saint barnabas high school - answers for textbook …...2015/09/14 · a line segment is a set of...
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Answers for Textbook Exercises
227
11. a. 1x � xb. Identity property of multiplication
12. a. 6(4 � b) � 6(4) � 6(b)b. Distributive property
13. a. 17 � 1b. Inverse property of multiplication
14. a. (24) � 24b. Commutative property of multiplication
15. a. 12(4 � 10) � 12(10 � 4)b. Commutative property of addition
16. a. � 1b. Inverse property of multiplication
17. x �
18. If a � 0, then b � 0 by the multiplicationproperty of zero. Therefore, a � b � a � 0 � a.
1-3 Definitions, Lines, and Line Segments(pages 10–11)
Writing About Mathematics1. “A hammer is a tool” is not a good definition
because it does not distinguish “hammers” fromother “tools.”
2. “A hammer is used to drive nails” is not a gooddefinition because it is not reversible. There aremany other objects that are not hammers thatcan be used to drive nails.
Developing Skills3. a. A noncollinear set of points is a set of three or
more points that do not all lie on the samestraight line.
b. The set of all pointsc. Noncollinear points do not all lie on the same
straight line.4. a. The distance between any two points on the
number line is the absolute value of thedifference of the coordinates of the points.
b. The set of real numbersc. The distance between any two points must be
a positive real number or 0.5. a. A line segment is a set of points consisting of
two points on a line, called endpoints, and allof the points on the line between theendpoints.
13
p A 1p B
A 14 B1
4
A 117 B
1-1 Undefined Terms (page 3)Writing About Mathematics
1. It may be called a plane because it is flat. It is nota mathematical plane because it does not extendendlessly in all directions. Also, when taking intoaccount the curvature of the earth or localgeographic features such as hills, it is not really aflat surface.
2. A stretched string does not go on forever. It alsohas a measurable thickness.
Developing Skills3. True 4. False 5. False6. True 7. False 8. True
Applying Skills9. Answers will vary.
10. Answers will vary.11. Answers will vary.
1-2 The Real Numbers and TheirProperties (page 6)
Writing About Mathematics1. a. The set of positive real numbers is not closed
under subtraction because, if the value of theminuend is less than or equal to the value ofthe subtrahend, then the difference is anegative real number or zero.
b. Yes. Subtracting any two real numbers resultsin another real number.
2. The set of negative real numbers is not closedunder multiplication because multiplying anytwo negative real numbers results in a positivereal number.
Developing Skills3. 0 4. 15. �a � 11 6. 07. a. 7 � 14 � 14 � 7
b. Commutative property of addition8. a. 7 � (�7) � 0
b. Inverse property of addition9. a. 3(4 � 6) � (3 � 4)6
b. Associative property of multiplication10. a. 11(9) � 9(11)
b. Commutative property of multiplication
For each proof in this Answer Key, one method is provided. Valid alternative proofs may exist and should beconsidered acceptable.
Chapter 1. Essentials of Geometry
Answers for Textbook Exercises
b. The set of all pointsc. The points in a line segment all lie on the same
line and are included between two endpoints.6. a. The measure of a line segment is the distance
between its endpoints.b. The set of all real numbersc. The measure of a line segment must be a
positive real number or 0.7. a. Congruent segments are segments that have
the same measure.b. The set of all pairs of line segmentsc. Congruent pairs of segments have the same
measure.8. AB � 2 9. BD � 3 10. CD � 2
11. FH � 4 12. GJ � 5 13. EJ � 814. Answers will vary. Example: AB � CD � DE15. P, Q, and R are noncollinear. If the points were
collinear, then PR � PQ � QR, but PR � 18while PQ � QR � 10 � 12 � 22.
Applying Skills16. a. 12 in.
b. He could either mark the board at 13 inchesor at 5 inches.
17. No. If Troy, Albany, and Schenectady were in astraight line, then the distance traveled on thereturn trip would equal the distance traveled onthe first trip.
1-4 Midpoints and Bisectors (pages 13–14)Writing About Mathematics
1. Yes. The notation is valid when and name the same line. In this case, A, B, C, and Dare collinear.
2. No. It is possible that A, M, and B arenoncollinear.
Developing Skills
3. AT � TC;4. RN � NS;5. BC � CD;6. SP � PT;7. a.
b. Let x � AB � CD and y � BC.Then AC � AB � BC � x � y and BD � BC � CD � y + x.
A B C D
SP > PTBC > CDRN > NSAT > TC
CDg
ABg
P
Q
R
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8. 2.5 9. �1010. a. Yes; PQ = 16, QR � 2, PR � 18, and
16 � 2 � 18.b. No; �1
Applying Skills11. 30512. a. 16 in.
b. At 16 in.
Hands-On Activity1–2.
3. D bisects ;
1-5 Rays and Angles (pages 18–19)Writing About Mathematics
1. A half-line consists of the set of points that lie toone side of a point on a line. A ray consists of apoint on a line and all the points to one side ofthis endpoint. Therefore, a ray is a half-line plusan endpoint.
2. and are the same ray: they both have endpoint P, and R and S are on the same side ofP on the line.
Developing Skills
3. a. and b. E
c. and d. �y, �E, �FED, �DEF
4. a. A
b. and c. �CAD, �BAD, �CABd. �CAB
e.f. �CAB and �BADg. �BADh. No; A, B, and D are not collinear.i. No; �BAC is not the union of opposite rays.
5. a. �EAB, �CAB, �BAC, �BAEb. �DEC and �CEDc. �CBA and �ABCd. E
e. and ; and f. �DEB and �AECg. �ABE and �EBC
ECh
EAh
EBh
EDh
ABh
ABh
ADh
EDh
EFh
EDh
EFh
PSh
PRh
AD > DBAB
C
A BD
E
1-6 More Angle Definitions (pages 21–23)Writing About Mathematics
1. Disagree; a bisector is a ray. However, the angle
bisector is on .2. Acute; since an obtuse angle has a measure
greater than 90° and less than 180°, the twoangles formed have measures greater than 45°and less than 90°.
Developing Skills3. a. Acute 4. a. Obtuse
b. 12° b. 49°5. a. Obtuse 6. a. Right
b. 63° b. 45°7. a. Acute 8. a. Straight
b. 41° b. 90°9. a. Acute 10. a. Acute
b. 28.5° b. 1.5°11. 45° 12. 30° 13. 72°14. b. �ACD � �DCB
c. m�ACD � m�DCB15. b. �DAC � �CAB
c. m�DAC � m�CAB16. Two right angles17. Two acute, 45° angles18. m�LMN � m�LMP � m�NMP19. m�LMP � m�LMN � m�NMP20. m�LMN � m�LMP � m�NMP21. m�ABE � m�EBC � m�ABC22. m�BEC � m�CED � m�DEB23. m�ADC � m�CDE � m�ADE24. m�AEC � m�AEB � m�CEB25. a. �ACD, �ACF, �DCB, �BCF
b. 90° c. and d. and 26. x � 15 27. m�ABC � 6828. m�QRS = 15029. Yes. An angle bisector divides an angle into two
congruent angles, which have equal measure.Let x � m�CBD � m�PMN. Then m�ABC � 2x and m�LMN � 2x. Therefore,m�ABC � m�LMN and �ABC � �LMN.
Hands-On Activity1–2.
3. The angles have equal measure and arecongruent:m�ABD = m�ABE � m�CBD � m�CBE � 90;�ABD � �ABE � �CBD � �CBE
4. and are perpendicular.ACDE
D
A CB
E
CBh
CAh
CFh
CDh
RSTg
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1-7 Triangles (pages 27–28)Writing About Mathematics
1. No. The statement does not specify that a triangleis a polygon. It is possible that three linesegments form a figure that is not a triangle.
2. The legs of an isosceles triangle must becongruent while the legs of a right triangle maynot be congruent.
Developing Skills
3. Legs: , Hypotenuse:4. Legs: , Hypotenuse:5. Legs: , Base:
Vertex angle: �N Base angles: �L, �M6. Legs: , Base:
Vertex angle: �R Base angles: �S, �T7–10. Answers will vary.
11. �BAC and �BCA12. �EDF and �DEF13. and
14. a. Answers will vary. Examples: is on ,
�ECB is adjacent to �BCG,b. Answers will vary. Examples: �FCE is
isosceles, AC � CB, C is the midpoint of .
Applying Skills15. 57°, 25° 16. 22 17. 21, 10, 2118. (1) With legs x � 5 and 4x � 11: 3, 7, 3
(2) With legs 3x � 13 and 4x � 11: 7, 19, 19
Review Exercises (pages 31–32)1. Set, point, line, plane2. It does not clearly define the class to which “line”
belongs.3. Collinear set of points4. Distance between two points on the real number
line5. Triangle 6. Bisector7. Opposite rays 8. Congruent angles9. Isosceles triangle 10. Line segment
11. 12. a. b. �S13. a. DE = 4, EF � 8, DF � 12
b. 3 c. 514. The midpoint of a line is a single point while a
bisector is any line that passes through themidpoint, and there are infinitely many linespassing through a point.
15. �BED and �AEC 16. �ADC17. and 18.19. �ABD and �DBC 20. 45°21. �BDA and �BDC
ACEDBE
RTLN
DE
FCG
ABg
AB
STSR
TSRTRS
LMNMNLJKLKJLCBABCA
22. No. In order for this equality to be true, A, B, andC would have to be collinear.
Exploration (pages 32–33)1. The points are the same. Each represents a
location on the surface.2. A geodesic in Euclidean geometry extends
infinitely in each direction. The geodesic on thesphere is a circle.
3. Two intersecting geodesics in Euclideangeometry form four angles. Two intersecting
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geodesics on a sphere form eight angles.The sides of these angles do not extend infinitely.
4. The sum of the measures of the angles of atriangle in Euclidean geometry is 180°.Drawing three geodesics on a sphere forms eight triangles. The sum of the measures of the angles of a triangle on sphere is greater than 180°.
Chapter 2. Logic
2-1 Sentences, Statements, and TruthValues (pages 39–41)
Writing About Mathematics1. A sentence in grammar need only have a subject
and a predicate. A mathematical sentence used inlogic must state a fact or a complete idea that canbe judged to be true or false.
2. Answers will vary.a. Example: Today is Wednesday.b. Example: I am six feet tall.c. Example: This country is in the southern
hemisphere.Developing Skills
3. Mathematical sentence4. Mathematical sentence5. Not a mathematical sentence6. Not a mathematical sentence7. Not a mathematical sentence8. Not a mathematical sentence9. Mathematical sentence
10. Not a mathematical sentence11. She 12. We 13. y14. x 15. This 16. He17. It 18. It 19. a. True20. a. Open 21. a. False 22. a. True
b. They23. a. Open 24. a. False 25. a. True
b. x26. a. False 27. {New York}28. {Nevada, Illinois}29. {Massachusetts, New York}30. {Nevada, Illinois, Massachusetts, Alaska,
New York}31. {Alaska} 32. {triangle} 33. �34. {triangle} 35. {square, rectangle}36. {square, rhombus}37. {square, rectangle, parallelogram, rhombus}38. {trapezoid}
39. {square, rectangle, parallelogram, rhombus,trapezoid}
40. The school does not have an auditorium.41. A stop sign is not painted red.42. The measure of an obtuse angle is not greater
than 90°.43. There are not 1,760 yards in a mile.44. Michigan is a city. 45. 14 � 2 � 16 � 1246. 3 � 4 � 5 � 6 47. Today is Wednesday.48. a. p 49. a. �p 50. a. q
b. True b. False b. False51. a. �q 52. a. r 53. a. �r
b. True b. Open b. Open54. a. �q 55. a. p 56. a. q
b. True b. True b. False57. a. Summer does not follow spring. b. False58. a. August is not a summer month. b. False59. a. A year does not have 12 months. b. False60. a. She does not like spring. b. Open61. a. Summer follows spring. b. True62. a. August is a summer month. b. True63. a. A year has 12 months. b. True64. a. She likes spring. b. Open
2-2 Conjunctions (pages 46–48)Writing About Mathematics
1. No.As shown in the following truth table, �(p ∧ q)and �p ∧ �q have different truth values.
p q (p ∧ q) �(p ∧ q) �p �q �p ∧ �q
T T T F F F F
T F F T F T F
F T F T T F F
F F F T T T T
2. p, q, and r must all be true because, for aconjunction to be true, each of the conjunctsmust be true.
Developing Skills3. p ∧ q 4. p ∧ r 5. �p6. �p ∧ r 7. q ∧ �r 8. �p ∧ �q9. �r ∧ �p 10. �r ∧ p 11. �(p ∧ q)
12. �(q ∧ �p) 13. False 14. False15. True 16. True 17. False18. False 19. True 20. False21. True 22. False 23. False24. True, True 25. True, False 26. False, True27. True 28. TrueApplying Skills29. True 30. False 31. Uncertain32. False 33. True 34. False35. Uncertain 36. False37. a. True 38. a. True
b. True b. Truec. False c. False
2-3 Disjunctions (pages 51–53)Writing About Mathematics
1. The truth set of the negation of a statementcontains the elements of the replacement set thatmake the statement false. The complement of aset contains the elements of the universe that arenot members of the set.
2. When two statements are connected by theinclusive or, the disjunction is true when one ofthe statements is true or both of the statementsare true. When two statements are connected bythe exclusive or, the disjunction is true only in thecase that exactly one of the statements is true.
Developing Skills3. a. m ∨ k 4. a. c ∨ l 5. a. c ∨ m
b. True b. True b. True6. a. �k ∨ �c 7. a. l ∨ k 8. a. l ∧ c
b. False b. True b. False9. a. �(c ∨ m) 10. a. �(�k ∨ l)
b. False b. True11. a. c ∧ k 12. a. (�c ∨ �m) ∧ l
b. True b. False13. a. Spring is a season or Halloween is a season.
b. True14. a. Breakfast is a meal and spring is a season.
b. True15. a. Spring is not a season or Halloween is a season.
b. False16. a. Breakfast is a meal and Halloween is not a
season.b. True
17. a. Breakfast is not a meal or spring is not a season.b. False
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18. a. It is not the case that spring is a season andHalloween is a season.
b. True19. a. It is not the case that breakfast is a meal or
spring is not a season.b. False
20. a. Breakfast is not a meal and spring is not aseason.
b. False21. True 22. True 23. False24. False, True 25. True, False 26. False27. TrueApplying Skills28. True 29. True 30. False31. Uncertain 32. True
2-4 Conditionals (pages 58–60)Writing About Mathematics
1. a. (1) Let p be 8 is divisible by 4.(True: 8 4 � 2)Let q be 8 is divisible by 2.(True: 8 2 � 4)p → q (T → T) is true.
(2) Let p be 6 is divisible by 4.(False: 6 4 � 1.5)Let q be 6 is divisible by 2.(True: 6 2 � 3)p → q (F → T) is true.
(3) Let p be 7 is divisible by 4.(False: 7 4 � 1.75)Let q be 7 is divisible by 2.(False: 7 2 � 3.5)p → q (F → F) is true.
b. No. Every number that is divisible by 4 is alsodivisible by 2 because the prime factorizationof 4 is 2 � 2.
2. The truth value is the same when p and q areboth true or when p and q are both false.
p q p → q q → p
T T T T
T F F T
F T T F
F F T T
Developing Skills3. a. A polygon is a square.
b. A polygon has four right angles.4. a. It is noon.
b. It is time for lunch.
5. a. You want help.b. You ask a friend for help.
6. a. You are not interrupted.b. You will finish more quickly.
7. a. The length of one side of a square is s.b. The perimeter of a square is 4s.
8. a. Many people work at a task.b. The task will be completed quickly.
9. a. 2x � 7 � 11b. x � 2
10. a. You do not get enough sleep.b. You will not be alert.
11. p → r 12. q → r 13. �p → �r14. �q → �r 15. q → p 16. p → r17. a. t → l 18. a. l → t 19. a. l → b
b. True b. False b. True20. a. �b → l 21. a. �t → �l 22. a. b → t
b. True b. False b. False23. a. (�b ∧ t) → l 24. a. (b ∧ �t) → �l
b. True b. False25. True 26. True 27. True28. True 29. True 30. True31. False32. a. If July is a warm month, then I work in my
garden.b. True
33. a. If I am busy every day, then I do not work inmy garden.
b. True34. a. If I like flowers, then I work in my garden.
b. True35. a. If I do not work in my garden, then July is not
a warm month.b. True
36. a. If July is a warm month and I like flowers,then I am busy every day.
b. False37. a. If July is a warm month and I work in my
garden, then I like flowers.b. True
38. a. If July is not a warm month, then I am busyevery day and I like flowers.
b. True39. a. If I work in my garden, then July is a warm
month or I am not busy every day.b. True
40. p → q 41. q → p 42. True, False43. True 44. True 45. True
Applying Skills46. True 47. False 48. Uncertain49. Uncertain 50. True
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2-5 Inverses, Converses, andContrapositives (pages 67–69)
Writing About Mathematics1. No. A conditional (p → q) can be true when p is
false and q is true. However, in the converse (q → p) of this case, F → T is false.
2. Yes. A conditional and its contrapositive arelogical equivalents, so they have the same truthvalues. Also, the converse of a statement and theinverse of a statement are logical equivalents, sothey have the same truth values.
