then the correspondence i s a congruence.

Theorem 7-4. (P. 200) if two sides of a t r iangle are no t congruent, then the angles opposite these two s i d e s are no t congruent, and the larger angle I s opposite the longer aide.

Theorem 7-5. (P. 201) If two angles of a triangle are not congruent, then the sides opposi te them are not congruent, and t he longer side is opposite the larger angle.

Theorem 7-6. (P. 206) The shortest segment Jolnlng a point to a l i n e is the perpendicular segment.

Theorem 7-7. (P. 206) (The Triangle Inequality.) The sum of the lengths of any two sides of a triangle is greater than the

of the third side.

Theorem 7-8. (P. 210) If two sides of one triangle are congruent respectively to two sides of a second triangle, and the included angle of the first triangle I s larger than the

included angle of the second, then the opposite slde of the first triangle Is longer than the opposite side of the second.

Theorem 7-9" [P. 211) If two sides of one triangle are congruent respectively to two sides of a second triangle, and the t h i r d s i d e of the first triangle I s longer than the t h i r d

s ide of the second, then the included angle of the f i r s t triangle

is larger than the included angle of the second.

Theorem 8-1. (P. 222) If each of two points of a line I s

equidistant f r o m two given points , then every point of the line is equidistant f r o m the given points.

Theorem 8-2. (P. 225) If each o f three non-colltnear points of a plane is equidistant from two points, then every point of the plane is equidistant f r o m these two points.

Theorem 8-3. (P. 226) If a l ine l a perpendicular to each

of two intersecting lines at their point of intersection, then It Is perpendicular to the plane of these lines.

Theorem 8-4. (P. 230) Through a given point on a given line there passes a plane perpendicular to the line.

Theorem 8-5. (P. 231) If a line and a plane are perpendicular, then the plane contains every line perpendicular to the

given line at I t s point of Intersection with the given plane.

Theorem 8-6. (P. 233 Through a given point on a given line there I s at most one plane perpendicular to the line.

Theorem 8-7. (P. 232) The perpendicular bisecting plane of a segment I s the s e t of a l l points equidistant f r o m the end- points of the segment.

Theorem 8-8. (P. 234) Two lines perpendicular to the same

plane are coplanar.

Theorem 8-9, (P. 235 ) Through a given point there passes one and only one plane perpendicular to a e v e n - line.

Theorem 8-10. ( P . 235 ) Through a given point there passes one and only one lme perpendicular t o a given plane,

Theorem 8-11. ( P . 2 3 5 ) The shortest segment to a plane f r o m an external po in t I s the perpendicular segment.

Theorem. (P. 242) Two parallel lines l i e in exactly one plane.

-, (F. 242 Two lmes l.n a plane are parallel if they are both perpendicular to the same l ine .

m. (P. 244) Wt L be a line, and l e t P be a point not on L. Then there is at least one line through P, parallel to L.

Theorem. (P. 246) If two l ines are cut by a transversal, and if one pair of alternate interior angles are congruent, then the other pair of alternate Interior angles are also congruent.

Theorem 9-5. (P. 246) If t w o lines are cut by a transversal, and if a pair of alternate interior angles are congruent, then the lines are parallel.

Theorem 9-6. (P. 252) If two lines are c u t by a transversal, and if one pair of corresponding angles are congruent, then the other three pairs of corresponding angles have the same property.

Theorem 9-7. (P. 252) If two lines are cut by a transversal, and if a pair of corresponding angles are congruent, then the lines are parallel.

Theorem 9-8. (P. 253 ) ~f two parallel l ines are cut by a t ransversal , then then alternate Interior angles are congruent.

Theorem 9-9. (P . 253 ) If two parallel lines are c u t by a transversal, each p a i r of corresponding angles are congruent.

Theorem 9-10. ( P . 254) If two parallel lines are c u t by a

transversal, I n t e r i o r angles on the same s ide of the transversal

are supplementary.

Theorem 9-11. ( P . 255) In a plane, two lines parallel to the same line are parallel to each other.

Theorem 9-12. ( P . 2 5 5 ) In a plane, If a l i n e is perpendlcular to one of two parallel lines It l a perpendicular t o the other.

Theorem 9-13. (P. 238) The sum of the measures of t h e

angles of .a t r i ang le is 180.

Corollary 9-13-1. (P. 2 5 9 ) Given a correspondence between

two triangles . If two pairs of corresponding angles are congruent, then the t h i rd pair of corresponding angles are also congruent.

Corollary 9-13-2. (P. 260) The acute angles of a right

triangle are complementary.

Corollary 9-13-3. (P. 260 ) For any triangle, the measure of

an exterior angle is the sum of the measures of the two remote interior angles.

Theorem 9-14. (P . 265 } Either diagonal divides a parallelo- ; gram i n to two congruent triangles.

Theorem 9-15. ( P . 265 ) In a parallelogram, any two opposite : sides are congruent.

1

Corollary 9-15-1. ( p . 2 6 6 ) If L, 1 1 L, and if P and Q are any two points on L,, then the distances of P and Q from

Ly are equal.

Theorem 9-16. (P. 2 6 6 ) in a parallelogram, any two opposite

angles are congruent.

Theorem 9-17. ( P . 266) In a parallelogram, any two consecutive angles are supplementary.

Theorem 9-18. ( P . 266) The diagonals of a parallelogram bisect each other.

Theorem 9-19. ( P . 266) Given a quadrilateral In which both pairs of opposite ides are congruent. Then the quadrilateral is a parallelogram.

Theorem 5-20. (P. 266) If two sides of a quadrilateral are parallel and congruent, then the quadrilateral is a parallelogram.

Theorem 3-21. (P. 266) If the diagonals of a quadrilateral bisect each o t h e r , then the quadrilateral is a parallelogram.

Theorem 9-22. (P. 267 ) The segment between the mid-points of two sides of a triangle is parallel to the th i rd s ide and half as long as the t h i rd s i d e ,

Theorem 9-23. (P. 268) If a parallelogram has one right

angle, then It has four right angles, and the parallelogram I s a rectangle.

Theorem 9-24. (P. 268 ) In a rhombus, the diagonals are

perpendicular to one another.

Theorem 9-25., ( P . 268 ) If the diagonals of a quadrilateral bisect each o the r and are perpendicular, then the quadrilateral is a rhombus.

w. (P. 276) If three paral le l lines i n t e r c e p t congruent segments on one transversal, then they Intercept congruent segments on any other transversal.

Corollary 9-26-1. (P. 277 ) If three or more parallel lines intercept c o n w e n t segnents on one transversal, then they in tercept congruent segments on any o t h e r transversal.

m. (P. 279 ) The medians of a triangle are concurrent In a point two-thirds the way from any vertex t o the

mid-point of the opposite side.

Theorem 10-1. (P. 292) If a plane intersects two parallel , planes, then it. intersects them In two parallel lines.

Theorem 10-2. (P. 292) If a line is perpendicular to one of two parallel planes it is perpendicular t o the o the r .

Theorem 10-3. (P. 293) Two planes perpendicular t o the same line are parallel.

Corollary 10-3-1. (P. 294) If t w o planes are each parallel t o a third plane, they are parallel to each other.

Theorem 10-4. (P. 294) Two lines perpendicular to the same

plane are parallel.

Corollary 10-4-1. ( I ? . 2 9 4 ) A plane perpendicular to one of two p a r a l l e l l ines is perpendicular to the o the r ,

Corollary 10-4-2. (P. 294) If two lines are each parallel to a third they are parallel to each other.

Theorem 10-5. (P . 295 ) Two parallel planes are everywhere

equidistant . Theorem 10-6. (p.501) Any two plane angles of a given

dihedral angle are congruent.

Corollary 10-6-1. (P. 302 ) if a l i n e is perpendicular to a

plane, then any plane containing this line is perpendicular to the given plane.

Corollary 10-6-2. ( p . 3 0 2 ) if two planes are perpendicular, then any line in one of them perpendicular to t h e i r line of

intersection is perpendicular to the other plane.

Theorem 10-7. (P. 307 ) The projection of a line i n t o a

plane is a l ine , unless the line and the plane are perpendicular.

Theorem 11-1. (P. 328 ) The area of a right t r i a n g l e is ha l f

the product of Its legs . Theorem 11-2. (P. 328) The area of a t r i a n g l e is half the

product of any base and the altitude to that base.