Developing Skills3. a. �p → �q 4. a. �t → w
b. q → p b. �w → tc. �q → �p c. w → �t
5. a. m → �p 6. a. p → qb. p → �m b. �q → �pc. �p → m c. q → p
7. a. If 6 is not greater than 3, then �6 is notgreater than �3.
b. False c. True8. a. If a trapezoid is not a parallelogram, then a
trapezoid does not have exactly two pairs ofparallel sides.
b. True c. True9. a. If 3(3) � 9, then 3(4) � 12.
b. True c. True10. a. If 22 � 4, then 32 � 6.
b. False c. True11. If you eat Quirky oatmeal, then you lower your
cholesterol.12. If you get rich, then you enter the Grand Prize
drawing.13. If your hair curls, then you use Shiny’s hair
cream.14. If your pet grows three inches, then you feed him
Krazy Kibble.15. a. If a number is exactly divisible by 2, then the
number is even.b. True c. True
16. a. If 0.75 is rational, then it is an integer.b. True c. False
17. a. If 82 � 12 � 72, then 8 � 1 � 7.b. False c. True
18. a. If 4(5) � 6 � 14, then 4(5) � 6 � 20 � 6.b. True c. True
19. a. If Rochester is not the capital of New York,then Rochester is not a city.
b. False c. False20. a. If two angles are not supplementary, then they
do not form a linear pair.b. True c. True
21. a. If 4 � 3 � 2, then 3 � 2 � 1.b. False c. False
22. a. If a triangle is not equiangular, then all anglesof the triangle are not equal in measure.
b. True c. True23. a. If is not a counting number, then is not
greater than 0.b. False c. False
24. (4) 25. (3) 26. (2) 27. (3) 28. (2)Applying Skills29. a. If Derek lives in Nevada, then he lives in Las
Vegas; sometimes true.b. If Derek does not live in Las Vegas, then he
does not live in Nevada; sometimes true.c. If Derek does not live in Nevada, then he does
not live in Las Vegas; always true.30. a. If the probability of picking a red marble from
a bin is , then the bin contains 3 red marbles and 3 blue marbles; sometimes true.
b. If a bin does not contain 3 red marbles and 3blue marbles, then the probability of picking a red marble from the bin is not ; sometimestrue.
c. If the probability of picking a red marble from a bin is not , then the bin does not contain 3 red marbles and 3 blue marbles; always true.
31. a. If a polygon is an octagon, then it has eightsides; always true.
b. If a polygon does not have eight sides, then itis not an octagon; always true.
c. If a polygon is not an octagon, then it does nothave eight sides; always true.
32. a. If a garden grows vegetables, then it growscarrots; sometimes true.
b. If a garden does not grow carrots, then it doesnot grow vegetables; sometimes true.
c. If a garden does not grow vegetables, then itdoes not grow carrots; always true.
33. a. If the area of a rectangle is 48 square feet,then the dimensions of the rectangle are 8 feetby 6 feet; sometimes true.
b. If the dimensions of a rectangle are not 8 feetby 6 feet, then the area of the rectangle is not48 square feet; sometimes true.
c. If the area of a rectangle is not 48 square feet,then the dimensions of the rectangle are not 8feet by 6 feet; always true.
34. a. If a number is divisible by 7, then it has 7 as afactor; always true.
b. If a number does not have 7 as a factor, then itis not divisible by 7; always true.
12
12
12
12
12
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c. If a number is not divisible by 7, then it doesnot have 7 as a factor; always true.
2-6 Biconditionals (pages 73–74)Writing About Mathematics
1. A prime number is a whole number greater than1 if and only if the prime number has exactly twofactors.
2. Yes. A biconditional is false when the truth valueof p is not the same as the truth value of q. Whenthis is the case, the conditional p → q has adifferent truth value from its converse.
Developing Skills3. True 4. False 5. True6. True 7. True 8. True9. True 10. True 11. False
12. False 13. True 14. False15. False 16. TrueApplying Skills17. a. x is divisible by 2 and 3 if and only if x is
divisible by 6.b. Let x � 2; 2 is divisible by 2 but not by 3 or 6;
(p ∧ q) ↔ r is F ↔ F, which is true.Let x � 3; 3 is divisible by 3 but not by 2 or 6;(p ∧ q) ↔ r is F ↔ F, which is true.Let x � 5; 5 is not divisible by 2, 3, or 6;(p ∧ q) ↔ r is F ↔ F, which is true.Let x = 6; 6 is divisible by 2, 3, and 6;(p ∧ q) ↔ r is T ↔ T, which is true.Let x � 8; 8 is divisible by 2 but not by 3 or 6;(p ∧ q) ↔ r is F ↔ F, which is true.Let x � 9; 9 is divisible by 3 but not by 2 or 6;(p ∧ q) ↔ r is F ↔ F, which is true.Let x � 11; 11 is not divisible by 2, 3, or 6;(p ∧ q) ↔ r is F ↔ F, which is true.Let x � 12; 12 is divisible by 2, 3, and 6;(p ∧ q) ↔ r is T ↔ T, which is true.
c. Yes. The prime factorization of 6 is 2 � 3. If acounting number is divisible by both of thesefactors, then it is divisible by 6. If a countingnumber is not divisible by both of thesefactors, then it is not divisible by 6.
18. a. Yes b. Yes c. Yes d. Yes e. No19. A triangle is isosceles if and only if it has two
congruent sides; true.20. Two angles are both right angles if and only if
they are congruent; false.21. Today is Thursday if and only if tomorrow is not
Saturday; false.22. Today is not Friday if and only if tomorrow is not
Saturday; true.
2-7 The Laws of Logic (pages 79–80)Writing About Mathematics
1. Yes. If p → q is false, then q must be false. If q ∨ ris true, and q is false, then by the Law ofDisjunctive Inference, r must be true.
2. Yes. When �q is true, then q is false. If p ∨ q istrue, and q is false, then by the Law ofDisjunctive Inference, p must be true. Since p istrue and �q is true, p ∧ �q is also true.
Developing Skills3. True by the Law of Disjunctive Inference4. True by the Law of Detachment5. Cannot be found to be true or false because one
disjunct is known to be true6. True by the Law of Disjunctive Inference7. True by the Law of Detachment8. Cannot be found to be true or false because the
hypothesis is false9. True by the Law of Detachment
10. Cannot be found to be true or false because thehypothesis is false
11. False; the truth values of both statements in atrue biconditional must be the same.
12. Cannot be found to be true or false because theconclusion is true
13. True by applying the Law of Detachment to thecontrapositive of the conditional
14. Cannot be found to be true or false because thehypothesis is false
Applying Skills15. I take band; the Law of Detachment.
16. is irrational; the Law of Disjunctive
Inference.17. b = 4; the Law of Detachment.18. It is not 8:15 A.M.; the Law of Detachment can be
applied to the contrapositive of the conditional.19. No conclusion; a conditional with a true
conclusion may have either a true hypothesis or afalse hypothesis.
20. x is even and a prime; a biconditional with a trueconclusion has a true hypothesis.
21. It is February, and it is not summer; in a trueconjunction, both statements must be true. TheLaw of Disjunctive Inference may be applied tothe disjunction.
22. No conclusion; a conditional with a trueconclusion may have either a true hypothesis or afalse hypothesis.
23. Last Saturday we flew kites; the Law ofDisjunctive Inference.
24. I study computer science and I take welding; the
"6
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Law of Disjunctive Inference.25. Five has exactly two factors; a biconditional with
a true hypothesis has a true conclusion.26. x is not an integer greater than 2 and prime; in a
true conjunction, both statements must be true.The Law of Detachment can be applied to thecontrapositive of the conditional.
27. Ray DF does not bisect angle CDE; thehypothesis of the contrapositive is true, so itsconclusion must be true, that is, the hypothesis ofthe conditional must be false.
2-8 Drawing Conclusions (pages 83–85)Writing About Mathematics
1. Yes. Both members of the conjunction must betrue. Since p is false, q is true by the Law ofDisjunctive Inference. Also, r is true.
2. No. Both members of the conjunction must betrue. Since the conjunct r is false, the truth valuesof p and q cannot be determined.
Developing Skills3. True 4. True 5. True6. p is true and q is true.7. p is true and q is false.8. p is false, and q cannot be determined.9. True
Applying Skills10. Laura is an investment manager. Marta is a
doctor. Shanti is a lawyer.11. Alex is a plumber. Tony is a bookkeeper. Kevin is
a teacher.12. Gamma13. Ren had chicken pot pie. Logan had pizza.
Kadoogan had a ham sandwich.14. Zach plays baseball. Steve plays soccer. David
plays basketball.15. Taylor studies Latin. Melissa studies French.
Lauren studies Spanish.16. Augustus is a truthteller. Brutus is a liar. Caesar
is a liar.
Review Exercises (pages 87–89)In 1 and 2, answers will vary. Examples are given.
1. I go to school. I play basketball. If I go to school,then I play basketball. If I play basketball, then Igo to school. If I do not play basketball, then I goto school.
2. I do not play basketball. I do not go to school. If Ido not go to school, then I play basketball. I go toschool and I do not play basketball. I do not go toschool and I play basketball.
3. No. The biconditional p ↔ q is true when p → qand q → p are both false or both true. Thecontrapositives �q → �p and �p → �q have the same truth values as the conditionals because they are logically equivalent. Therefore,�p ↔ �q has the same truth values as p ↔ q.
4. a. At first you don’t succeed.b. You should try again.
5. a. You are late one more time.b. You will get a detention.
6. True 7. False 8. True9. False 10. True 11. False
12. True 13. 5 14. p15. True 16. False, True 17. True18. False 19. False 20. False21. True 22. True23. {6, 7, 8, 9, 10} 24. {1, 2, 3, 4, 5}25. {2, 3, 5, 7} 26. {1, 4, 6, 8, 9, 10}27. {2, 3, 5, 6, 7, 8, 9, 10} 28. {7}29. {2, 3, 5} 30. {1, 2, 3, 4, 5, 6, 8, 9, 10}31. {1, 2, 3, 4, 5, 7} 32. {7}33. a. If I do not live in Oregon, then I do not live in
the Northwest.b. If I live in the Northwest, then I live in Oregon.c. If I do not live in the Northwest, then I do not
live in Oregon.d. I live in Oregon if and only if I live in the
Northwest.34. �A is not the vertex angle of isosceles �ABC.35. Janice, Sarah, Laurie36. Judy, Janice, Sue, Sarah, Laurie37. 1st Virginia, 2nd Kay, 3rd Janice38. Peter plays the violin and tennis. Carlos plays the
cello and soccer. Ralph plays the flute andbaseball.
39. Answers will vary. Examples are given.a. (1) 6 (2) Not possible
(3) 4 (4) 5b. Since there does not exist a number that is
divisible by 6 but not divisible by 2, p → q isalways true.
40. Peter
Exploration (page 90)1. a. 1 and 3 2. a. 2 and 3
b. Answers will vary. b. Answers will vary.3. a. Answers will vary. b. Answers will vary.
c. Because the hypothesis is always false, theconditional is true whether the conclusion istrue or false.
235
Cumulative Review (pages 90–92)Part I
1. 3 2. 2 3. 3 4. 3 5. 16. 3 7. 4 8. 4 9. 3 10. 1
Part II11. “I win the ring toss game” must be true and “If I
win the ring toss game, then I get a goldfish”must be true. The Law of Detachment states thatif the conditional is true and the hypothesis istrue, then the conclusion is true.
12. 3; RS is half the length of RT. Since RS � ��5 � (�1)� � 4, T is located 4 units from S.T is located at �1 � 4 � 3.
Part III
13. refers to line AB. refers to the ray that
has endpoint A and contains point B. refers
to the line segment that has endpoints A and B.
AB refers to the length of the segment .
14. By definition, the measure of straight angle ABC
is 180°. bisects �ABC, so it splits the angle
into two congruent parts, each of which measures
180 2 � 90°. Since intersects �ABC to
form 90°, or right, angles, we can say that is
perpendicular to �ABC.
Part IV15. a(2b � 1) � a(2b) � a(1) Distributive
property� (a � 2) � b � a(1) Associate
property� (2 � a) � b � a(1) Commutative
property� 2ab � a Identity property
16. a. If x is divisible by 4, then x is divisible by 12.b. If x is not divisible by 12, then x is not divisible
by 4.c. The converse can be false: 8 is divisible by 4
but not by 12.The inverse is always true. Since 4 is a factorof 12, every number that is divisible by 12 isalso divisible by 4.
d. If x is not divisible by 4, then x is not divisibleby 12.
BDh
BDh
BDh
AB
AB
ABh
ABg
3-1 Inductive Reasoning (pages 95–97)Writing About Mathematics
1. No; triangles may contain a right or obtuse angle.2. Answers will vary. Example: 3 � 5 � �2
Developing Skills3. a. Answers will vary.
b. 90°c. The sum of the measures of the acute angles
of a right triangle is 90°.4. a. Answers will vary.
b. The quadrilaterals formed are parallelograms.c. A quadrilateral with vertices at the midpoints
of the sides of another quadrilateral is aparallelogram.
5. a. Answers will vary.b. The four triangles are equilateral.c. The four triangles formed by connecting
midpoints of the sides of an equilateraltriangle are equilateral.
6. Probably true; the experiment must includelines intersecting at acute, right, and obtuseangles.
7. Probably true; the experiment must includequadrilaterals with acute, obtuse, and rightangles.
8. False; pairs of different types of numbers must beincluded: natural numbers, positive and negativeintegers, rational numbers, and irrationalnumbers.
9. Probably true; draw parallelograms of differentsizes and angle measures.
10. False; draw quadrilaterals of different shapes:trapezoids, parallelograms, rectangles, squares,rhombuses, and others.
11. Probably true; draw triangles of different shapes:acute, right, obtuse, isosceles, equilateral, andscalene.
12. False; let n � 40, then (40)2 � 40 � 41 � 1,681.This is not a prime number since 1,681 � 412.
Applying Skills13. No 14. Yes 15. No16. No 17. Yes
Hands-On Activitya. 1, 4, and 9 are perfect squares.b. 1, 4, 9, 16, and 25. The numbers are perfect
squares.c. The cards facing up will all be perfect squares.
236
3-2 Definitions as Biconditionals (pages 99–100)
Writing About Mathematics1. Doug is correct. “If a container can be used to
carry food, then it is a lunchbox” is false. Forexample, a paper bag can be used to carry food.
2. a. , but 1 � �2 is false.b. Positive real numbers
Developing Skills3. a. If a triangle is equiangular, then it has three
congruent angles.b. If a triangle has three congruent angles, then it
is equiangular.c. A triangle is equiangular if and only if it has
three congruent angles.4. a. If a line or a subset of a line is a bisector of a
line segment, then it intersects the segment atits midpoint.
b. If a line or a subset of a line intersects a linesegment at its midpoint, then it is a bisector ofthe segment.
c. A line or a subset of a line is a bisector of aline segment if and only if it intersects thesegment at its midpoint.
5. a. If an angle is acute, then its degree measure isgreater than 0 and less than 90.
b. If an angle has a degree measure greater than0 and less than 90, then it is acute.
c. An angle is acute if and only if its degreemeasure is greater than 0 and less then 90.
6. a. If a triangle is obtuse, then it has one obtuseangle.
b. If a triangle has one obtuse angle, then it isobtuse.
c. A triangle is obtuse if and only if it has oneobtuse angle.
7. a. If a set of points is noncollinear, then the setcontains three or more points that do not alllie on the same straight line.
b. If a set of points contains three or more points that do not all lie on the same straight line, then the set of points isnoncollinear.
c. A set of points is noncollinear if and only ifthe set contains three or more points that donot all lie on the same straight line.
8. a. If a part of a line is a ray, then it consists of apoint on the line, called an endpoint, and allthe points on one side of the endpoint.
122 , 1
Chapter 3. Proving Statements in Geometry
b. If a part of a line consists of a point, called anendpoint, and all the points on one side of theendpoint, then it is a ray.
c. A part of a line is a ray if and only if it consistsof a point, called an endpoint, and all thepoints on one side of the endpoint.
9. A point B is between A and C if and only if A, B, and C are distinct collinear points and AB � BC � AC.
10. Two segments are congruent if and only if theyhave the same length.
11. A point is the midpoint of a line segment if andonly if it divides the segment into two congruentsegments.
12. A triangle is right if and only if it has a rightangle.
13. An angle is straight if and only if it is the union oftwo opposite rays and its degree measure is 180.
14. Two rays are opposite rays if and only if they aretwo rays of the same line with a commonendpoint and no other point in common.
Applying Skills15. A triangle is equilateral if and only if it has three
congruent sides.16. Two angles are congruent if and only if they have
the same measure.17. Two lines are perpendicular if and only if they
intersect to form right angles.
3-3 Deductive Reasoning (pages 103–105)Writing About Mathematics
1. Yes. An equilateral triangle has three congruentsides, so it satisfies the definition for an isoscelestriangle, which is that a triangle must have twocongruent sides.
2. Yes. If B is not between A and C, then AB � BC � AC.
Developing Skills3. b. If a ray bisects an angle, then it divides the
angle into two congruent angles.4. b. If a triangle is scalene, then no two of its sides
are congruent. If two segments are notcongruent, then they do not have the samemeasure.
5. b. If two lines are perpendicular, then theyintersect to form right angles.
6. b. If A, B, and C are distinct collinear points andAB � BC � AC, then B is between A and C.
7. b. If two lines are perpendicular, then theyintersect to form right angles.
c. �LMN is a right angle.
237
8. b. If a line bisects a segment, then it divides thesegment into two congruent segments.
c. or DF � FE9. b. If points P, Q, R are collinear with
PQ � QR � PR, then Q is between P and R.c. Q is between P and R.
10. b. If two rays are opposite rays, then they form astraight angle.
c. �TSR is a straight angle.11. b. If a point is the midpoint of a line segment,
then it divides the segment into two congruentsegments.
c. or LM � MN12. b. If the degree measure of an angle is greater
than 0 and less than 90, then the angle is acute.c. �a is acute.
13. a. Statements Reasons1. M is the midpoint 1. Given.
of .2. � 2. Definition of
midpoint.3. AM � MB 3. Definition of
congruentsegments.
b. We are given that M is the midpoint of .Therefore, M divides into twocongruent segments, and . Sincecongruent segments have the same length,AM � MB.
14. a. Statements Reasons1. RS � ST 1. Given.2. 2. Definition of
congruentsegments.
3. �RST is isosceles. 3. Definition ofisosceles triangle.
b. We are given �RST with RS � ST. Segmentswith the same length are congruent, so
. An isosceles triangle is a trianglewith two congruent sides. Therefore, �RST isisosceles.
15. a. Statements Reasons
1. bisects �ACB. 1. Given.2. �ACE � �ECB 2. Definition of angle
bisector.3. m�ACE 3. Definition of
� m�ECB congruent angles.