Theorem 11-3. (P. 530) The area of a parallelogram I s the product of any base and the corresponding altitude.

Theorem 11-4. (P. 333) The area of a trapezoid is half the

product of i t s altitude and the sum of its bases.

Theorem 11-5. (P. 332) If two triangles have the same

altitude, then the ratio of their areas is equal to the r a t i o of their bases.

Theorem 11-6. (P. 332) If two triangles have equal altitudes and equal bases, then they have equal areas.

,Theorem 11-7. (P. 339) h he Pythagorean Theorem.) In a right triangle, the square of the hypotenuse is equal to the sum

of the squares of the legs.

Theorem 11-8. (P. 340) If the square of one s ide of a triangle I s equal to the sum of the squares of the other two, then the triangle is a right triangle, with a right angle opposite

the first s ide .

Theorem 11-9. (P. 346) (The 30 - 60 Triangle Theorem.) The

hypotenuse of a right triangle I s twice as long as the shorter l eg if and only if the acute angles are 30' and 60".

Theorem 11-10. ( P . 346 ) (The Isosceles R i g h t Triangle

he or em.) A right triangle I s isosceles if and only If the

hypotenuse I s -/;T times as long as a leg.

Theorem 12-1. (P . 368 ) he Basic Proportionality heo or em . ) If a line parallel to one side of a triangle Intersects the other two sides in distinct points, then I t cuts off segments which are proportional to these sides.

Theorem 12-2. (P. 369 ) If a line intersects two sides of a triangle, and cuts o f f segments proportional to these two sides, then it I s parallel to the third side.

Ã

Theorem 12-3. ( p . 3 T 4 ) h he A.A.A . Similarity he or em.) Given a correspondence between two triangles. If corresponding angles are congruent, then the correspondence is a similarity.

Corollary 12-3-1. (P ,376 ) h he A .A. Corollary. ) Given a

correspondence between two triangles. If two pairs of corresponding angles are congruent, then the correspondence is a similarity.

Corollary 12-3-2. (P. 376) If a line parallel to one side of a triangle intersects the other two sides in distinct points, then it cuts of f a triangle similar t o t h e given triangle.

Theorem 12-4. (P , 376) (The S .A .S . Similarity heo or ern.) Given a correspondence between two triangles. If two corresponding angles are congruent, and the including sides are proportional, then the correspondence is a similarity.

Theorem 12-5. (P. 3788) h he S .S .S. Similarity Theorem.) Given a correspondence between two triangles. If corresponding sides are proportional, then the correspondence i s a similarity.

Theorem 12-6. (P. 391) In any right tr iangle, the altitude to the hypotenuse separates the triangle i n t o two triangles which

are similar both to each other and to the original triangle.

Corollary 12-6-1. (P. 392) Given a r ight triangle and the

altitude from the r i gh t angle to the hypotenuse:

(1) The altitude is the geometric mean of the segments in to which it separates the hypotenuse.

(2) Either leg Is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to the leg.

Theorem 12-7. f P . 5 ~ 5 ) The ratio of the areas of two

similar triangles I s the square of the ra t io of any two corresponding sides.

Theorem 15-1. (P . 410) The intersection of a sphere w i t h

a plane th rough Its c e n t e r is a c i r c l e w i t h the same center and radius.

Theorem 13-2. ( P . 414) Given a l i n e and a c irc le in the

same p lane . L e t P be t he center of the c i r c l e , and l e t F be

the foot of t h e perpendicular from P to the lineÃ Then e i t h e r

(1) Every po in t of the l i n e is outside the circle, or

(2) F is on t h e c i r c l e , and the l i n e is tangent to t h e circle a t F, or

( 3 ) F is inside t h e c i r c l e , and the l i n e intersects t h e

c i r c l e in exactly two p o i n t s , which are equidistant from F.

Corollary 13-2-1. (? . 416) Every line tangent to C is perpendicular to the radius drawn to t he po in t of contact .

Corollary 13-2-2. (P. 416) Any line in E, perpendicular to a radius a t its outer end, is t angent t o t h e c i r c l e .

Corollary 13-2-3. (P . 416) Any perpendicular from t he cen t e r of C t o a chord bisects the chord.

Corollary 15-2-4. ( P . 416) The segment joining the center of C to the mid-point of a chord is perpendicular to the chord.

Corollary 13-2-5. ( P . 416) I n the plane of a circle, the

perpendicular b i s e c t o r of a chord passes through the center of the circle.

Corollary 13-2-6. (P. 417) If a line in the plane of a circle Intersects the interior of t h e c i r c l e , then i t In te rsec t s

t h e c i r c l e in exactly two p o i n t s ,

Theorem 13-3. ( P . 417) In the same circle or in congruent circles, chords equidistant from t h e center are congruent.

Theorem 13-4. (P. '417) In the same c i r c l e or in congruent c i r c l e s , any two congruent chords are equ id i s t an t from t h e c e n t e r .

Theorem 13-5. (F. 424) Given a p l a n e E and a sphere S with center P. Let P be t h e foot of the perpendicular segment from P to E. Then either

(1) Every po in t of E is o u t s i d e S, or

(2) F is on S, and E is tangent t o 3 at F, or

( 3 ) F is Ins ide S, and E i n t e r s e c t s S in a c i r c l e w i t h center P.

Corol lary 13-5-1. (P. 426) A plane tangent to S I s

perpendicular to the radius drawn to the point of contact .

Corollary 13-5-2. (P. 426) A plane perpendicular t o a rad ius at i t s ou te r end is tangent to S .

Corollary 13-5-3. (P. 426) A perpendicular from P to a chord of S bisects the chord.

Corollary 15-5-4. (P. 426) The segment joining the center of S to the midpoint of a chord is perpendicular to the chord.

Theorem 13-6. (P . 431 ) If a and @ are arcs of the same circle having only the point B In common, and if their union is an arc "St?, then m'ftS + r&! = ma.

Theorem 13-7. (P . 434) The measure of an inscribed angle I s h a l f the measure of its i n t e rcep ted arc .

Corollary 13-7-1 . (P . 437 ) An angle inscribed In a semi-

c i r c l e is a right angle.

Corollary 13-7-2. (P. 437) Angles inscribed in the same

arc are congruent.

Theorem 13-8. (P. 441) In the same c i r c l e or in congruent circles, If two chords are congruent, then so a l so are the corresponding minor arcs .

Theorem 13-9. ( P . w1) In t h e same circle o r i n congruent c i rc les , If two arcs are congruent, then so are the corresponding

chords . -Theorem 15-10. ( P . 442) Given an angle with vertex on the

c i r c l e formed by a secant ray and a tangent ray. The measure of the angle I s ha l f the measure of the Intercepted a r c .

Theorem 13-11. (P. 448) The two tangent segments to a

ci rc le f rom an external point are congruent, and form congruent angles with the l i n e Joining the external point t o the center of the c i r c l e .

Theorem 15-12. ( P . 449) Given a circle C and an external point Q, let L1 be a secant line through Q, intersecting C i n p o i n t s R and S; and l e t L p be another secant line through Q, intersecting C in points T and U. Then QR QS = QU QT.

Theorem 13-13. (P. 450) Given a tangent segment t o a

c i r c l e , and a secant l i n e through Q, intersecting the c i r c l e in

points R and 3 . Then QR QS = QT.

Theorem 13-14, (P. 451) If two chords i n t e r s e c t w i th in a

circle, t h e product of t he lengths of t he segments of one equals

t he product o f t h e lengths of the segments of t h e other.

Theorem 14-1. (P . 467) The bisector of an angle, minus its end-point, is t h e s e t of points in t h e interior of the angle equidistant from t h e sides of t h e angle.

Theorem 14-2. (P . 469) The perpendicular "bisectors of the s ides of a t r i a n g l e are concurrent in a point equ id i s t an t from the three vertices of t h e t r i ang l e .

Corollary 14-2-1. (P. 470) There is one and only one c i r c l e through th ree non-coll lnear polnfcs .

Corollary 14-2-2. (P. 470) Two d i s t i n c t c i r c l e s can in te r sec t in a t most two points.

Theorem 14-3. (P. 470) The three altitudes of a triangle are concur ren t .

Theorem 14-4. ( P . 471) The angle bisectors of a t r iangle are concurrent in a point e q u i d i s t a n t from t h e t h r e e sides.