CEh
RS > ST
RS > ST
MBAMAMB
AMB
MBAMAMB
LM > MN
DF > FE
b. We are given that bisects �ACB. If a ray bisects an angle, then it divides the angle intotwo congruent angles, so �ACE � �ECB.Congruent angles have the same measure.Therefore, m�ACE � m�ECB.
16. Statements Reasons1. DE � EF 1. Given.2. 2. Definition of
congruent segments.3. E is the midpoint 3. Definition of
of . midpoint.
17. An obtuse angle is an angle whose degreemeasure is between 90 and 180. An obtusetriangle is triangle with an obtuse angle. Sincem�A � 90 and m�B � 90, �A and �B are notobtuse. Since �ABC is an obtuse triangle, �Cmust be obtuse. Therefore, m�C 90.
18. In isosceles �ABC with �A as the vertex angle,we can only infer that legs, and , arecongruent and have equal measure. The length of
is unknown.
3-4 Direct and Indirect Proofs (pages 108–109)
Writing About Mathematics
1. Yes. �ABC is made up of the rays and ,
which intersect at vertex B. B is a point on both
lines and . Therefore, and
intersect at B.
2. Yes. The notation means that A, B, and Care collinear and that B is between A and C.
Developing Skills3. b. Statements Reasons
1. LM � MN 1. Given.2. 2. Definition of
congruentsegments.
c. Statements Reasons1. is not 1. Assumption.
congruent to .2. LM � MN 2. It two segments are
not congruent, thenthey do not havethe same measure.
3. LM � MN 3. Given.4. 4. Contradiction.LM > MN
MNLM
LM > MN
ABCg
BCg
BAg
BCg
BAg
BCh
BAh
BC
ACAB
DEF
DE > EF
CEh
238
4. b. Statements Reasons1. �PQR is a 1. Given.
straight angle.2. m�PQR � 180 2. A straight angle
measures 180°.
c. Statements Reasons1. m�PQR � 180 1. Assumption.2. �PQR is not a 2. If an angle does not
straight angle. measure 180°, thenit is not a straightangle.
3. �PQR is a straight 3. Given.angle.
4. m�PQR � 180 4. Contradiction.
5. b. Statements Reasons1. �PQR is a straight 1. Given.
angle.
2. and are 2. Definition of opposite rays. straight angle.
c. Statements Reasons
1. and are 1. Assumption.not opposite rays.
2. �PQR is not a 2. If two rays are not straight angle. opposite rays, then
their union is not astraight angle.
3. �PQR is a straight 3. Given.angle.
4. and are 4. Contradiction.two opposite rays.
6. b. Statements Reasons
1. and are 1. Given.opposite rays.
2. P, Q, and R are on 2. Definition of the same line. opposite rays.
c. Statements Reasons1. P, Q, and R are not 1. Assumption.
on the same line.
2. and are 2. If two rays are not not opposite rays. of the same line,
then they are notopposite rays.
3. and are 3. Given.opposite rays.
4. P, Q, and R are on 4. Contradiction.the same line.
QRh
QPh
QRh
QPh
QRh
QPh
QRh
QPh
QRh
QPh
QRh
QPh
7. b. Statements Reasons1. �PQR is a straight 1. Given.
angle.
2. and are 2. Definition of opposite rays. straight angle.
3. P, Q, and R are on 3. Definition of the same line. opposite rays.
c. Statements Reasons1. P, Q, and R are not 1. Assumption.
on the same line.
2. and are 2. If two rays are not not opposite rays. of the same line,
then they are notopposite rays.
3. �PQR is not a 3. If two rays are not straight angle. opposite rays, then
their union is not astraight angle.
4. �PQR is a straight 4. Given.angle.
5. P, Q, and R are on 5. Contradiction.the same line.
8. b. Statements Reasons
1. bisects �DEF. 1. Given.2. �DEG � �GEF 2. Definition of angle
bisector.3. m�DEG 3. Definition of
� m�GEF congruent angles.
c. Statements Reasons1. m�DEG 1. Assumption.
� m�GEF2. �DEG is not 2. If two angles do not
congruent to have the same �GEF. measure, then they
are not congruent.
3. does not 3. If a ray does not bisect �DEF. divide an angle into
two congruentangles, then it doesnot bisect the angle.
4. bisects �DEF. 4. Given.5. m�DEG 5. Contradiction.
� m�GEF
9. The indirect proofs were longer than the direct proofs because the statements to beproved followed directly from definitions ingeometry.
EGh
EGh
EGh
QRh
QPh
QRh
QPh
239
10. Statements Reasons1. �DEG � �GEF 1. Assumption.
2. bisects �DEF. 2. Definition of angle bisector.
3. does not bisect 3. Given.�DEF.
4. �DEG is not 4. Contradiction.congruent to �GEF.
Applying Skills
11. a. Given: is perpendicular to .
b. Prove: is the bisector of �ABD.
c. Statements Reasons
1. ⊥ 1. Given.2. �ABD and �CBD 2. Definition of
are right angles. perpendicular lines.3. m�ABD � 90 and 3. The measure of a
m�CBD � 90 right angle is 90°.4. �ABD � �CBD 4. Congruent angles
have the samemeasure.
5. is the bisector 5. Definition of angle of �ABD. bisector.
12. a. Given: m�EFG � 180
b. Prove: and are not opposite rays.
c. Statements Reasons
1. and are 1. Assumption.opposite rays.
2. �EFG is a straight 2. Definition of angle. straight angle.
3. m�EFG � 180 3. The measure of astraight angle is180°.
4. m�EFG � 180 4. Given.
5. and are 5. Contradiction.not opposite rays.
3-5 Postulates,Theorems, and Proof (pages 113–115)
Writing About Mathematics1. Yes. An angle is congruent to itself. Congruent
angles can be expressed in either order. Anglescongruent to the same angle are congruent toeach other.
2. No. A line is not perpendicular to itself.Developing Skills
3. Reflexive property of equality
FGh
FEh
FGh
FEh
FGh
FEh
BDh
ABCg
BDh
BDh
ABCg
BDh
EGh
EGh
4. Transitive property of equality5. Symmetric property of equality6. Transitive property of equality
Applying Skills7. Statements Reasons
1. y � x � 4 1. Given.2. x � 4 � y 2. Symmetric property.3. y � 7 3. Given.4. x � 4 � 7 4. Transitive property.
8. Statements Reasons1. AB � BC � AC 1. Given.2. AC � AB � BC 2. Symmetric property.3. AB � BC � 12 3. Given.4. AC � 12 4. Transitive property.
9. Statements Reasons1. M is the midpoint 1. Given.
of .2. 2. Definition of
midpoint.3. LM � MN 3. Definition of
congruent segments.4. N is the midpoint 4. Given.
of .5. 5. Definition of
midpoint.6. MN � NP 6. Definition of
congruent segments.7. LM � NP 7. Transitive property.
10. Statements Reasons1. m�FGH � m�JGK 1. Given.2. m�HGJ � m�JGK 2. Given.3. m�FGH � m�HGJ 3. Transitive property.4. �FGH � �HGJ 4. Definition of
congruent angles.
5. is the bisector of 5. Definition of angle �FGJ. bisector.
11. It is not given that �ADB and �ADC are rightangles. This may not be assumed from thediagram.
Hands-On Activitya. Definition of an equilateral triangle, transitive
property of congruence, and definition ofcongruent segments
GHh
MN > NPMP
LM > MNLN
240
b. Statements Reasons1. �ABC is 1. Given.
equilateral.2. �BCD is 2. Given.
equilateral.3. 3. Definition of
equilateral triangle.4. 4. Definition of
equilateral triangle.5. 5. Transitive property.6. AB � CD 6. Definition of
congruent segments.
3-6 The Substitution Postulate (page 117)Writing About Mathematics
1. Yes. Segments congruent to the same segmentare congruent to each other.
2. No. While and are congruent, may belocated anywhere in relation to ; that is,may not be perpendicular to .
Developing Skills3. Statements Reasons
1. MT � 1. Given.2. RM � MT 2. Given.3. RM � 3. Substitution postulate.
4. Statements Reasons1. AD � DE � AE 1. Given.2. AD � EB 2. Given.3. EB � DE � AE 3. Substitution postulate.
5. Statements Reasons1. m�a � m�b � 180 1. Given.2. m�a � m�c 2. Given.3. m�c � m�b � 180 3. Substitution postulate.
6. Statements Reasons1. y � 7 � 2x 1. Given.2. y � x � 5 2. Given.3. x � 5 � 7 � 2x 3. Substitution postulate.
7. Statements Reasons1. 12 � x � y 1. Given.2. x � 8 2. Given.3. 12 � 8 � y 3. Substitution postulate.
12RT
12RT
STPQST
PQRSPQ
AB > CD
BC > CD
AB > BC
8. Statements Reason1. BC2 � AB2 � AC2 1. Given.2. AB � DE 2. Given.3. BC2 � DE2 � AC2 3. Substitution postulate.
9. Statements Reason1. AB � 1. Given.2. = EF 2. Given.3. AB � EF 3. Transitive property.4. = EF 4. Given.5. AB � 5. Transitive property.
10. Statements Reason1. m�Q � m�R 1. Given.
� m�S � 752. m�Q � m�S 2. Given.
� m�T3. m�T � m�R � 75 3. Substitution postulate.4. m�R � m�T 4. Given.
� m�U5. m�U � 75 5. Substitution postulate.
3-7 The Addition and SubtractionPostulates (pages 122–123)
Writing About Mathematics1. Cassie is incorrect. The definition of subtraction
in terms of addition depends on the definition ofnegative numbers. However, there is no conceptof a “negative line segment,” so this definitionwould be invalid for line segments.
2. a. Yes. We are given that m�ABC � 30,m�CBD � 45, and m�DBE � 15. Then,m�ABC � m�DBE � 30 � 15 � 45 �m�CBD.
b. No. Since �ABC and �DBE are not adjacentangles, �ABC � �DBE does not representan angle.
Developing Skills3. Statements Reasons
1. and 1. Given.2. AE � ED � AD 2. Partition postulate.
BF � FC � BC3. AE � BF and 3. Given.
ED � FC4. BF � FC � AD 4. Substitution postulate.5. AD � BC 5. Transitive property.
BFCAED
12GH
12GH
"CD"CD
241
4. Statements Reasons1. �SPR � �QRP 1. Given.
and �RPQ � �PRS2. �SPR � �RPQ 2. Addition postulate.
� �QRP � �PRS3. �SPR � �RPQ 3. Partition postulate.
� �SPQ4. �QRP � �PRS 4. Partition postulate.
� �QRS5. �SPQ � �QRS 5. Substitution
postulate.
5. Statements Reasons1. and 1. Given.
2. 2. Subtraction postulate.
3. 3. Partition postulate.
4. 4. Substitution postulate.
or
6. Statements Reasons1. 1. Given.2. 2. Reflexive property.3. 3. Addition postulate.
4. 4. Given.5. 5. Partition postulate.
6. 6. Substitutionpostulate.
7. Statements Reasons1. �LMN � �PMQ 1. Given.2. �NMQ � �NMQ 2. Reflexive property.3. �LMN � �NMQ 3. Addition postulate.
� �NMQ � �QMP4. �LMQ 4. Partition postulate.
� �LMN � �NMQ�NMP� �NMQ � �QMP
5. �LMQ � �NMP 5. Substitutionpostulate.
AC > BDBD > BC 1 CDAC > AB 1 BCABCD> BC 1 CDAB 1 BCBC > BCAB > CD
AM > BN2 NC> BN 1 NC
AM 1 MC 2 MCBC > BN1NCAC > AM1MC
> BC 2 NCAC 2 MC MC > NCAC > BC
8. Statements Reasons1. m�AEB � 180 and 1. Given.
m�CED � 1802. �AEB � �CED 2. Definition of
congruent angles.3. �AEB 3. Partition postulate.
� �AEC � �CEB�CED� �CEB � �BED
4. �AEC � �CEB 4. Substitution � �CEB � �BED postulate.
5. �CEB � �CEB 5. Reflexive property.6. �AEC � �BED 6. Subtraction postulate.7. m�AEC 7. Definition of
� m�BED congruent angles.
3-8 The Multiplication and DivisionPostulates (pages 126–127)
Writing About Mathematics1. The word “positive” is needed because only
positive quantities have square roots, and anumber has both a positive and a negative squareroot.
2. Yes. The conditions of the postulate require thatc � d. Therefore, if one of these variables is notequal to 0, then both are not equal to 0 and thefractions and are defined. If “c � 0” and “d � 0” are both removed, it is possible that c � 0 and d � 0, and that the fractions and areundefined.
Developing Skills3. Statements Reasons
1. AB � 1. Given.2. 4 � 4 2. Reflexive property.3. 4AB � BC 3. Multiplication
postulate.4. BC � CD 4. Given.5. 4AB � CD 5. Substitution
postulate.6. CD � 4AB 6. Symmetric property.
4. Statements Reasons1. m�a � 3m�b 1. Given.2. m�b � 20 2. Given.3. 3m�b � 60 3. Substitution
postulate.4. m�a � 60 4. Transitive property.
14BC
cd
ab
cd
ab
242
5. Statements Reasons1. 1. Given.2. 2MN = NP 2. Doubles of equal
quantities are equal.3. LM � 2MN 3. Given.4. LM � NP 4. Transitive property.5. 5. Definition of
congruent segments.
6. Statements Reasons1. 2(3a � 4) � 16 1. Given.2. 3a � 4 � 8 2. Halves of equal
quantities are equal.3. 4 � 4 3. Reflexive property.4. 3a � 12 4. Addition postulate.5. 3 � 3 5. Reflexive property.6. a � 4 6. Division postulate.
7. Statements Reasons1. 1. Given.2. 3 � 3 2. Reflexive property.3. 3QR � RS 3. Multiplication
postulate.4. PQ � 3QR 4. Given.5. PQ � RS 5. Transitive property.6. 6. Definition of
congruent segments.
Alternative Proof:
Statements Reasons1. PQ � 3QR 1. Given.
2. 2. Given.
3. PQ � or 3. Substitution
PQ � RS postulate.4. 4. Definition of
congruent segments.
Applying Skills8. Let m � Monday’s distance, t � Tuesday’s
distance, w � Wednesday’s distance, and f � Friday’s distance.We are given w � and w = . Then by the transitive property of equality. By themultiplication postulate, m � 2f. We are alsogiven that m � 2t, so by the transitive property ofequality, 2f � 2t. Since halves of equal quantitiesare equal, f � t or the distance Melanie walkedon Friday is equal to the distance she walked onTuesday.
13m 5 2
3f23f1
3m
PQ > RS
3 A 13RS B
QR 5 13RS
PQ > RS
QR 5 13RS
LM > NP
MN 5 12NP
9. Let L � library, P � post office, G � grocerystore, and B � bank.We are given , with LP � 4PG and GB � 3PG. By the partition postulate,PB � PG � GB. Using the substitutionpostulate, PB � PG � 3PG or PB � 4PG. By thetransitive property, LP � PB or the distancefrom the library to the post office is equal to thedistance from the post office to the bank.
10. To show that doubles of congruent segments arecongruent:
Given: B is the midpoint of , M is themidpoint of , and .
Prove:
Statements Reasons1. B is the midpoint 1. Given.
of , M is the midpoint of .
2. AC � 2AB, 2. Definition ofLN � 2LM midpoint.
3. 3. Given.4. AB � LM 4. Definition of
congruent segments.5. 2AB � 2LM 5. Doubles of equal
quantities are equal.6. AC � LM 6. Substitution postulate.7. 7. Definition of
congruent segments.
To show that doubles of congruent angles arecongruent:
Given: is the angle bisector of �ABD,is the angle bisector of �JKM, and �ABC � �JKL.
Prove: �ABD � �JKM
Statements Reasons1. is the angle 1. Given.
bisector of �ABD,
is the angle bisector of �JKM.
2. m�ABD 2. Definition of angle� 2m�ABC, bisector.m�JKM � 2m�JKL
3. �ABC � �JKL 3. Given.4. m�ABC � m�JKL 4. Definition of
congruent angles.5. 2m�ABC 5. Doubles of equal
� 2m�JKL quantities are equal.6. m�ABD � m�JKM 6. Substitution postulate.7. �ABD � �JKM 7. Definition of
congruent angles.
KLh
BCh
KLh
BCh
AC > LN
AB > LM
LNAC
AC > LNAB > LMLN
AC
LPGB
243
To show that halves of congruent segments arecongruent:
Given: B is the midpoint of , M is themidpoint of , and .
Prove:
Statements Reasons1. B is the midpoint of 1. Given.
, M is themidpoint of .
2. , 2. Definition of
midpoint.
3. 3. Given.4. AC � LN 4. Definition of
congruent segments.
5. 5. Halves of equal quantities are equal.
6. AB � LM 6. Substitutionpostulate.
7. 7. Definition ofcongruent segments.
Given: is the angle bisector of �ABD,
is the angle bisector of �JKM, and
�ABD � �JKM.Prove: �ABC � �JKL
Statements Reasons1. is the angle 1. Given.
bisector of �ABD,
is the angle
bisector of �JKM.
2. , 2. Definition of angle
bisector.3. �ABD � �JKM 3. Given.4. m�ABD � m�JKM 4. Definition of
congruent angles.
5. 5. Halves of equal quantities are equal.
6. m�ABC � m�JKL 6. Substitutionpostulate.
7. �ABC � �JKL 7. Definition ofcongruent angles.
Review Exercises (pages 129–130)1. a. If a triangle is obtuse, then it has one obtuse
angle.
5 12m/JKM
12m/ABD
m/JKL 5 12m/JKM
m/ABC 5 12m/ABD
KLh
BCh
KLh
BCh
AB > LM
12AC 5 1
2LN
AC > LN
LM 5 12LMN
AB 5 12ABC
LNAC
AB > LMAC > LNLN
AC
b. If a triangle has one obtuse angle, then it isobtuse.
c. A triangle is obtuse if and only if it has oneobtuse angle.
2. a. If two angles are congruent, then they havethe same measure.
b. If two angles have the same measure, thenthey are congruent.
c. Two angles are congruent if and only if theyhave the same measure.