Theorem 14-5. (P. 476) (The Two Ci rc l e he or em) If two circles have radii a and b, and if c is the distance between t h e i r centers, then the c i r c l e s intersect in two p o i n t s , one on each side of the l i n e of centers, provided each of of a,

b , c I s less than t h e sum of t h e other two.

Construction 14-6. (P. 477) TO copy a given triangle.

Construct ion 14-7. ( P . 479) To copy a given angle.

Construe tJ .on 14-8. (P . 481) To construct the perpendicular b i s e c t o r of a given segment.

Corollary 14-8-1. (P . 481 ) To b i s e c t a given segment . Construction 14-9. (P . 482) To cons t ruc t a perpendicular

to a given line through a given point.

Construction 14-10. (P. 484) To construct a p a r a l l e l t o a

given line, through a given external point.

Construction 14-11. (F . 484) To div ide a segment into a given number of congruent segments.

Cons t ruc t ion 14-12. ( P . 491) To circumscribe a c i r c l e about

a given t r i a n g l e .

Construct ion 14-15. (P . 491 ) To b i s e c t a given ang le .

Construction 1k-14. (P. 492) To inscribe a c i r c l e in a

given t r i ang le .

of the circumference Theorem 15-1. (P. 517) The ratio F, to t he diameter, Is tlie same for a l l c i r c l e s .

Theorem 15-2. (P. 522) The area of a c i r c l e of radius r 2 is Trr

Theorem 15-3. (P. 526) If two arcs have equal radii, their l eng ths are proportional t o the i r measures.

Theorem 15-4. (P . 526) A n arc of measure q and radius r has length &qr .

Theorem 15-5. (P. 527) The area of a sec to r is ha l f t he

product of i t s radius by the l ength of i t s a rc .

Theorem 15-6. (P. 527) The area of a sector of radius r 2 and arc measure q is ^ y i p .

Theorem 16-1. (P. 535) All cross-sections of a t r iangular p r i sm are congruent to t h e base.

Corollary 16-1-1. (P. 536) The upper and lower bases of a t r i a n g u l a r prism are congruent.

Theorem 16-2. ( P . 536) ( p r i s m Cross-Section heo or em . ) Al l

cross-sections of a prism have t h e same area.

Corollary 16-2-1. ( P . 537) The two bases of a prism have equal areas.

Theorem 16-3. (P. 537) The l a te ra l faces o f a prism are

parallelogram regions, and the l a t e r a l faces of a right prism

are rec tangular regions.

Theorem 16-4. (P . 540) -A cross-section of a triangular

pyramid, by a plane between the vertex and the base, Is a t r iangular region similar to the base. If the distance from the

ver tex to the cross-section plane is k and the altitude is h,

then the r a t io of the area of t h e cross-section t o the area of k the base I s .

Theorem 16-5, (P. 542) In m y pyramid, t h e ratio of the

a of a cross-section and the area of the base Is ($) , where h i s the a l t i tude of the pyramid and k is the distance from the vertex to the plane of the cross-section.

Theorem 16-6, (P . 543) (The Pyramid Cross-Section heo or em. ) Given two pyramids w i t h the same a l t i t u d e . If the bases have the

same area, then cross-sections equidistant from the bases a l so have t h e same area.

Theorem 16-7. (P . 548) The volume of any prism is the product of the altitude and the area of the base.

Theorem 16-8. (P. 549) If two pyramids have t h e same altl-

tude and the same base area, then they have the same volume.

Theorem 16-9. (P. 550) The volume of a t r i a n g u l a r pyramid

is one-third the product of its a l t i t u d e and i t s base area.

Theorem 16-10. (P . 551) The volume of a pyramid is one-third

t he product of its a l t i t ude and i t s base area.

Theorem 16-11. ( P . 555) A cross-section of a c i r c u l a r cylinder is a c i r c u l a r region congruent t o t h e base.

Theoren 16-12. ( P . 555) The area of a cross-section of a

c i r c u l a r cy l inder is equal to t h e area of the base,

Theorem 16-13. (P. 555) A cross-section of a cone of a l t i t u d e h, made by a plane at a distance k from t h e vertex,

is a circula$ region whose area has a r a t i o to the area of the k base of (,-) .

Theorem 16-14. (P . 557) The volume of a c i r cu l a r cylinder is the product of the altitude and the area o f the base.

Theorem 16-15. (P . 557) The volume o f a c i r c u l a r cone Is one-third the product of the a l t i t u d e and the area of t h e base.

Theorem 16-16. (P. 559) The volume of a sphere of radius P

is +2. Theorem 16-17. (P. 562) The surface area of a sphere of

radius r is S = h2 . Theorem 17-1. (P. 579) On a non-vertical l i n e , all segments

have the same slope.

Theorem 17-2. (P. 584) Two non-vertical lines are paral le l

if and only I f they have the same slope.

Theorem 17-3. (P . 586) Two non-vert ical l i n e s a re perpen- d i cu l a r if and o n l y if their slopes are the negative reciprocals of each o t h e r .

Theorem 17-4. (P. 589) (The Distance p or mu la.) The distance between the po in t s (x,,y-,) and (xpy2) is equal t o

Theorem 17-5. (P. 593) h he Mid-Point Fo??nIulaO) ~ e t P^ = ( x f i ) and l e t Pg = ( x g 9 y 2 ) . Then the mid-point - x + X^

of PIPn i s t h e poin t P = ( 1 Y! + y2 2 1 2 1

Theorem 17-6. ( P . 605) L e t L be a non-vertlcal l i n e wi th

w i t h slope m, and let P be a point of L, w i t h coordinates

(xl,yl), For every po in t Q = ( x , ~ ) of L, t h e equat ion

y - yi = m(x - x,) is satisfied.

Theorem 17-7. ( P . 607) The graph of the equation

y - y- = m ( x - x.,) is the line that passes through the point (x,,y.,) and has slope m.

Theorem 17-8. (P . 611) The graph of the equation y = mx + b is the line with slope m and y-intercept b.

Theorem 17-9. (P. 613) Every l i n e I n the plane is the graph of a l inear equation in x and y.

Theorem 17-10. ( P . 615) The graph of a linear equa t ion in x and y is always a l i n e .

Theorem 17-U. (P. 623) The graph of t h e equa t ion

( x - a)2 + (y - b)* = r2 Is the c i r c l e w i t h c e n t e r at ( a , b ) and radius r .

Theorem 17-12. (P. 624) Every c i r c l e I s t h e graph of an 2 equation of t h e form s + y2 + Ax + By + C = 0.

Theorem 17-13. (P. 625) Given t h e equation

x2 + y2 + Ax + By + C = 0. The graph of this equation Is (1) a c i r c l e , (2) a poin t or (3) the empty s e t .

Index of Definitions -- For precisely defined geometric terms the reference I s to

the formal definition. For other terms the reference is to an informal definition or t o t h e most prominent d iscuss ion .

absolute value, 27 acute angles, 86 al ternate i n t e r i o r angles, 245 altitude

of prism, 535 of pyramid, 540 of triangle, 214, 215

angleb), 71 acute, 86 alternate Interior, 245 bisector o f , 129 central , 429 complementary, 86 congruent, 86, 109 consecutive, 264 corresponding, 251 d i h e d r a l , 299 exterior, 293 exterior of, 75 inscr ibed , 432 intercepts an arc, 433 i n t e r i o r o f , 73 measure of, 79, 80 obtuse, 86 or polygon, 506 opposite, 264 reflex, 78 remote i n t e r i o r , 195 r i g h t , 85 right dihedra l , 301 sides of, 71 straight, 78 supplementary, 82 vertex of, 71 vertical, 88

apothem, 512 a r c ( s ) , 429

center of, 437 congruent , 441 degree measure of, 430 end-points of, 429 length of, 525 major, 429 minor, 429 of sector, 527

area, 320 ,, circle, 521, 522 parallelogram, 330 polygonal region, 320 rectangle, 322 right t r iangle , 328 sphere, 562 trapezoid, 351 triangle, 328 unit of, 321

arithmetic mean, 364 auxiliary sets , 176 base of pyramid, 540 between, 41, 182 bisector of an angle, 129 bisector of a segment, 169 bisects, 47, 129 Cavalleri~s Principle, 548 center of