3. a. If two lines are perpendicular, then theyintersect to form right angles.
b. If two lines intersect to form right angles, thenthey are perpendicular.
c. Two lines are perpendicular if and only if theyintersect to form right angles
4. We assume a postulate to be true without proof. A theorem is a statement that has beenproven.
5. Symmetric property6. No. The reflexive property does not hold true.
Example: 1 1 is false. The symmetric propertydoes not hold true. Example: If 2 1, then 1 2 is false. The transitive property holds true. Example: If 3 2 and 2 1, then 3 1 is true.
7.
Statements Reasons1. bisects 1. Given.
at M.2. M is the midpoint 2. Definition of
of . bisector.3. 3. Definition of midpoint.4. CM � MD 4. Definition of
congruent segments.
8.
Statements Reasons1. 1. Given.2. 2. Given.3. 3. Transitive property.RM > ST
MS > STRM > MS
R M S T
CM > MDCD
CDABg
B
A
M DC
244
9.
Statements Reasons1. 1. Given.2. 2. Partition postulate.
3. 3. Substitution postulate.
4. 4. Reflexive property.5. 5. Subtraction
postulate.
10.
Statements Reasons1. SQ � RP 1. Given.2. QR � QR 2. Reflexive property.3. SQ � QR 3. Addition postulate.
� QR � RP4. SR � SQ � QR 4. Partition postulate.
QP � QR � RP5. SR � QP 5. Substitution
postulate.
11.
Statements Reasons
1. bisects �ABD. 1. Given.2. �ABC � �CBD 2. Definition of angle
bisector.3. m�ABC � m�CBD 3. Definition of
congruent angles.4. m�CBD � m�PQR 4. Given.5. m�ABC � m�PQR 5. Transitive property.
12.
Statements Reasons
1. and bisect 1. Given.each other at E.
ABCD
A D
E
C B
BCh
A
B D
CP
QR
S Q R P
AB > CDBC > BC> BC 1 CD
AB 1 BCBD > BC 1 CDAC > AB 1 BCAC > BD
A B C D
(Cont.)
Statements Reasons2. E is the midpoint of 2. Definition of a
and of . bisector.
3. and 3. Definition of
midpoint.4. AE � EB and 4. Definition of
CE � ED congruentsegments.
5. AE � CE � EB � ED 5. Addition postulate.6. CE � BE 6. Given.7. AE � BE � CE � ED 7. Substitution
postulate.8. AB � AE � EB 8. Partition postulate.
CD � CE � ED9. AB � CD 9. Substitution
postulate.
13. A quantity may be substituted for its equal in any statement of equality. is not astatement of equality; therefore, the substitutionpostulate is not valid.
14. a. If a sequence of numbers or letters is apalindrome, then the sequence reads the samefrom left to right as from right to left.
b. If a sequence of numbers or letters reads thesame from left to right as from right to left,then the sequence is a palindrome.
c. A sequence of numbers or letters is apalindrome if and only if it reads the samefrom left to right as from right to left.
Exploration (page 130)Step 5 is not valid. Since a � b, the quantity a � b isequal to 0. Thus, we cannot apply the divisionpostulate.
Cumulative Review (pages 206–208)Part I
1. 3 2. 4 3. 4 4. 1 5. 46. 2 7. 3 8. 2 9. 4 10. 3
Part II11. Given
� 7 � �7 Reflexive property3x � 6 Addition postulate
x � 2 Division postulate
3x 1 7 5 13
CD ' BC
CE > ED
AE > EB
CDAB
245
12. Statements Reasons1. 1. Given.2. �EDF is a right angle. 2. Definition of
perpendicular lines.3. �DEF is a right 3. Definition of right
triangle. triangle.
Part III13. Statements Reasons
1. �ABC with D a point 1. Given.on .
2. AB � AD � DB 2. Partition postulate.3. AC � AD � DB 3. Given.4. AC � AB 4. Transitive property.5. 5. Definition of
congruentsegments.
6. �ABC is isosceles. 6. Definition ofisosceles triangle.
14. Yes. By the partition postulate, PQ � QR � PR.Thus,
Therefore, PQ � 4(5) � 3 � 17 and QR � 3(5) � 2 � 17. Thus, and Q isthe midpoint of .
Part IV15. Yes. By the partition postulate,
Therefore, m�ABD � 3(24) � 18 � 90 andm�DBC � 5(24) � 30 � 90. Thus,m�ABC � 90 � 90 � 180. An angle whosedegree measure is 180 is a straight angle.
16. a. Addition postulateb. Reflexive property of equalityc. Partition postulated. Substitution postulatee. Division postulate
x 5 24
8x 2 12 5 7x 1 12
3x 1 18 1 5x 2 30 5 7x 1 12
m/ABD 1 m/DBC 5 m/ABC
PQRPQ > QR
5 5 a
7a 2 1 5 8a 2 6
(4a 2 3) 1 (3a 1 2) 5 8a 2 6
AC > AB
AB
DE ' EF
Chapter 4. Congruence of Line Segments,Angles, and Triangles
246
7. a. Given: and Prove:
b. Statements Reasons1. 1. Given.2. 2. Given.3. 3. Transitive property of
congruence.
8. a. Given: 2EF � DB and Prove: EF � GH
b. Statements Reasons1. 2EF � DB 1. Given.2. EF � 2. Halves of equal
quantities are equal.3. 3. Given.4. EF � GH 4. Transitive property of
equality.
9. a. Given: CE � CF, CD � 2CE, CB � 2CFProve: CD � CB
b. Statements Reasons1. CD � 2CE 1. Given.2. CE � CF 2. Given.3. CD � 2CF 3. Substitution postulate.4. CB � 2CF 4. Given.5. CD � CB 5. Transitive property of
equality.
10. a. Given: RT � RS, ,Prove: RD � RE
b. Statements Reasons1. 1. Given.2. RT � RS 2. Given.3. 3. Substitution postulate.
4. 4. Given.5. RD � RE 5. Substitution postulate.
11. a. Given: AD � BE and BC � CD Prove: AC � CE
b. Statements Reasons1. AD � BE 1. Given.2. BC � CD 2. Given.3. AD � CD 3. Subtraction postulate.
� BE – BC4. AD � AC � CD, 4. Partition postulate.
BE � BC � CE5. AC � CE 5. Substitution postulate.
RE 5 12RS
RD 5 12RS
RD 5 12RT
RE 5 12RSRD 5 1
2RT
GH 5 12DB
12DB
GH 5 12DB
AB > EFEF > CDAB > CD
AB > EFEF > CDAB > CD4-1 Postulates of Lines, Line Segments,
and Angles (pages 139–140)Writing About Mathematics
1. Points E and F must name the same point.2. No. There is only one positive real number
that is the length of a given line segment. A different number is the length of a different line segment.
Developing Skills3. a. Given: AB � AD and DC � AD
Prove: AB � DC
b. Statements Reasons1. AB � AD 1. Given.2. DC � AD 2. Given.3. AB � DC 3. Transitive property
of equality.
4. a. Given: and Prove:
b. Statements Reasons1. 1. Given.2. 2. Given.3. 3. Transitive property
of equality.
5. a. Given: m�1 � m�2 � 90, m�A � m�2Prove: m�1 � m�A � 90
b. Statements Reasons1. m�1 � m�2 � 90 1. Given.2. m�A � m�2 2. Given.3. m�1 � m�A � 90 3. Substitution
postulate.
6. a. Given: m�A � m�B, m�1 � m�B,m�2 � m�A
Prove: m�1 � m�2
b. Statements Reasons1. m�1 � m�B 1. Given.2. m�A � m�B 2. Given.3. m�1 � m�A 3. Transitive property
of equality.4. m�2 � m�A 4. Given.5. m�1 � m�2 5. Transitive property
of equality.
AD > BDBD > CDAD > CD
AD > BDBD > CDAD > CD
12. a. Given: �CDB � �CBD and �ADB � �ABD
Prove: �CDA � �CBA
b. Statements Reasons1. �CDB � �CBD 1. Given.2. �ADB � �ABD 2. Given.3. �CDB � �ADB 3. Addition postulate.
� �CBD � �ABD4. �CDA 4. Partition postulate.
� �CDB � �BDA,�CBA� �CBD � �DBA
5. �CDA � �CBA 5. Substitutionpostulate.
13. m�QSR � 37 14. m�PST � 5015. m�PST � 45Applying Skills16. a. Answers will vary.
b. Given: A triangle is equilateral.Prove: The measures of the sides are equal.
c. If triangle is equilateral then all of its sides arecongruent. Congruent sides have equalmeasures.
17. a. Answers will vary.b. Given: Points E and F are distinct and two
lines intersect at E.Prove: The lines do not intersect at F.
c. Two lines can intersect at only one point.If they intersect at both E and F, then E and F must name the same point. But we are given that E and F are distinct. This is a con-tradiction, so the lines do not intersect at F.
18. a. Answers will vary.b. Given: A line through a vertex of a triangle is
perpendicular to the opposite side.Prove: The line separates the triangle into two
right triangles.c. A line through a vertex of a triangle that is
perpendicular to the opposite side separates itinto two triangles. Perpendicular linesintersect to form right angles, so each of thetriangles has a right angle. A triangle with aright angle is a right triangle.
19. a. Answers will vary.b. Given: Two points on a circle are endpoints of
a line segment.Prove: The length of the line segment is
shorter than the length of the portionof the circle with the same endpoints.
247
c. The shortest distance between two points isthe line segment between them. Hence, itslength is shorter than the length of the arc.
20. Yes. Through two distinct points, one and onlyone line can be drawn.
4-2 Using Postulates and Definitions inProofs (pages 142–144)
Writing About Mathematics
1. The symbol refers to a line segment withendpoints A and C with point B lying betweenthem. The symbol refers to a line segment with endpoints A and B with point C lyingbetween them. These symbols could not refer tosegments of the same line.
2. Yes. Since m�ABD � 90, then:m�ABD � m�DBC � 180 or m�DBC � 90m�DBC � m�CBE � 180 or m�CBE � 90m�ABE � m�CBE � 180 or m�ABE � 90
Developing Skills
3. a. Given: , bisects , and bisects .
Prove:
b. Statements Reasons1. 1. Given.2. AB � CB 2. Definition of
congruent segments.3. bisects , 3. Given.
bisects .4. D is the midpoint 4. Definition of
of , E is the bisector.midpoint of .
5. , 5. Definition of midpoint.
6. AD � CE 6. Halves of equalquantities are equal.
7. 7. Definition ofcongruent segments.
AD > CE
CE 5 12CB
AD 5 12AB
CBAB
CBFEABFD
AB > CB
AD > CECB
FEABFDAB > CB
ACB
ABC
4. a. Given: bisects �DCB, bisects �DAB,
and �DCB � �DAB.Prove: �CAB � �DCA
b. Statements Reasons1. �DCB � �DAB 1. Given.2. m�DCB 2. Definition of
� m�DAB congruent angles.
3. bisects �DCB, 3. Given.
bisects �DAB4. m�CAB 4. Definition of angle
� , bisector.m�DCA�
5. m�CAB 5. Halves of equal � m�DCA quantities are
equal.6. �CAB � �DCA 6. Definition of
congruent angles.
5. a. Given:Prove:
b. Statements Reasons1. 1. Given.2. , 2. Partition postulate.
3. 3. Substitutionpostulate.
4. 4. Substitutionpostulate.
6. a. Given: is a segment and AB � CD.Prove: AC � BD
b. Statements Reasons1. AB � CD 1. Given.2. AC � AB � BC,
BD � BC � CD 2. Partition postulate.3. AC � BC � CD 3. Substitution
postulate.4. AC � BD 4. Substitution
postulate.
7. a. Given: is a segment, B is the midpointof , and C is the midpoint of .
Prove: AB � BC � CD
b. Statements Reasons1. B is the midpoint 1. Given.
of .AC
BDACABCD
ABCD
AE > BD
AE > BE 1 DEBD > BE 1 EDAE > AD 1 DEAD > BE
AE > BDAD > BE
12m/DCB
12m/DAB
AChCAh
ACh
CAh
248
Statements ReasonsC is the midpoint of .
2. 2. Definition of midpoint.
3. AB � BC, 3. Definition of BC � CD congruent segments.
4. AB � BC � CD 4. Transitive propertyof equality.
8. a. Given: P and T are distinct points, P is themidpoint of .
Prove: T is not the midpoint of .b. Postulate 4.9 states that a line segment has
only one midpoint. P is the midpoint of ,and T does not name P. Therefore, T is not themidpoint.
9. a. Given:Prove:
b. Statements Reasons1. 1. Given.2. 2. Symmetric property.3. 3. Subtraction
postulate.4. , 4. Partition postulate.
5. 5. Substitutionpostulate.
10. a. Given: , E is the midpoint of ,and F is the midpoint of .
Prove:
b. Statements Reasons1. 1. Given.2. AD � BC 2. Definition of
congruent segments.3. E is the midpoint 3. Given.
of .F is the midpointof .
4. , 4. Definition of
midpoint.5. AE � FC 5. Halves of equal
quantities are equal.6. 6. Definition of
congruent segments.AE > FC
FC 5 12BC
AE 5 12AD
BC
AD
AD > BC
AE > FCBC
ADAD > BC
DE > BFBE > BF1FEDF > DE1EF> BE 2 FE
DF 2 EFEF > FEDF > BE
DE > BFDF > BE
RS
RSRS
BC > CDAB > BC,
BD
11. a. Given: , and and bisect eachother at E.
Prove:
b. Statements Reasons1. 1. Given.2. AC � DB 2. Definition of
congruent segments.3. and bisect 3. Given.
each other at E.4. E is the midpoint 4. Definition of
of and of . bisector.5. , 5. Definition of
midpoint.
6. AE � EB 6. Halves of equalquantities are equal.
7. 7. Definition ofcongruent segments.
12. a. Given: bisects �CDA, �3 � �1, �4 � �2Prove: �3 � �4
b. Statements Reasons
1. bisects �CDA. 1. Given.2. �1 � �2 2. Definition of angle
bisector.3. �3 � �1, 3. Given.
�4 � �24. �3 � �4 4. Substitution
postulate.
Applying Skills13. a. m�CDE � 114
b. m�FDE � 76c. m�CDG � 76d. �CDE is obtuse.
14. a. Answers will vary.b. AB � BC � 10c. The distance from A to is AB. The
distance from a point to a segment is thelength of a perpendicular from the point to thesegment. In this case, it is AB � 10.
15. a. Answers will vary.b. Given: and bisect each other at N,
, and RN � LN.
c. Statements Reasons1. and bisect 1. Given.
each other at N.2. N is the midpoint 2. Definition of
of and of . bisector.LMRS
LMRS
RS ' LMLMRS
BD
DRh
DRh
AE > EB
EB 5 12DB
AE 5 12AC
DBAC
DBAC
AC > DB
AE > EB
DBACAC > DB
249
3. RS � 2RN, 3. Definition of LM � 2LN midpoint.
4. RN � LN 4. Given.5. RS � LM 5. Doubles of equal
quantities are equal.
d. RS � LM � 48e. The distance from L to is LN. The distance
from a point to a segment is the length of aperpendicular from the point to the segment.In this case, it is LN � � 24.
4-3 Proving Theorems About Angles (pages 152–154)
Writing About Mathematics1. Yes. �AEC and �BED are supplements of
congruent angles �BEC and �AED. If twoangles are congruent, then their supplements arecongruent.
2. No. The converse is “If two angles aresupplementary, then the angles form a linearpair.” Supplementary angles do not need to beadjacent.
Developing Skills3. Statements Reasons
1. m�ACD 1. Given.� m�DCB � 90
2. �B � �DCA, 2. Given.�A � �DCB
3. m�B � m�DCB, 3. Definition of m�A � m�DCB congruent angles.
4. m�B � m�A � 90 4. Substitution postulate.5. �A and �B are 5. Definition of
complements. complementaryangles.
4. Statements Reasons1. �BFC and �BFG 1. Definition of a
form a linear pair. linear pair.�ADE and �ADCform a linear pair.
2. �BFC and �BFG 2. If two angles form aare supplementary. linear pair, then they�ADE and �ADC are supplementary.are also supple-mentary.
3. �ADC � �BFG 3. Given.4. �ADE � �BFG 4. If two angles are
congruent, then theirsupplements arecongruent.
12(48)
RS
5. Statements Reasons1. �ADB is a right 1. Given.
angle.2. 2. Given.3. �CEB is a right 3. Definition of
angle. perpendicular lines.4. �ADB � �CEB 4. If two angles are right
angles, then they arecongruent.
6. Statements Reasons1. �ABC and �BCD 1. Given.
are right angles.2. m�ABC � 90 and 2. Definition of right
m�BCD � 90 angles.3. m�EBA � m�EBC 3. Partition postulate.
� m�ABCm�ECB � m�ECD� m�BCD
4. m�EBA � m�EBC 4. Substitution postulate.� 90m�ECB � m�ECD� 90
5. �EBA and �EBC 5. Definition of are complements. complementary�ECB and �ECD angles.are complements.
6. �EBC � �ECB 6. Given.7. �EBA � �ECD 7. If two angles are
congruent, then theircomplements arecongruent.
7. Statements Reasons1. �ABC and �DBF 1. Given.
are right angles.2. m�ABC � 90 and 2. Definition of right
m�DBF � 90 angles.3. m�ABC � m�DBC 3. Partition postulate.
� m�ABCm�DBC � m�CBF� m�DBF
4. m�ABC � m�DBC 4. Substitution� 90 postulate.m�DBC � m�CBF� 90
5. �ABC and �DBC 5. Definition of are complements. complementary �DBC and m�CBF angles.are complements.
6. �DBC � �DBC 6. Reflexive property ofcongruence.
CE ' DBE
250
Statements Reasons7. �ABC � �CBF 7. If two angles are
congruent, then theircomplements arecongruent.
8. Statements Reasons
1. intersects 1. Given.at G.
2. �CGH and �DGE 2. Definition of are vertical angles. vertical angles.
3. �CGH � �DGE 3. Vertical angles arecongruent.
4. m�BHG 4. Given.� m�CGH
5. �BHG � �CGH 5. Definition ofcongruent angles.