arc, 437 c i r c l e , 409 sphere, 409

central angle, 429 centroid, 280, 621 chord. 410 c irc le(s ) , 409

area of, 521, 522 circumference of, 516 congruent, 417 equation o f , 623, 624, 625 exterior of, 412 great, 410 Interior of, 412 segment of, 528 tangent, 417

c i r cu l a r cone, 554 cylinder, 553 reasoning, 119 region, 520

area of, 521 circumference , 51 6 circumscribed

circle, 490 triangle, 490

colllnear, 54 complement, 86 complementary angles, 86 concentr ic

circles, 409 spheres, 409

conclusion, 60

concurrent sets, 278, 469 cone,

c i r c u l a r , 554 r i g h t c i r c u l a r , 555 volume of, 557

congruence , 97 congruent

angles, 86, 109 arcs, 441 c i r c l e s , 43.7 segments, 109 triangles, 98, 111

consecutive angles, 264 consecutive s i d e s , 264 cons t r u c t ions, 477 converse, 202 convex polygon, $07 convex sets, 62 coordinate system, 37, 571 coordinates of a point, 37, 569 co-planar, 6 4 corollary, 128 correspondence, 97 corresponding angles, 251 cross-section

of a prism, 535 of a pyramid, 540

cube, 229 cylinder

circular, 553 volume o f , 557

diagonal, 264, 509 diameter, 410 dihedra l angle, 299

edge of, 299 face of, 299 measure of, 301 plane angle o f , 300

distance, 34 distance between

a point and a l i n e , 206 a point and a plane, 235 two parallel l i n e s , 266

distance formula, 589 edge of half plane, 64 end-point ( s )

of arc, 429 of ray, 46 of segment, 45

empty set , 18

equat ion of c i r c l e , 623 of l i n e , 605, 611

equiangular triangle, 128 equi la teral t r i a n g l e , 128 Euler, 327 existence proofs , 165 e x t e r i o r angle, 193 ex'ce-'-Lor

o r an angle , 73 of a c i r c l e , 412 of a t r iangle , 7 G

f ~ c e of half-space, 66 frustum, 559 Garfield's Proof, 344 geometric mean, 561 graph, 600 great c i r c l e , 410 half-plane, 6Q

edge of 64 ha1 f -space, 66

face o f , 66 h o r i z o n t a l l i n e s , 576 hypotenuse, 172 hypothesis , 60 identity congruence, 100, 109 i f and only if, 203 i f- then, 60 inconsistent equat ions, 618 i nd i r ec t proof, 160 Inequalities, 24 i n f i n i t e r u l e r , 37 inscribed

angle, 432 measure of , 434

c i r c l e , 490 polygon, ,511 quadrilateral, 438 t r i a n g l e , 490

in tegers , 22 i n t e r c e p t , 275, 433 interior

of ang le , 73 of circle, 412 of triangle, 7 4

i n t e r s ec t , 18 intersection of sets, 16, 18, 4-73 i r r a t i o n a l numbers, 23 isosceles triangle, 127 kite, 272

l a t e r a l edge, 53'7 face, 537 sur face , 537

lemma, 196 l eng th

of arc , 525 of segment, 45

linear equation, 613 linear pair, 82 line(s), 10

oblique, 216 para l l e l , 241 perpendicular, 86 skew, 241 t ransversal , 244

major arc, 429 mean

arithmetic, 364 geometric, 361

measure of angle, 79, 80 of dihedral angle, 301 of distance, 50, 54, 36

median of t r apezo id , 272 of triangle, 130

mid-point, 47 formula of, 595

minor arc, 429 Non-Euclidean geometries, 255 negat ive real numbers, 191 numbers

Irrational , 23 negat ive , 191 p o s i t i v e , 191 rational, 22 r e a l , 25 whole, 22

obl ique l i n e s , 216 obtuse angle, 86 on oppos i t e sides, 64 on the same s i d e , 64 one-to-one correspondence, 97 opposi te

angles 264 rays, 46 sides, 264

order, 24 order pos tu la tes , 191, 192

ordered pair, 571 origin, 568 parallel

l i n e s , 241 slopes of, 584

l i n e s and planes , 291 planes, 291

parallelepiped, 538 parallelogram, 265

area of, 330 perimeter

of triangle, 287 of polygon, 512"

perpendicular l i n e s . 86

slopes of, 586 line and plane, 219 planes, 301

perpendicu la r bisector, 169 PI, T T , 518 plane(s), 10

p a r a l l el, 291 perpendicular, 301

plane angle, 300 point, 10 poin t - s lope form, 605 point of ';angency

of c i r c l e s , 413 of spheres, 423

polygon, 506 angle o f , 506 apothem of, 512 convex, 507 diagonal of, 509 inscribed, 511 perimeter of, 512 regular, 511 sides of, 506 ver t ices of, 506

polygonal region, 317 polyhedral regions, 546 positive real numbers, 191 postulate ( s ), 9

of order, 191, 192 power* of a point , 451 prism, 554

a l t i t u d e o f , 535 cross-section of, 535 l a t e r a l edge, 537 l a t e r a l f ace , 537 1 a t e pal surface , 537 lower base, 555 rec tangula r . 555

prism (continued) r i g h t , 535 to ta l surface, 537 tr iangular , 535 upper base, 535

projection of a l i n e , 306 of a point;, 506

proof converse, 202 double-column form of, 118 existence, 165 i nd i rec t , 160 uniqueness , 165 writing of, 117

propor t iona l sequences, 360 pyramid, 540

altitude of, 540 base o f , 540 regular, 544 vertex of, 540 volume of , 551

Pythagorean Theorem, 339 quadrant , 571 quadr i la tera l , 263

consecutive angles of, 264 consecutive sides of, 264 cyc l LC, 473 diagonal of, 264 inscr ibed , 438 opposite angles of, 264

radius, 409, 410 of sector, 527

ra t iona l numbers, 22 ray, 46

end-point of, 46 opposite, 46

real numbers, 25 rectangle, 268

area o f , 322 rectangular para l le lep iped , 538 reflex angle, 78 region

c i r c u l a r , 520 polygonal, 517 polyhedral, 546 t r i a n g u l a r , 317

regular polygon, 511 pyramid, 544

remote i n t e r i o r angle, 193

rhombus, 268 r igh t angle, 85 ,

right dihedral angle, 301 - -

right prism, 535 right t r iang le , 172 sca lene triangle, 128 sector , 527

arc of, 527 r a d i u s of, 527

segment ( s ), 45 bisec to r , 169 congruent , 109

segment of a c i r c l e , 528 semi-circle, 429 separation, 182 sefc(s), 15

auxiliary, 176 concurrent, 278 convex, 62 element of , 15 empty, 18 intersection of, 16, 473 member of, 15 union, of, 17

side ( a ) consecut ive , 364 of angle, 71 of d ihedra l angle, 299 of polygon, 506 of t r i ang l e , 72 opposite, 264

similarity, 365 skew l i ne s , 241 slope, 577

of parallel lines, 5811 of perpendicular lines, 586

slope-intercept form, 611 space, 53 sphere, 409

exter ior of, 423 interior of, 425 surface area of, 562 volume of, 559

square, 268 square root, 25 straight angle, 78 subse t , 15 supplement, 82 supplementary angles, 82

tangent circles, 417 common external, 454 common i n t e r n a l , 454 externally, 417 internally, 417 l i n e and circle , 413 plane and s here, 423 segment., 44

theorem, 9 1

total surface o f a prism, 537 transversal, 244 trapezoid, 265

area of, 331 t r iangle ( s ) , 72

alt i tude of, 214 a n g l e bisector of, 130 area of, 328 cen t ro id of, 280 congruent, 98, 3-11 equiangular, 128 equilateral, 128 exter ior of , 74 Interior of, 74 isosceles, 127, 128, 346 median of , 130 overlapping, 123 perimeter of , 287 right, 172 scalene, 128 s i d e s of, 72 s i p l a g , 365 30 -60 , 346 vertex of, 72

triangular region, 317 undefined terns, 9, 10 union o f s e t s , 17 uniqueness proofs, 165 vertex

of angle, 71 of polygon, 506 of pyramid, 540 of triangle,

vertical angles, 2 ver t i ca l l i n e , 576 volume

of cone, 557 of cylinder 557 of prism, 5 k of pyramid, 551 of sphere, 559

whole numbers, 22 x-axis, 568 y-axis, 568 y-intercept, 611

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