6. �BHG � �DGE 6. Transitive property ofcongruency.
9. Statements Reasons1. �ABC and �DBC 1. Definition of a linear
form a linear pair. pair.�ACB and �ECBform a linear pair.
2. �ABC and �DBC 2. If two angles form aare supplementary. linear pair, then they�ACB and �ECB are supplementary.are supplementary.
3. �ABC � ACB 3. Given.4. �DBC � �ECB 4. If two angles are
congruent, theirsupplements arecongruent.
10. Statements Reasons
1. and 1. Given.intersect at E.
2. �AEC � �CEB 2. Given.
3. 3. If two angles intersect to formcongruent adjacentangles, then they areperpendicular.
11. Statements Reasons1. �ABC is a right 1. Given.
angle.2. m�ABC � 90 2. Definition of right
angles.3. m�DBA � m�CBD 3. Partition postulate.
� m�ABC
AEBg
' CEDg
CEDg
AEBg
DCg
EFg
(Cont.)
Statements Reasons4. m�DBA � m�CBD 4. Substitution postulate.
� 905. �DBA and �CBD 5. Definition of
are complementary. complementary angles.6. �BAC and �DBA 6. Given.
are complementary.7. �BAC � �CBD 7. If two angles are
complements of thesame angle, then theyare congruent.
12. m�AED � 70, m�DEB � 110, m�AEC � 11013. m�DEB � m�AEC � 120, m�AED
� m�CEB � 6014. m�BEC � m�DEA � 75,
m�AEC � m�DEB � 10515. m�CEB � m�DEA � 45,
m�BED � m�AEC � 135 16. a. y � 20, x � 30
b. m�RPL � 50, m�LPS � 130,m�MPS � 50
17. If two angles are straight angles, then they bothmeasure 180 degrees. Hence, they both have thesame measure. By definition of congruent angles,the two angles are congruent.
18. If two angles, �1 and �2, are both supplementsof the same angle, �3, then m�1 � m�3 � 180and m�2 � m�3 � 180. By the substitutionpostulate, m�1 � m�3 � m�2 � m�3. By thesubtraction postulate, m�1 � m�2 so �1 � �2.
19. If two angles, �1 and �2, are congruent and �3 and �4 are their respective supplements, thenm�1 � m�3 � 180 and m�2 � m�4 � 180.By the substitution postulate, m�1 � m�3 �m�2 � m�4. Since m�1 � m�2, by thesubtraction postulate, m�3 � m�4. Therefore,�3 � �4.
Applying Skills20. 120° 21. 40° 22. 130°23. 75° 24. 65° 25. 33°
4-4 Congruent Polygons andCorresponding Parts (page 157)
Writing About Mathematics1. No. One pair of congruent corresponding sides is
not sufficient to prove two triangles congruent.
251
2. No. �RST � �STR implies that �R � �S,�S � �T, �T � �R, , , and
. However, this is not necessarily true.
Developing Skills
3. , , , �A � �C,�ABD � �DBC, �ADB � �CDB
4. , , , �A � �C,�ADB � �CBD, �ABD � �CDB
5. , , , �A � �E,�ABD � �EBC, �D � �C
6. Reflexive property7. Symmetric property8. Transitive property9. Symmetric property
10. Symmetric property
4-5 Proving Triangles Congruent UsingSide, Angle, Side (pages 160–161)
Writing About Mathematics
1. No. Suppose that they are. Let t1 represent thetop of the first pole, b1 represent the point wherethe first pole meets the ground, and w1 representthe point where the wire of the first pole meetsthe ground. Define t2, b2, and w2 similarly for thesecond pole. If the poles have the same height,then . Since the poles are bothperpendicular to the ground, �t1b1w1 � �t2b2w2.Since the point where the wires meet the groundis 5 feet away from the foot of the poles,b1w1 � b2w2 � 5 feet, and so .Therefore, �t1b1w1 � �t2b2w2 by SAS. Inparticular, t1w1 � t2w2, but these represent thelengths of the wires. The assumption is false, andthe heights of the poles are unequal.
2. No. Two non-congruent triangles can have one,two, or three angles congruent. Also, two non-congruent triangles can have one side congruent,or two sides congruent if the angle between themis not congruent.
Developing Skills3. Yes 4. Yes 5. No6. Yes 7. No 8. Yes9. �D � �C
10.11. �EDA � �CDB
AD > DB
b1w1 > b2w2
t1b1 > t2b2
AD > ECBD > BCAB > EB
AB > DCDB > DBAD > CB
DA > DCBD > BDAB > CB
TR > RSST > TRRS > ST
Applying Skills12. a. Answers will vary;
Given: and bisect each other.b. Prove: �ABE � �CBD
c. Statements Reasons1. and 1. Given.
bisect each other.2. B is the midpoint 2. Definition of
of and of bisector..
3. , 3. Definition of midpoint.
4. �ABE � �DBC 4. Vertical angles arecongruent.
5. �ABE � �CBD 5. SAS.
13. a. Answers will vary;Given: ABCD is a quadrilateral; AB � CD,
BC � DA; �DAB, �ABC, �BCD, and�CDA are right angles.
b. Prove: Diagonal AC separates thequadrilateral into two congruenttriangles.
c. Statements Reasons1. AB � CD and 1. Given.
BC � DA2. and 2. Definition of
congruent segments.3. �ABC and �ADC 3. Given.
are right angles.4. �ABC � �ADC 4. Right angles are
congruent.5. �ABC � �CDA 5. SAS.
14. a. Answers will vary;Given: �PQR and �RQS form a linear pair,�PQR � �RQS, and PQ � QS.
b. Prove: �PQR � �RQS
c. Statements Reason1. PQ � QS 1. Given.2. 2. Definition of
congruent segments.3. �PQR � �RQS 3. Given.4. 4. Reflexive property.5. �PQR � �RQS 5. SAS.
RQ > RQ
PQ > QS
BC > DAAB > CD
DB > BEAB > BCDBE
ABC
DBEABC
DBEABC
252
4-6 Proving Triangles Congruent UsingAngle, Side, Angle (pages 163–164)
Writing About Mathematics1. Yes. Since the two triangles are congruent,
corresponding angles are congruent. Since one ofthe angles in the first triangle is a right angle, oneof the angles in the second triangle must also be aright angle.
2. a. No. The two triangles are not necessarilycongruent, as the figure shows.
b. �A cannot be congruent to �D. cannotbe congruent to .
Developing Skills3. Yes. Two angles and the side between them in
one triangle are congruent to two angles and theside between them in the other triangle.
4. Yes. Two angles and the side between them inone triangle are congruent to two angles and sidebetween them in the other triangle.
5. No. Only one pair of congruent angles are given.6. �ACD � �CAB7. �AED � �CEB8.
Applying Skills9. Statements Reasons
1. �E � �C, 1. Given.�EDA � �CDB
2. D is the midpoint 2. Given.of .
3. 3. Definition of midpoint.
4. �DAE � �DBC 4. ASA.
ED > DCEC
DB > DB
FECB
C
A B
D
F
E
10. Statements Reasons
1. bisects �ADC 1. Given.
and bisects �ABC.
2. �ADB � �BDC, 2. Definition of angle �CBD � �DBA bisector.
3. 3. Reflexive property ofcongruence.
4. �ABD � �CBD 4. ASA.
11. Statements Reasons1. 1. Given.2. �CDA � �BDA 2. If two lines are
perpendicular, thenthey intersect to formcongruent adjacentangles.
3. 3. Reflexive property ofcongruency.
4. bisects �BAC. 4. Given.5. �CAD � �BAD 5. Definition of angle
bisector.6. �ADC � �ADB 6. ASA.
12. Statements Reasons1. �DBC � �GFD 1. Given.2. �DBC and �ABD 2. Definition of a linear
form a linear pair; pair.�GFD and �DFEform a linear pair.
3. �DBC and �ABD 3. Linear pairs of anglesare supplementary; are supplementary.�GFD and �DFEare supplementary.
4. �DFE � �ABD 4. The supplements ofcongruent angles arecongruent.
5. bisects at D. 5. Given.6. D is the midpoint 6. Definition of bisector.
of .7. 7. Definition of
midpoint.8. �FDE � �ADB 8. Vertical angles are
congruent.9. �DFE � �DBA 9. ASA (steps 4, 7, 8).
4-7 Proving Triangles Congruent UsingSide, Side, Side (pages 166–167)
Writing About Mathematics1. Yes. If two triangles are congruent, then
corresponding sides are congruent. Therefore, if
FD > DBFB
FBAE
ADh
AD > AD
AD ' BC
DB > DB
BDh
DBh
253
the two sides of one triangle are congruent, thentwo sides of the other triangle must be alsocongruent and that triangle is isosceles.
2. Yes. If all corresponding sides are congruent,then by SSS, the two triangles are congruent.
Developing Skills3. Yes 4. Yes 5. No6. 7. 8.9. SAS
10. Not enough information11. ASA or SAS 12. SSS13. Not enough information14. Not enough information15. No. One triangle can be acute and the other can
be obtuse.
Applying Skills16. Statements Reasons
1. and 1. Given.
2. �ACE and �BDE 2. Perpendicular are right angles. segments meet to
form right angles.3. �ACE � �BDE 3. Right angles are
congruent.4. �AEC � �BED 4. Vertical angles are
congruent.5. bisects . 5. Given.6. E is the midpoint 6. Definition of bisector.
of .7. 7. Definition of
midpoint.8. �EAC � �EBD 8. ASA (steps 3, 7, 4).
CE > EDCED
CEDAEB
BD ' CEDAC ' CED
C
A B
D
F
E
FB > ECAD > BCBC > DC
17. Statements Reasons1. �ABC is equilateral. 1. Given.2. 2. In an equilateral
triangle, all sides arecongruent.
3. D is the midpoint 3. Given.of .
4. 4. Definition ofmidpoint.
5. 5. Reflexive property ofcongruence.
6. �ACD � �BCD 6. SSS.
18. Statements Reasons1. PS � QS 1. Assumption.2. 2. Definition of
congruent segments.3. 3. Given.4. �PSR � �QSR 4. If two lines are
perpendicular, thenthey intersect to formcongruent adjacentangles.
5. 5. Reflexive property ofcongruence.
6. �PSR � �QSR 6. SAS.7. �PSR is not 7. Given.
congruent to �QSR.8. PS � QS 8. Contradiction.
19. Statements Reasons1. AC � AB � 15 1. Given.2. 2. Definition of
congruent segments.3. 3. Reflexive property of
congruence.
4. is the bisector 4. Given.
of �BAC.5. �BAD � �CAD 5. Definition of angle
bisector.6. �ABD � �ACD 6. SAS.
Review Exercises (pages 169–170)1. The measure of the angle is 125° and its
complement is 55°.2. m�PMN � 48, m�LMP � 1323. m�K � m�Q � 634. B � F; two lines cannot intersect in more than
one point.
ADh
AD > AD
AC > AB
SR > SR
RS ' PQ
PS > QS
DC > DC
AD > DBAB
AC > BC
254
5. and coincide; at a given point on a given line, one and only one perpendicular can bedrawn to the line.
6. LM � MN � LR � RN; the shortest distancebetween two points is the length of the linesegment joining these two points.
7. Yes. An angle has one and only one bisector.8. Not necessarily. Two lines cannot intersect in
more than one point. However, we are only given
that and intersect at M. If they are
distinct lines, then P does not lie on . If they coincide, then it does.
9. Yes. At a given point on a given line, only oneperpendicular can be drawn to the line.
10. Statements Reasons1. m�A � m�D � 50 1. Given.2. �A � �D 2. Definition of
congruent angles.3. AB � DE � 10 cm 3. Given.4. 4. Definition of
congruent segments.5. m�B � m�E � 45 5. Given.6. �B � �E 6. Definition of
congruent angles.7. �ABC � �DEF 7. ASA.
11. Statements Reasons1. bisects . 1. Given.2. E is the midpoint 2. Definition of bisector.
of .3. 3. Definition of
midpoint.4. m�D � m�F 4. Given.5. �D � �F 5. Definition of
congruent angles.6. �DEH � �GEF 6. Vertical angles are
congruent.7. �GFE � �HDE 7. ASA.
12. Statements Reasons1. �B � �E 1. Assumption.2. 2. Given.3. 3. Given.4. �ABC � �DEF 4. SAS.5. �ABC is not 5. Given.
congruent to �DEF.6. �B is not congruent 6. Contradiction.
to �E.
BC > EFAB > DE
DE > EFDEF
DEFGEH
AB > DE
MNg
PMg
MNg
KMg
LMg
Exploration (page 170)1.
2. We are given that STUVWXYZ is a cube. In acube, all edges are congruent. Therefore,
. In a cube, all faces are squares.All angles in a square are right angles.Therefore, �STX, �UTX, and �STU are rightangles. Right angles are congruent, so �STX � �UTX � �STU. Then �STX � �UTX � �STU by SAS.
Cumulative Review (pages 170–173)Part I
1. 4 2. 2 3. 1 4. 45. 4 6. 2 7. 3 8. 19. 4 10. 2
Part II11. Statements Reasons
1. bisects at M. 1. Given.2. M is the midpoint 2. Definition of bisector.
of .3. 3. Definition of
midpoint.4. �R � �S 4. Given.5. �RMQ � �SMP 5. Vertical angles are
congruent.6. �RMQ � �SMP 6. ASA.
12. Statements Reasons1. DE � DG, 1. Given.
EF � GF2. , 2. Definition of
congruent segments.3. 3. Reflexive property of
congruence.4. �DEF � �DGF 4. SSS.
Part III13. Yes. Let w � “Our team wins,” c � “We
celebrate,” and p � “We practice.” Since both c ∨ p and �p are true, by the Law of DisjunctiveInference, c is true. Since �w → �c implies c → w, w is true by the Law of Detachment.Therefore, our team won.
DF > DFEF > GFDE > DG
RM > MSRS
RSPQ
ST > TX > UT
255
14. We have a system of two equations:
Solving for x in the second equation, x � 5y.Substituting for x in the first equation:
Therefore, y � 10 and x � 5(10) � 50.Part IV15. Let x � the measure of one angle, then
� the measure of the other angle.
Therefore, the angles measure 52° and 38°.16. a. �DML � �ENM � �FLN
b. We are given that �DEF is equilateral andequiangular. Thus, and
. We are also given that M, N,and L are the midpoints of the sides of thetriangle, and so ,
, and .Since halves of congruent segments are congruent,
. Therefore, �DML ��ENM � �FLN by SAS.
c. In part b, we showed that �DML � �ENM ��FLN. Since corresponding parts ofcongruent triangles are congruent,
. Therefore, �MNL isequilateral.
d. By the partition postulate,
m�DLF � m�DLM � m�MLN � m�FLNm�FNE � m�FNL � m�LNM � m�ENMm�DME � m�EMN � m�LMN � m�DML
Since all straight angles are congruent and�DLF, �FNE, and �DME are straightangles, m�DLF � m�FNE � m�DME.Thus, by the substitution postulate,
m�DLM � m�MLN � m�FLN� m�FNL � m�LNM � m�ENM� m�EMN � m�LMN � m�DML
LM > MN > LN
NF > FL > LDDM > ME > EN >
12FD 5 FL 5 LD1
2EF 5 EN 5 NF
12DE 5 DM 5 ME
DE > EF > FD/D > /E > /F
x 5 52
32x 5 78
x 1 x2 1 12 5 90
x2 1 12
y 5 10
18y 5 180
2(5y) 2 y 1 5y 1 4y 5 180
2x 2 y 5 x 1 4y
(2x 2 y) 1 (x 1 4y) 5 180
In part b, we showed that �DML � �ENM � �FLN. Sincecorresponding parts of congruent triangles arecongruent, m�DML � m�FLN � m�ENM andm�DLM � m�FNL � m�EMN. Thus, by thesubstitution postulate,
m�DLM � m�MLN � m�FLN � m�DLM � m�LNM � m�FLN� m�DLM � m�LMN � m�FLN
256
Therefore, by the subtraction postulate, we havethat m�MLN � m�LNM � m�LMN or �MLN � �LNM � �LMN, and so �NLM isequiangular.
Chapter 5. Congruence Based on Triangles
5-1 Line Segments Associated withTriangles (pages 177–178)
Writing About Mathematics
1. In �ABC, let �B be a right angle. The altitudedrawn to vertex B is . The altitudes drawn to vertices A and C are the sides of the triangle,
and , and therefore intersect with vertex B.
2. The altitudes intersect outside of the triangle.
Developing Skills3. a. Answers will vary.
b. �ACE � �BCEc.d. �ADC and �BDC are right angles.
4. a. Answers will vary.b. In acute triangles, the altitudes intersect inside
of the triangle. In obtuse triangles, thealtitudes intersect outside of the triangle. Inright triangles, the altitudes intersect at thevertex of the right angle.
5. a. Answers will vary.b. In all triangles, the angle bisectors intersect
inside of the triangle.6. a. Answers will vary.
b. In all triangles, the medians intersect inside ofthe triangle.
7.
Statements Reasons1. 1. Given.2. �P � �Q 2. Given.
PR > QR
R
P S Q
AF > BF
BCBA
DB
3. is a median. 3. Given.4. S is the midpoint 4. Definition of
of . median.5. 5. Definition of midpoint.6. �PSR � �QSR 6. SAS (steps 1, 2, 5).
8.
Statements Reasons1. is an angle 1. Given.
bisector of �DEF.2. bisects �DEF. 2. Definition of angle
bisector of a triangle.3. �DEG � �FEG 3. Definition of angle
bisector.4. 4. Reflexive property.5. is an altitude. 5. Given.6. 6. Definition of an
altitude of a triangle.7. �DGE � �FGE 7. If two lines are
perpendicular, thenthey intersect to formcongruent adjacentangles.
8. �DEF � �FEG 8. ASA (steps 3, 4, 7).
9. C
A D B
EG ' DFEGEG > EG
EG
EG
E
D G F
PS > QSPQ
RS
(Cont.)
Statements Reasons1. is an altitude 1. Assumption.
of �ABC.2. 2. Definition of an
altitude of a triangle.3. �ADC � �BDC 3. If two lines are
perpendicular, thenthey intersect to formcongruent adjacentangles.
4. 4. Reflexive property.5. is a median 5. Given.
of �ABC.6. D is the midpoint 6. Definition of a
of . median.7. 7. Definition of a
midpoint.8. �ADC � �BDC 8. SAS (steps 4, 3, 7).9. �ADC is not 9. Given.
congruent to �BDC.
10. is not an 10. Contradiction altitude of �ABC. (steps 8, 9).
Applying Skills
10. Assume that is an angle bisector of �LMO.Then bisects �LNM and �LNO � �MNO.
by the reflexive property ofcongruence. It is given that is an altitude of�LNM, so by definition, . Therefore,�LON and �MON are right angles andcongruent. �LON � �MON by SAS.
since corresponding parts of congruent triangles are congruent. However,�LMN is scalene, so this is a contradiction.Therefore, the assumption is false and is not an angle bisector.
11. Let the telephone pole and the two wires form apair of triangles with the ground, �ACD and �BCD. Side , the pole, is congruent to itself.If the pole is perpendicular to the ground, then
. This forms two congruent rightangles, �ADC and �BDC. Since D is themidpoint of , . Therefore, �ACD ��BCD by SAS. The wires, and , are corresponding parts of congruent triangles, so arecongruent and have equal length.
BCACAD > BDAB
CD ' AB
CD
NO
LN > MN
NO ' LMNO
NO > NONO
NO
CD
AD > BDAB
CDCD > CD
CD ' AB
CD
257
12. Area (�AMC) � �
Area (�BMC) � �
Since M is the midpoint of , AM � MB. By thesubstitution postulate, � .Therefore, the median divides �ABC into twotriangles with equal areas.
13. The farmer can determine the midpoint of one side of the piece of land and then build afence from that point to the opposite vertex ofthe plot.
5-2 Using Congruent Triangles to ProveLine Segments Congruent and AnglesCongruent (pages 179–181)
Writing About Mathematics1. Yes. Since �ABE � �DEF, �A � �D and
�B � �E. Congruent angles have equalmeasures, so if m�A � m�B � 90, then m�D � m�E � 90 and �D and �E are alsocomplementary.
2. Yes. The legs of an isosceles triangle arecongruent. Therefore, since the leg of oneisosceles triangle is congruent to the leg ofanother isosceles triangle, then the other legs arealso congruent. As the vertex angles arecongruent, the triangles are congruent by SAS.
Developing Skills3. a. �RPM � �SPM
b. SASc. �RPM � �SPM, �R � �S,
4. a. �ABD � �CDBb. SSSc. �ABD � �CDB, �BDA � �DBC,
�A � �C5. a. �ABE � �CDE
b. SASc. �A � �C, �B � �D,
6. a. �ABE � �CDEb. ASAc. �B � �D, ,
7. a. �PQR � �RSPb. SASc. �QRP � �SPR, �RPQ � �PRS,
8. a. �ABE � �CDEb. SSSc. �A � �C, �B � �D, �BEA � �DEC
PR > RP
BE > DEAB > CD
AB > CD
RP > SP
12MB(DC)1
2AM(DC)AB
12MB(DC)1
2bh
12AM(DC)1
2bh
9. Statements Reasons1. 1. Given.2. D is the midpoint 2. Given.
of .3. 3. Definition of midpoint.4. 4. Reflexive property.5. �CAD � �CBD 5. SSS (steps 1, 3, 4).6. �A � �B 6. Corresponding parts of
congruent triangles arecongruent.
10. Statements Reasons1. 1. Given.2. �CAB � �ACD 2. Given.3. 3. Reflexive property.4. �ABC � �CDA 4. SAS.5. 5. Corresponding parts of
congruent triangles arecongruent.
11. Statements Reasons1. and 1. Given.
bisect each other.2. E is the midpoint 2. Definition of bisector.
of and of .3. and 3. Definition of midpoint.
4. �AEC � �BED 4. Vertical angles arecongruent.
5. �AEC � �BED 5. SAS.6. �C � �D 6. Corresponding parts of
congruent angles arecongruent.
12. Statements Reasons1. KL � NM 1. Given.2. 2. Segments of equal
measure are congruent.3. �KLM and 3. Given.
�NML are right angles.
4. �KLM � �NML 4. Right angles arecongruent.
5. 5. Reflexive property.6. �KLM � �NML 6. SAS (steps 2, 4, 5).7. �K � �N 7. Corresponding parts of
congruent angles arecongruent.
13. a. x � 8 b. AB � 31c. BC � 32 d. EF � 32
14. a. a � 5 b. m�P � 35c. m�Q � 55 d. m�M � 55
LM > LM
KL > NM
CE > DEAE > BE
CDAB
CEDAEB
AD > CB
AC > AC
AB > CD
CD > CDAD > BD
AB
CA > CB
258
Applying Skills
15. a. Consider isosceles �ABC with vertex angle�B. The median forms �ABD and�CBD. By the definition of an isoscelestriangle, . By the reflexive propertyof congruence, . Since is amedian, . Therefore, �ABD ��CBD by SSS.
b. Since corresponding parts of congruenttriangles are congruent, �BDA � �BDC. Iftwo lines intersect to form congruent adjacentangles, then the two lines are perpendicular.Therefore, and �BDA and �BDCare right angles.
16. a. Consider quadrilateral ABCD in whichand . Since either
diagonal, or , is congruent to itself,�ABC � �CDA or �ABD � �CDB by SSS.
b. Since corresponding parts of congruenttriangles are congruent, �B � �D incongruent triangle ABC and CDA, or �A � �C in congruent triangles ABD and CDB.
17. a. Given: with ,, and
b. Statements Reasons1. , 1. Given.
,
2. �ABP � �DCP 2. SSS.3. �APB � �DPC 3. Corresponding
parts of congruenttriangles arecongruent.
c. Statements Reasons1. , 1. Given.
,
2. 2. Reflexive property.3. 3. Addition postulate.
> BC 1 CDAB 1 BCBC > BCAB > CDPB > PCPA > PD
PB > PCPA > PDAB > CD
A B C D
P
PB > PCPA > PDAB > BC > CDABCD
BDACAD > CBAB > CD
BD ' AC
AD > CDBDBD > BD
AB > CB
BD
Statements Reasons4. , 4. Partition postulate.
5. 5. Substitutionpostulate.
6. �ACP � �DBP 6. SSS (steps 1, 5).7. �APC � �DPB 7. Corresponding
parts of congruenttriangles arecongruent.
18. a. Statements Reasons1. is the median 1. Given.
from P in �LNP.2. M is the midpoint 2. Definition of
of . median.3. 3. Definition of
midpoint.4. is the altitude 4. Given.
from P in �LNP.5. 5. Definition of
altitude.6. �LMP � �NMP 6. If two lines are
perpendicular,then they intersectto form congruentadjacent angles.
7. 7. Reflexive property.
8. �LMP � �NMP 8. SAS (steps 3, 6, 7).9. 9. Corresponding
parts of congruenttriangles arecongruent.
10. �LNP is isosceles. 10. Definition ofisosceles triangle.
b. Statements Reasons1. �LNP is isosceles. 1. Proved in part a.2. 2. Definition of
isosceles triangle.3. is the median 3. Given.
from P to .4. 4. Definition of a
median.5. 5. Reflexive
property.6. �LMP � �NMP 6. SSS.
PM > PM
LM > MNLN
PM
LP > NP
PL > PN
PM > PM
PM ' LN
PM
LM > PMLN
PM
AC > BDBD > BC1CDAC > AB1BC
259
7. �LPM � �NPM 7. Correspondingangles of congruenttriangles arecongruent.
8. is the angle 8. Definition of an bisector from P in angle bisector.�LNP.
5-3 Isosceles and Equilateral Triangles(pages 184–185)
Writing About Mathematics
1. Yes. and are also corresponding sides ofcongruent triangles ACE and DBE.
2. By drawing median , Abel can only show thatand . Since �ABC is
equiangular, �A � �C. However, this is notsufficient for SSS, ASA, or SAS trianglecongruence. Therefore he cannot show that�ABD � �CBD or that �ABC is equilateral.
Developing Skills3. m�B � m�C � 304. m�S � 90, m�R � m�T � 455. x � 10, y � 30, m�D � m�E � m�F � 60
6. Statements Reasons1. � 1. Given.2. �CAB � �CBA 2. Isosceles triangle
theorem.3. �CAD � �CBE 3. If two angles are
congruent, then theirsupplements arecongruent.
7. Statements Reasons1. , 1. Given.
2. �BAC � �BCA, 2. Isosceles triangle �DAC � �DCA theorem.
3. �BAC � �DAC 3. Addition postulate.� �BCA � �DCA
4. �BAD � �BAC 4. Partition postulate.� �CAD,
�BCD � �BCA� �ACD
5. �BAD � �BCD 5. Substitution postulate.
AD > CDAB > CB
CBCA
BD > BDAD > CDBD
DEAE
PM
8. Statements Reasons1. , 1. Given.
2. 2. Subtraction postulate.
3. ,3. Partition postulate.
4. 4. Substitution postulate.5. �CDE � �CED 5. Isosceles triangle
theorem.
9. Statements Reasons1. 1. Given.2. �DAB � �DBA 2. Given.3. �CAB � �CBA 3. Isosceles triangle
theorem.4. �CAB � �DAB 4. Subtraction postulate.
� �CBA � �DBA5. �CAB � �CAD 5. Partition postulate.
� �DAB,�CBA � �CBD
� �DBA6. �CAD � �CBD 6. Substitution postulate.
10. a. Statements Reasons1. 1. Given.2. �A � �B 2. Isosceles triangle
theorem.3. AC � BC 3. Definition of
congruentsegments.
4. D is the midpoint 4. Given.of .E is the midpoint of .F is the midpoint of .
5. , 5. Definition of
, midpoint.
6. 6. Halves ofcongruent segmentsare congruent.
7. �ADF � �BEF 7. SAS (steps 5, 2, 6).
AD > BEAF > BF
BE 5 12BC
AD 5 12AC
AB
BC
AC
AC > BC
AC > BC
CD > CECB > CE 1 EBCA > CD 1 DA> CB 2 EBAC 2 DADA > EBAC > CB
260
b. Statements Reasons1–7. �ADF � �BEF 1–7. Steps from
part a.8. 8. Corresponding
parts of congruenttriangles arecongruent.
9. �ADF is isosceles. 9. Definition of anisosceles triangle.
11. Statements Reasons1. AB � AC 1. Given.2. 2. Definition of
congruent segments3. �B � �C 3. Isosceles triangle
theorem.4. BG � EC 4. Given.5. EG � EG 5. Reflexive property of
equality.6. BG � EG 6. Subtraction postulate.
� EC � EG7. BG � BE � EG, 7. Partition postulate.
EC � EG � GC8. BE � GC 8. Substitution postulate
(steps 6, 7).9. 9. Definition of
congruent segments.10. , 10. Given.
11. �BED � �CGF 11. If two lines areperpendicular, thenthey intersect to formcongruent adjacentangles.
12. �BED � �CGF 12. ASA (steps 3, 9, 11).13. 13. Corresponding parts
of congruent trianglesare congruent.
12. Statements Reasons1. 1. Assumption.2. �A � �D 2. Isosceles triangle
theorem.3. 3. Given.4. �AEB � �DEC 4. SAS.5. 5. Corresponding parts of
congruent triangles arecongruent.
EB > EC
AB > CD
AE > DE
BD > CF
CG ' GFBE ' DE
BE > GC
AB > AC
DF > EF
(Cont.)
Statements Reasons6. is not 6. Given.
congruent to .7. is not 7. Contradiction
congruent to . (steps 5, 6).
Applying Skills
13. Let �ABC be an isosceles triangle with vertexangle B and . Draw , the anglebisector of B. By the definition of angle bisector,�ABD � �CBD. Also, by the reflexive property of congruence. Therefore,�ABD � �CBD by SAS. Since correspondingparts of congruent triangles are congruent,�A � �C.
14. Let �ABC be an equilateral triangle. Bydefinition, . Let D, E, and F bethe midpoints of , , and , respectively.The midpoint divides a segment in half. Sincehalves of congruent segments are congruent,
.Every equilateral triangle is equiangular,so �A � �B � �C. By SAS, �AEF � �BDE� �CFD. Since corresponding parts ofcongruent triangles are congruent,
.
15. We are given that BC � DC, so . Bythe isosceles triangle theorem, �CBD � �CDB.Since we are given , �CBD and �FBC are supplements and �CDB and �GDC aresupplements. Supplements of congruent anglesare congruent. Therefore, �FBC � �GDC.
16. The contrapositive of the isosceles triangletheorem states if two angles of a triangle are notcongruent, then the sides opposite these anglesare not congruent. In �PQR, m�R � m�Q.Then �R is not congruent to �Q, so is notcongruent to and PQ � PR.
5-4 Using Two Pairs of Congruent Triangles(pages 187–188)
Writing About Mathematics
1. No. We cannot prove that �BCE � �BDE. It isgiven that , and we know that
by the reflexive property ofcongruence, but this is not sufficient to show thatthe triangles are congruent.
2. Yes. Consider �ABC � �DEF with medians and to sides and , respectively. ThisDEABFH
CG
EB > EBCB > DB
PRPQ
‹FBDG
›
BC > DC
DE > EF > FD
BD > CD > BE > AE > AF > CF
ABCABCAB > BC > CA
BD > BD
BDAB > CB
DEAE
ECEB
261
creates triangles �ACG and �DFH. Sincecorresponding parts of congruent triangles arecongruent, , and �A � �D. Themedians cut and in half. Halves ofcongruent segments are congruent. Therefore,
and �ACG � �DFH by SAS.Therefore, since corresponding partsof congruent triangles are congruent.
Developing Skills3. Statements Reasons
1. �ABC � �DEF 1. Given.2. , 2. Corresponding parts of
�A � �D, congruent triangles are congruent.
3. AC � DF, 3. Definition of congruentAB � DE segments.
4. M is the midpoint 4. Given.of , N is the midpoint of .
5. , 5. Definition of midpoint.
6. 6. Halves of congruentsegments arecongruent.
7. �AMC � �DNF 7. SAS (steps 2, 6).
4. Statements Reasons1. �ABC � �DEF 1. Given.2. , 2. Corresponding
�A � �D, parts of congruent �ACB � �DFE triangles are
congruent.3. bisects �ACB, 3. Given.
bisects �DFE.4. m�ACG 4. Definition of angle
� , bisector.
m�DFH�
5. �ACG � �DFH 5. Halves of congruentangles are congruent.
6. �ACG � �DFH 6. ASA (steps 2, 5).7. 7. Corresponding parts
of congruent trianglesare congruent.
CG > FH
12m/DFE
12m/ACB
FHCG
AC > DF
AM > DN
DN 5 12DE
AM 5 12AB
DEAB
AB > DE
AC > DF
CG > FHAG > DH
DEABAC > DF
5. Statements Reasons1. and 1. Given.
bisect each other.2. E is the midpoint 2. Definition of bisector.
of and of .
3. , 3. Definition of midpoint.
4. �AEB � �CED 4. Vertical angles arecongruent.
5. �ABE � �CED 5. SAS.6. �A � �C 6. Corresponding parts
of congruent trianglesare congruent.
7. �AEG � �CEF 7. Vertical angles arecongruent.
8. �AEG � �CEF 8. ASA (steps 6, 3, 7).9. 9. Corresponding parts
of congruent trianglesare congruent.
10. E is the midpoint 10. Definition of of . midpoint.
6. Statements Reasons1. �AME � �BMF 1. Given.2. , 2. Corresponding parts
�AEM � �BFM of congruent trianglesare congruent.
3. �AEM and �AED 3. If two angles form a are supplements, linear pair, then they�BFM and �BFC are supplementary.are supplements.
4. �AED � �BFC 4. Supplements ofcongruent angles arecongruent.
5. 5. Given.6. �AED � �BFC 6. SAS (steps 2, 4, 5).7. 7. Corresponding parts
of congruent trianglesare congruent.
7. Statements Reasons1. 1. Given.
2. bisects �CBA. 2. Given.3. �CBD � �ABD 3. Definition of angle
bisector.4. 4. Reflexive property.5. �CBD � �ABD 5. SAS.6. �CDB � �CDA 6. Corresponding parts
of congruent trianglesare congruent.
7. bisects �CDA. 7. Definition of anglebisector.
DBh
BD > BD
BDhBC > BA
AD > BC
DE > CF
AE > BF
FEG
FE > EG
DE > BEAE > CDDEB
AEC
DEBAEC
262
8. Statements Reasons1. RP � RQ 1. Given.2. 2. Definition of
congruent segments.3. �RPT � �RQT 3. Isosceles triangle
theorem.4. 4. Reflexive property.5. �RPT � �RQT 5. SAS.6. �RTP � �RTQ 6. Corresponding parts
of congruent trianglesare congruent.
7. 7. If two lines intersectto form congruentadjacent angles, thenthey areperpendicular.
Applying Skills
9. We are given that in quadrilateral ABCD, AB �
CD and BC � DA. It follows that and. Also, by the reflexive
property of congruence. �ABD � CDB by SSS.Corresponding parts of congruent triangles arecongruent, so �ABD � �CDB. Since M is the midpoint of , . �BME � �DMFbecause vertical angles are congruent. Therefore,�BME � �DMF by ASA. Corresponding partsof congruent triangle are congruent, so
. By definition, M is the midpoint of, so bisects at M.
10. a. We are given that l intersects at M, and Pand S are points on l on the same side of .Also, PA � PB and SA � SB, so and . by the reflexive property of congruence, so �PAS � �PBS bySSS. Corresponding parts of congruenttriangles are congruent, so �APM � �BPM.
by the reflexive property of congruence, so �PAM � �PBM by SAS.Corresponding parts of congruent trianglesare congruent, so �PMA � �PMB. If twolines intersect to form congruent adjacent angles, then they are congruent, so l ⊥ .
b. Same as part a.c. Both methods require proving two pairs of
triangles congruent and then usingcorresponding angles to showperpendicularity.
AB
PM > PM
PS > PSSA > SBPA > PB
ABAB
EMFBMDEMFEM > MF
BM > DMBD
BD > BDBC > DAAB > CD
RT ' PQ
RT > RT
RP > RQ
5-5 Proving Overlapping TrianglesCongruent (pages 189–190)
Writing About Mathematics
1. Yes. By the reflexive property of congruence,�B � �B, , and �C � �C, so �DBC � �ECB by SAS. Then the medians tothe legs, and , are corresponding parts ofcongruent triangles and, therefore, congruent.
2. The triangle must be equilateral.Developing Skills
3. Statements Reasons1. , , 1. Given.
and �A and �B are right angles.
2. �A � �B 2. Right angles arecongruent.
3. 3. Reflexive property.4. 4. Addition property.5. , 5. Partition postulate.
6. 6. Substitutionpostulate.
7. �DAF � �CBE 7. SAS (steps 1, 2, 6).8. 8. Corresponding
parts of congruenttriangles arecongruent.
4. Statements Reasons1. 1. Given.2. �S � �S 2. Reflexive property.3. 3. Given.4. 4. Subtraction
postulate.5. 5. Partition postulate.
6. 6. Substitutionpostulate.
7. �SRQ � �STP 7. SAS (steps 1, 2, 6).8. �R � �T 8. Corresponding
parts of congruenttriangles arecongruent.
5. Statements Reasons1. , , 1. Given.
2. �DAB and �CBA 2. Perpendicular lines are right angles. intersect to form
right angles.
CB ' ABDA ' ABDA > CB
SP > SQSQT > SQ 1 QTSPR > SP 1 PR> SQT 2 QTSPR 2 PRPR > QT
SPR > SQT
DF > CE
AF > BEEB > EF 1 FBAF > AE 1 EFAE 1 EF > EF 1 FBEF > EF
AE > FBDA > CB
BECD
BC > BC
263
Statements Reasons3. �DAB � �CBA 3. Right angles are
congruent.4. 4. Reflexive property.5. �DAB � �CBA 5. SAS (steps 1, 3, 4).6. 6. Corresponding
parts of congruenttriangles arecongruent.
6. Statements Reasons1. �BAE � �CBF, 1. Given.
�BCE � �CDF,
2. 2. Reflexive property.3. 3. Addition postulate.4. 4. Partition postulate.
5. 5. Substitutionpostulate.
6. �ACE � �BDF 6. ASA (steps 1, 5).7. , �E � �F 7. Corresponding
parts of congruenttriangles arecongruent.
7. Statements Reasons1. M is the midpoint of 1. Given.
, N is the midpoint of .
2. , 2. Definition ofmidpoint.
3. 3. Given.4. 4. Substitution
postulate.5. �R � �S 5. Isosceles triangle
theorem.6. 6. Reflexive property.7. �RSM � �SRN 7. SAS (steps 4, 5, 6).8. 8. Corresponding
parts of congruenttriangles arecongruent.
RN > SM
RS > RS
MR > NSTM > TN
TN > NSTM > MRTS
TR
AE > BF
AC > BDBD > BC 1 CDAC > AB 1 BCAB 1 BC > BC 1 CDBC > BCAB > CD
AC > BD
AB > AB
8. Statements Reasons1. , 1. Given.2. 2. Addition postulate.
3. 3. Partition postulate.
4. 4. Substitutionpostulate.
5. �B � �B 5. Reflexive property.6. �ABE � �CBD 6. SAS (steps 1, 5, 4).7. �BDC � �BEA 7. Corresponding
parts of congruenttriangles arecongruent.
8. �ADC � �CEA 8. Supplements ofcongruent anglesare congruent.
Applying Skills
9. Let �ABC be an isosceles triangle with vertexangle B and . Let the angle bisectors ofthe base angles be and . By the isosceles triangle theorem, �BAC � �BCA. Since anangle bisector splits an angle in half, and halvesof congruent angles are congruent,�BAE � �BCD. By the reflexive property ofcongruence, �B � �B. Therefore,�ABE � �CBD by ASA. Becausecorresponding parts of congruent triangles are congruent, the angle bisectors, and , arecongruent.
10. Let �ABC be an isosceles triangle with vertex B,with D, E, and F the midpoints of , , and
, respectively. By the definition of midpoint,. Since �ABC is isosceles, .
The midpoints D and E split these sides in half.Since halves of congruent segments arecongruent, so . By the isosceles triangletheorem, �A � �C. Then �ADF � �CEF bySAS. Corresponding parts of congruent trianglesare congruent, so . Since and are congruent sides of �DEF, the triangle isisosceles.
11. We will use an indirect proof. Let be themedian and the altitude to side in scalenetriangle �ABC. Then D is the midpoint of , so
. Also, �ADB and �ADC are rightangles and therefore congruent. By the reflexiveproperty of congruence, . Therefore,�ABD � �ACD by SAS. Corresponding parts
AD > AD
BD > CDBC
BCAD
EFDFDF > EF
AD > CE
AB > BCAF > CFCA
BCAB
CDAE
CDAEAB > CB
AB > CBCB > CE 1 EBAB > AD 1 DB5 CE 1 EBAD 1 DB
DB > EBAD > CE
264
of congruent triangles are congruent, so �B � �C. However, that contradicts thestatement that �ABC is scalene, so is not an altitude.
5-6 Perpendicular Bisector of a LineSegment (pages 195–196)
Writing About Mathematics1. Perpendicular lines are two lines that intersect to
form right angles. If two lines intersect to formcongruent adjacent angles, then they areperpendicular. If two points are each equidistantfrom the endpoints of a line segment, then thetwo points determine the perpendicular bisectorof a line.
2. Corollary 5.1b and Example 1 are converses ofeach other. Corollary 5.1b can be rewritten as, “Ifa triangle is isosceles, then the vertex lies on theperpendicular bisector of the base.” Example 1can be rewritten as, “If a point lies on theperpendicular bisector of a line segment, then thepoint is the vertex of an isosceles triangle formedby the endpoints of the line segment and thepoint.”
Developing Skills3. Statements Reasons
1. is the ⊥ bisector 1. Given.of .
2. �RSA and �RSB 2. Definition of are right angles. perpendicular lines.
3. �RSA � �RSB 3. Right angles arecongruent.
4. S is the midpoint of 4. Definition of . bisector.
5. 5. Definition ofmidpoint.
6. 6. Reflexive property.7. �RSA � �RSB 7. SAS (steps 5, 3, 6).8. �ARS � �BRS 8. Corresponding parts
of congruent anglesare congruent.
4. Statements Reasons1. PR � PS, QR � QS 1. Given.2. is the ⊥ bisector 2. If two points are
of . each equidistant from the endpointsof a line segment,then the pointsdetermine the ⊥bisector of the linesegment.
RSPQ
RS > RS
AS > BSASB
ASBRS
AD
(Cont.)
Statements Reasons3. RT � ST 3. If a point is on the
perpendicular bisectorof a line segment, thenit is equidistant fromthe endpoints of theline segment.
5. Statements Reasons1. AB � BC 1. Given.
� CD � DA2. is the ⊥ 2. If two points are
bisector of , each equidistant is the ⊥ from the endpoints
bisector of . of a line segment, thenthe points determinethe ⊥ bisector of theline segment.
6. Statements Reasons1. �ACE � �BCE 1. Given.2. 2. Reflexive property.3. �AED � �BED 3. Given.4. �AEC � �BEC 4. Supplements of
congruent angles arecongruent.
5. �AEC � �BEC 5. ASA (steps 1, 2, 4).6. , 6. Corresponding parts
of congruent trianglesare congruent.
7. AC � BC, 7. Definition of AE � BE congruent segments.
8. is the ⊥ 8. If two points are bisector of . each equidistant
from the endpoints ofa line segment, thenthe points determinethe ⊥ bisector of theline segment.
Applying Skills
7. Let intersect at M. We are given that PA � PB and TA � TB, so and
. By the reflexive property ofcongruence, .Therefore, �APT � �BPTby SSS. Corresponding parts of congruenttriangles are congruent, so �APM � �BPM.
by the reflexive property ofcongruence, so �APM � �BPM by SAS.Therefore, since corresponding partsof congruent triangles are congruent. By
definition, M is the midpoint of , so is a PTg
AB
AM > MB
PM > PM
PT > PTTA > TB
PA > PBABPT
g
ADBCED
AE > BEAC > BC
CE > CE
ACBD
BDAC
265
bisector.Also, adjacent angles �AMP and �BMPare congruent because they are correspondingparts of congruent triangles. Therefore,
.
8. Let bisect �B in �ABC. Then �ABD � �CBD. Also, let . Then �ADB and �CDB are right angles andcongruent. By the reflexive property of congruence, . �ADB � �CDB byASA. Corresponding parts of congruent trianglesare congruent, so . Therefore, �ABCis isosceles.
9. Since the line forms a linear pair of angles withthe side of the triangle, the sum of the anglemeasures should be 180.
However, � 27 � 69 and � 15 � 111.Therefore, these two angles are not right angles,and so the line through the vertex and theopposite side are not perpendicular.
5-7 Basic Constructions (page 202–203)Writing About Mathematics
1. The perpendicular bisector to a side of a triangleis an altitude if it goes through the oppositevertex. The altitude from a vertex is aperpendicular bisector if it goes through themidpoint of the opposite side. A perpendicularbisector need not contain the opposite vertex andan altitude need not contain the midpoint of theopposite side.
2. In both constructions, a line is drawn through twopoints equidistant from the endpoints of asegment. However, in Construction 3, theendpoints of the segment are given and thepoints equidistant from them are constructed,while in Construction 6, a point is given and theendpoints of the segment and a second pointequidistant from those endpoints are constructed.
Developing Skills3. a. Use Construction 1.
b. Use Construction 1. Then, on the ray whoseendpoint is newly constructed, repeatConstruction 1 to create a second, non-overlapping segment.
c. Use Construction 3.d. Follow part b. Then use Construction 3 to
bisect the second line segment constructed.
32(84)1
2(84)a 5 84
A 12a 1 27 B 1 A 3
2a 2 15 B 5 180
AB > CB
BD > BD
BD ' ABBD
AB ' PTg
4. a. Use Construction 2.b. Use Construction 2. Then, on the newly
constructed side, repeat Construction 2 tocreate a second, adjacent angle.
c. Use Construction 4.d. Follow part b twice. Then, use Construction 4
to bisect the most recently constructed angle.5. a. Use Construction 1 with .
b. Use Construction 1 with . Open the compass to a radius equal to BC. With thepoint on an endpoint of the newly constructedsegment, draw an arc above the segment.Open the compass to a radius equal to CD.With the point on the other endpoint of thenewly constructed segment, draw a second arcabove the segment intersecting the first arc.Draw a segment from each endpoint to thisintersection.
c. Follow part b, substituting BC for CD.d. Follow part b using only .
6. a. Use Construction 4.b. Use Construction 4. Then, use Construction 4
to bisect one of the congruent halves.c. Same as part b.
7. a. Use Construction 3 with . Draw a segmentfrom C to the newly constructed midpoint.
b. Use Construction 6 with and point C.c. Use Construction 6 with and point A.d. Use Construction 4 with �A.
8. a. Answers will vary.b. Yes. Place the point of the compass at the
point at which the perpendicular bisectorsintersect. Place the pencil on any vertex of thetriangle and draw a circle. Since the point ofintersection is equidistant from each of thevertices, all three vertices lie on the circle.
Hands-On Activity
a. Construct a segment congruent to .Construct lines perpendicular to this segmentat each of the endpoints. Construct a segment congruent to at each endpoint on the perpendiculars, in the same direction. Draw asegment connecting the endpoints not on thefirst segment.
b. Construct a segment congruent to .Construct intersecting arcs with radius ABcentered at each of the endpoints. Draw asegment from each endpoint to thisintersection.
c. Construct a segment congruent to .Construct a line perpendicular to this segmentat an endpoint. Bisect the angle formed.
AB
AB
AB
AB
BCAB
AB
CD
ABBC
266
d. Follow part b. Bisect one of the angles of thetriangle.
e. Construct a segment congruent to . Constructa point C not on the line containing the segment.Connect the endpoints of the segment to point Cto form a triangle. Construct the perpendicularbisector of each side of the triangle. Construct acircle whose center is the intersection of theperpendicular bisectors and the radius is thedistance to any of the vertices of the triangle.
Review Exercises (pages 204–205)1. x � 60, y � 30 2. m�PQR � 50 3. 464. m�PQS � 65, m�SQR � 115
5. Statements Reasons1. �ABC is equilateral. 1. Given.2. 2. Definition of
equilateral triangle.3. D lies on the 3. Given.
perpendicular bisectors of , ,and .
4. DA � DB = DC 4.A point on theperpendicular bisectorof a segment is equi-distant from the end-points of that segment.
5. 5. Segments with equalmeasure arecongruent.
6. �ADC � �BDC 6. SSS (steps 2, 5).� �ADB
7. �ADC, �BDC, and 7. Definition of �ADB are isosceles. isosceles triangle.
6. Statements Reasons1. 1. Assumption.2. is a median of 2. Given.
�ABC.3. 3. The median from the
vertex of an isoscelestriangle is perpen-dicular to the base.
4. is the altitude 4. Definition of altitude.to side .
5. is not the 5. Given.altitude to side .
6. is not 6. Contradiction congruent to . (steps 1, 5).BCAC
ABCD
ABCD
CD ' AB
CDAC > BC
DA > DB > DC
CABCAB
AB > BC > CD
AB
7. Statements Reasons1. �ABC and �ABD 1. Given.
are isosceles with as base.
2. , 2. An isosceles triangle has two congruent sides.
3. AC � BC, 3. Definition of AD � BD congruent segments.
4. is the perpen- 4. If two points are dicular bisector each equidistant of . from the endpoints of a
line segment, then thetwo points determinethe ⊥ bisector of theline segment.
8. Statements Reasons1. is the median 1. Given.
to .2. D is the midpoint 2. Definition of median.
of .3. 3. Definition of midpoint.4. 4. Given.5. 5. Transitive property.6. �A � �ACD, 6. Isosceles triangle
�B � �BCD theorem.7. m�A � m�ACD, 7. Definition of
m�B � m�BCD congruent angles.8. m�ACB 8. Partition postulate.
� m�ACD� m�BCD
9. m�ACB 9. Substitution postulate.� m�A � m�B
9. a. Use Construction 5.b. Use Construction 4 to bisect �ADC.c. m�EDC � 45d. m�EDB � 135
10. a. Results will vary.
b. Draw opposite . Use Construction 6 to
construct a line through P perpendicular to
.
Exploration (page 205)1. Statements Reasons
1. 1. Given.
2. 2. Addition postulate.> MN 1 NBAL 1 LM> MN > NBAL > LM
SQRg
QRh
QSh
AD > CDCD > DBAD > DB
AB
ABCD
AB
CDg
AD > BDAC > BCAB
267
3. 3. Partition postulate.
4. 4. Substitutionpostulate.
5. is an altitude 5. Given.of �CLM.
6. 6. Definition of altitude.7. �CMA � �CMB 7. If two lines are
perpendicular, thenthey intersect to formcongruent adjacentangles.
8. 8. Reflexive property.9. �ACM � �BCM 9. SAS (steps 4, 7, 8).
10. 10. Corresponding partsof congruent trianglesare congruent.
2–5. Results will vary.
Cumulative Review (pages 206–208)Part I
1. 3 2. 4 3. 2 4. 15. 2 6. 1 7. 3 8. 49. 3 10. 4
Part II11. m�PQS � 67.2, m�SQR � 112.8
m�PQS � 3(16.4) � 18 � 67.2m�SQR � 7(16.4) � 2 � 112.8
12. 5(4 � x) � 32 � x Given.20 � 5x � 32 � x Distributive
property ofmultiplication.
20 � 5x � x � 32 � x � x Addition postulate.20 � 6x � 32 � 0 Additive inverse.20 � 6x � 32 Additive identity.
6x � 12 Subtractionpostulate.
x � 2 Division postulate.
a 5 16.4
10a 5 164
10a 1 16 5 180
3a 1 18 1 7a 2 2 5 180
CA > CB
CM > CM
CM ' AB
CM
AM > MBMB > MN 1 NBAM > AL 1 LM
Part III13. Statements Reasons
1. 1. Given.2. �ABF is the 2. Given.
supplement of �A.3. �FBD is the 3. If two angles form a
supplement of linear pair, then they �ABF. are supplementary.
4. �A � �FBD 4. If two angles aresupplements of thesame angle, then theyare congruent.
5. 5. Given.6. 6. Addition postulate.
7. 7. Partition postulate.
8. 8. Substitutionpostulate.
9. �AEC � �BFD 9. SAS (steps 1, 4, 9).
14. Statements Reasons1. 1. Given.2. is the bisector 2. Given.
of �ABC.3. �CBD � �ABD 3. Definition of angle
bisector.4. 4. Reflexive property.5. �CBE � �ABE 5. SAS.6. 6. Corresponding parts
of congruent trianglesare congruent.
Part IV15. Statements Reasons
1. Point P is not on 1. Given..
2. PB � PC 2. Given.
‹ABCD
›
AE > CE
BE > BE
BDhAB > CB
AC > BDBD > BC 1 CDAC > AB 1 BC> BC 1 CD
AB 1 BCAB > CD
AE > BF
268
Statements Reasons3. 3. Segments with equal
measures arecongruent.
4. �CBP � �BCP 4. Isosceles triangletheorem.
5. �ABP and �CBP 5. If two angles form aare supplements. linear pair, then they�DCP and �BCP are supplementary.are supplements.
6. �ABP � �DCP 6. If two angles arecongruent, then theirsupplements arecongruent.
16. Statements Reasons1. and are ⊥ 1. Given.
bisectors of each other.
2. Let M be the 2. Construction.intersection of and .
3. M is the midpoint of 3. Definition of bisector.and of .
4. , 4. Definition of midpoint.
5. �AMB � �CMB 5. If two lines are � �CMD � �AMD perpendicular, then
they intersect to formcongruent adjacentangles.
6. �AMB � �CMB 6. SAS.� �CMD � �AMD
7. 7. Corresponding parts of congruent trianglesare congruent.
8. AB � BC 8. Definition of � CD � DA congruent segments.
> CD > DAAB > BC
BM > DMAM > CM
BDAC
BDAC
BDAC
PB > PC
Chapter 6.Transformations and the Coordinate Plane
6-1 The Coordinates of a Point in a Plane(pages 213–214)
Writing About Mathematics
1. Since is a horizontal segment, must be a vertical segment with BC � AC � 7. Therefore,the coordinates of B are (5, 3) or (5, �11).
2. The area of the polygon is equal to the sum ofthe area of two triangles EFD and GFD. If O isthe origin, then the area of the triangle EFD is
BCAC
= . The area of a
triangle GFD is .Therefore, the area of a polygon FEDG equals 22.5 square units.
Developing SkillsIn 3–12, part a, answers will be graphs.
3. b. 14 sq units 4. b. 20 sq units5. b. 20 sq units 6. b. 16 sq units7. b. 21 sq units 8. b. 24 sq units
12FD 3 GO 5 1
2 3 5 3 4 5 10
12 3 5 3 5 5 12.51
2FD 3 EO
9. b. 20 sq units 10. b. 6 sq units11. b. 10 sq units 12. b. 16 sq units13. D(1, 4)14. R(2, �7), S(�1, �7) or R(2, �1), S(�1, �1)15. a. Graph 16. a. Graph
b. 24 sq units b. 6 sq units
6-2 Line Reflections (pages 220–221)Writing About Mathematics
1. Every real number corresponds to exactly onepoint on the number line. Every point on the realnumber line corresponds to exactly one number.Hence, there is a one-to-one correspondencebetween points on the number line and the realnumbers.
2. No. Two different points can have the sameimage. For example, points (1, 3) and (2, 3) bothmap to (2, 3).
Developing Skills3. Since distance is preserved under a line
reflection and PQ � QR, the images of thesesegments also have equal length, that is,P�Q� � Q�R� and . Therefore,�P�Q�R� is isosceles.
4. Since angle measure is preserved under a linereflection and m�N � 90, m�N� � 90, and�L�M�N� is a right triangle.
5. Yes. Since distance is preserved under a linereflection, AB � A�B�, BC � B�C�, and CA � C�A� so , , and
. By SSS, �ABC � �A�B�C�.6. Since , �ACB is a right angle, and
m�ACB � 90. Since angle measure is preservedunder a line reflection, m�A�C�B� � 90.Therefore, and intersect to form rightangles, and .
7. Yes. Midpoint is preserved under a linereflection.
8. Yes. Collinearity is preserved under linereflection.
9. Point D; a point is on the line reflection if it is itsown image.
10. Since the triangle has line symmetry with respectto the altitude from S, RS � ST and ,and the triangle is isosceles. However, since itdoes not have line symmetry with respect to the altitude from R, RS � RT and is notcongruent to , and the triangle is not equilateral. Therefore, �RST is an isoscelestriangle with exactly two congruent sides.
RTRS
RS > ST
ArCr ' BrCrBrCrArCr
AC ' BCCA > CrAr
BC > BrCrAB > ArBr
PrQr > QrRr
269
Applying Skills
11. (1) Let O be the point where intersects k.(2) By the definition of a line reflection, k is the
perpendicular bisector of and . LetM be the point where and k intersectand N be the point where and kintersect.
(3) Then, and at M and N, respectively. Therefore,m�AMO � m�A�MO � 90 and m�BMO � m�B�MO � 90. Also,
and by the reflexive property of congruence.
(4) Thus, by SAS, �AMO � �A�MO and�BNO � �B�NO.
(5) Since corresponding parts of congruenttriangles are congruent, and
, and so AO � A�O and OB � OB�.
(6) By the partition postulate,A�B� � A�O � OB� and AB � AO � OB.By the substitution postulate,A�B� � AO � OB � AB. Therefore,
.
12. The lines of symmetry of a rhombus are itsdiagonals. Therefore, the lines of symmetry of thebaseball diamond would be from home plate andsecond base and from first base and third base.
13. E, D, C14.
6-3 Line Reflections in the CoordinatePlane (pages 225–227)
Writing About Mathematics1. The distance is the absolute value of the
difference of the y-coordinates.
�b – a� � 1�b � a� � ��1� � �b � a�
� �(�1)(b � a)� � �a � b�Thus, �b � a� and �a – b� have the same value.
2. Second quadrantDeveloping SkillsIn 3–17, part a is a graph of the point indicated in b.
3. b. (2, �5) 4. b. (1, �3) 5. b. (�2, �3)6. b. (2, 4) 7. b. (0, �2) 8. b. (�3, 5)9. b. (�1, 4) 10. b. (�2, �3) 11. b. (2, 3)
H O X
AB > ArBr
OB > OBrAO > ArO
BN > BrNAM > ArM
BBr ' kAAr ' k
BBrAAr
BBrAAr
AB
12. b. (1, 0) 13. b. (5, 3) 14. b. (5, �3)15. b. (�2, 4) 16. b. (�5, �1) 17. b. (2, 2)Applying Skills18. (1) The line of reflection, the x-axis, is a
horizontal line. P(a, b) and P�(a, �b) have the same x-coordinates, so is a segment of a vertical line. Every horizontal isperpendicular to every vertical line.Therefore, the x-axis is perpendicular to .
(2) Let Q be the point at which intersects the x-axis, so the y-coordinate of Q is 0.The length of a vertical line segment is theabsolute value of the difference of the y-coordinates of the endpoints. Therefore,
and . Since PQ � P�Q, Q is the midpoint
of and the x-axis bisects .(3) Steps 1 and 2 prove that the x-axis is the
perpendicular bisector of . By the definition of a line reflection, P�(a, �b) is theimage of P(a, b).
19. a. A line is a line of symmetry of a figure if thefigure is its own image under a reflection inthat line. Since collinearity is preserved undera line reflection, it suffices to show that eachvertex of ABCD is the image of anothervertex under .
Therefore, y � x is a line of symmetry.b. Under :
Therefore, the y-axis is a line symmetry.20. Under :
Since F and G do not map to one of the verticesof EFGH, the y-axis is not a line of symmetry.
21. y � 0 and x � 3
H(0, 3) S H(0, 3)
G(5, 3) S Gr(25, 3)
F(5, 23) S Fr(25, 23)
E(0, 23) S E(0, 23)
ry-axis
D(0, 4) S Dr(0, 4) 5 D
C(4, 0) S Cr(24, 0) 5 A
B(0, 24) S Br(0, 24) 5 B
A(24, 0) S Ar(4, 0) 5 C
ry-axis
D(0, 4) S Dr(4, 0) 5 C
C(4, 0) S Cr(0, 4) 5 D
B(0, 24) S Br(24, 0) 5 A
A(24, 0) S Ar(0, 24) 5 B
ry5x
PPr
PPrPPr5 Zb Z
PrQ 5 Z2b 2 0 Z PQ 5 Zb 2 0 Z 5 Zb Z
PPrPPr
PPr
270
Hands-On Activity 1a.
,
b.
, ,
c.
, ,
d.
, , , Sr(9, 22)Rr(3, 24)Qr(5, 0)Pr(7, 26)
Q�
S
Q
O
R�
P
1
1R
P�
S�
U
T
q
Lr(2, 25)Kr(23, 25)Jr(0, 24)
y
x
J�
p
JK
M
N
L�K�
L
O 11
Gr(22, 22)Fr(22, 3)Er(1, 0)
y
O
1 x
1
E�
l F G
E
I
HF�
G�
Br(2, 6)Ar(0, 0)
y
O
1 x1A�
A
C
BDB� k
Hands-On Activity 2a.
b.
c.
d.
6-4 Point Reflections in the CoordinatePlane (pages 231–232)
Writing About Mathematics1. Yes. Since a point reflection is a transformation
and a transformation is a one-to-onecorrespondence, the endpoints of the segmentare in a one-to-one correspondence with theendpoints of the image of the segment. Thus,
, , , and. Since collinearity is preserved, the
image of is and the image of is .AB
ArBrArBrABRP(Br) 5 B
RP(B) 5 BrRP(Ar) 5 ARP(A) 5 Ar
A�
x
A
B
O
C�
1
1
C
B�
y
C�
x
C
A
O 1
1
BA�
y
B�
A
B B�
A�
A
A�
271
2. Not necessarily. For example, let P be the origin, l be the line y � x, and m be the x-axis. Then while
and . The two images are not
the same.Developing Skills
3. (�1, �5) 4. (2, �4) 5. (1, 0)6. (0, �3) 7. (�6, �6) 8. (1, 5)
Applying Skills9. Suppose that A � P. Then A is its own image
under , that is, A� � A. Let B� be the image ofB under . By the definition of a point reflection, P is the midpoint of , and so PB � PB�. By the substitution postulate,AB � A�B�. Therefore, distance is preservedunder a point reflection.
10. Given: �ABC and �A�B�C�, the image of�ABC under .
Prove: m�ABC � m�A�B�C�Proof: Consider �A�B�C�, the image of �ABCunder a reflection in point P. Since distance ispreserved, AB � A�B�, BC � B�C�, and CA � C�A�, so , , and
. Thus, �A�B�C� � �ABC by SSS.Since corresponding parts of congruent trianglesare congruent, �ABC � �A�B�C�. Therefore,m�ABC � m�A�B�C�, and angle measure ispreserved.
11. Given: and A�, D�, and B�, the images of A,D, and B, respectively, under .
Prove: D� is a point on .Proof: Since A, D, and C are collinear, AD � DB� AB. Since distance is preserved, AD � A�D�,DB � D�B�, and AB � A�B�. By substitution,A�D� � D�C� � A�B�. If D� were not on ,then A�D� � D�B� A�B� because a straight lineis the shortest distance between two points. ButA�D� � D�C� � A�B�. Therefore, A�, D�, and B�
are collinear, and collinearity is preserved undera point reflection.
12. Given: M, the midpoint of , and A�, M�, andB�, the images of A, M, and B,respectively, under .
Prove: M� is the midpoint of .Proof: Since distance is preserved under a pointreflection, A�M� � AM, and M�B� � MC. Since
ArBrRP
AB
ArBr
ArBrRP
ADB
CA > CrArBC > BrCrAB > ArBr
RP
BBrRP
RP
rx-axis(2b, 2a) 5 (2b, a)ry5x(2a, 2b) 5 (2b, 2a)
RO(a,b) 5 (2a,2b)
M is the midpoint of , AM � MB, and, by thesubstitution postulate, A�M� � M�B�. Therefore,M� is the midpoint of , and midpoint ispreserved under a point reflection.
13. H, I, O, X, Z14. Under :
Each vertex maps to another vertex, so the figureis its own image under a reflection in the origin.Therefore, it has point symmetry.
15. a. (2, �6) b. (1, �6)c. No. For example, while
. The two images are notthe same.
16. a. (4, �8) b. (7, �2)c. Yes. Point reflection preserves collinearity.
6-5 Translations in the Coordinate Plane(pages 236–238)
Writing About Mathematics1. A point P(x, y) is fixed if its coordinates are the
same before and after the translation. However,under a translation , the coordinates of its image P� are (x � a, y � b). Therefore, P � P� ifand only if a � b � 0. A translation other than
does not have any fixed points.2. Let A have the coordinates (x, y). Then A� is
ry-axis(x, y) � (�x, y). A is a reflection in the linex � 3, which is the same as translating �3 units inthe x-direction, reflecting in the y-axis, and thentranslating 3 units back in the x-direction.
T–3,0(�x, y) � (�x � 3, y)
ry-axis(�x � 3, y) � (x � 3, y)
T3,0(x � 3, y) � (x � 6, y)
The coordinates of A are the same as those obtained under the translation
.a. Yes b. Yes c. Yes
Developing Skills3. a. P b. H c. O
d. L e. K4. a. Graph b. A�(6, �1), B�(11, 0), C�(5, 3)
c. Graph5. a. Graph b. Graph
c. A�(�6,5), B�(�2, 7), C�(�1, 4)
(x, y) S (x 1 6, y)
T0,0
Ta,b
r(2, 0)(3, 6) 5 (1, 26)rx-axis(3, 6) 5 (3, 26)
S(0, 5) S Q(0, 25)
R(5, 0) S P(25, 0)
Q(0, 25) S S(0, 5)
P(25, 0) S R(5, 0)
RO
ArBr
AB
272
6. a � 6, b � 07.8. a. Graph b.
c. B�(7, 5), C�(8, 9) d. Graph9. a. Graph b. A�(�1, �5), B(1, �5)
c. Graph d. none
Applying Skills10. a. Given: �ABC and , the image of
�ABC under .Prove: m�ABC �
Proof: Consider �A�B�C�, the image of�ABC under a translation . Since distanceis preserved, AB � A�B�, BC � B�C�, and CA � C�A� or , , and
. Thus, �A�B�C� � �ABC by SSS.Since corresponding parts of congruenttriangles are congruent, �ABC � �A�B�C�.Therefore, m�ABC � m�A�B�C�, and anglemeasure is preserved.
b. Given: and A�, D�, and B�, the images ofA, D, and B, respectively, under .
Prove: D� is a point on .Proof: Since A, D, and C are collinear,AD � DB � AB. Since distance is preserved,
, , and . Bysubstitution, A�D� � D�C� � A�B�. If D� werenot on , then because a straight line is the shortest distancebetween two points. But, A�D� � D�C� � A�B�.Therefore, A�, D�, and B� are collinear, andcollinearity is preserved under a translation.
c. Given: M, the midpoint of , and A�, M�, andB�, the images of A, M, and B,respectively, under .
Prove: M� is the midpoint of .Proof: Since distance is preserved,A�M� � AM, and M�B� � MB. Since M is the midpoint of , AM � MB, and, by thesubstitution postulate, A�M� � M�B�.Therefore, M� is the midpoint of , andmidpoint is preserved under a translation.
In 11–13, part a, answers will be graphs.
11. b. L�(4, 0), M�(0, 0), N�(0, 2)c. L(4, 0), M(8, 0), N(8, 2)d. P(4, 0), Q(8, 0), R(8, 2)e. The two triangles coincide.
ArBr
AB
ArBr
Ta,b
AB
ArDr1DrBr . ArBrArBr
AB 5 ArBrDB 5 DrBrAD 5 ArDr
ArBrTa,b
ADB
CA > CrArBC > BrCrAB > ArBr
Ta,b
m/ArBrCrTa,b
/ArBrCr
(x, y) S (x 1 3, y 1 4)(x, y) S (x, y 2 4)
12. b. D�(�2, 3), E�(1, 3), F�(0, 0)c. D(�2, 3), E(1, 3), F(6, 0)d. R(�2, 3), S(1, 3), T(6, 0)e. The two triangles coincide.
13. b. A�(1, �2), B�(5, �2), C�(4, �5)c. A(1, �4), B(5, �4), C(4, �1)d. Yes. a � 0, b � �6
14. Let d be the distance between the two lines. Forvertical lines, x � 2d and y � 0. For horizontallines, x � 0 and y � 2d.
15. a. Graphb. A�(1, 0), B�(5, 0), C�(4, �3)c. A(1, �4), B(5, �4), C(4, �1)d. ; yes
6-6 Rotations in the Coordinate Plane(pages 242–243)
Writing About Mathematics1. a. The image of P(x, y) under this rotation is
P�(�x, �y).b. Reflection in the originc. Yes, the two transformations are equivalent.
2. Reflection in the origin and 180° rotation aboutthe origin.
Developing Skills3. a. D 4. a. J
b. E b. Kc. F c. Ld. J d. De. K e. Ef. A f. Gg. B g. Hh. C h. I
5. a. A�(1, 2), B�(1, 5), C�(�4, 5), D�(�4, 2)b. Yesc. The midpoint of is (3.5, �1), the midpoint
of is (1, 3.5). Yes, the image of themidpoint is the midpoint of .
6. a. Q(�3, 1) b. R(�1, �3)c. S(3, �1) d. P�(3, �1)e. S and P� are the same since rotating a point
90° clockwise is equivalent to rotating thepoint 270° counterclockwise.
6-7 Glide Reflections (page 246)Writing About Mathematics
1. No. Under a glide reflection, a translation isperformed in the direction of the line ofreflection. A translation has no fixed points.Since the translation is along the line ofreflection, the fixed points of the line reflection
ArBrArBr
AB
T0,26
273
are moved by the translation. Therefore, theresulting transformation has no fixed points.
2. No. For example, consider the translation that is in a direction perpendicular to the x-axis.Then, the image of (x, y) under followed by
is (x, �y � 1). However, the image of (x, y)under followed by is (x, �y � 1). Thetwo images do not coincide.
Developing SkillsIn 3–8, part b, answers will be graphs.
3. a. A�(5, �1), B�( 9, �4), C�(7, �5)c. Yes, since the translation is in the direction of
the line of reflection.d.
4. a. A�(�1, �5), B�(�5, �2), C�(�3, �1)c. Yes, since the translation is in the direction of
the line of reflection.d.
5. a. A�(6, �1), B�(10, �4), C�(8, �5)c. Yes, since the translation is in the direction of
the line of reflection.d.
6. a. A�(�4, 4) , B�(�8, 7), C�(�6, 8)c. No, since the translation is not in the direction
of the line of reflection.d.
7. a. A�(4, 4), B�(7, 8), C�(8, 6)c. Yes, since the translation is in the direction of
the line of reflection.d.
8. a. A�(0, 3), B�(�3, 7), C�(�4, 5)c. No, since a rotation is not equivalent to a line
reflection.d.
Applying Skills9. A(�2, �5)
10. a. The line of reflection is x � 0. The translationis .
b. C�(0, �4)11. a. The line of reflection is the x-axis or y � 0.
The translation is .b. C�(0, �2)
12. a. Translations and glide reflectionsb. Rotations and point reflectionsc. Reflections
6-8 Dilations in the Coordinate Plane(pages 249–250)
Writing About Mathematics1. The length of each side of the image of the
triangle is k times the length of its corresponding
T24,0
T0,211
(a, b) S (2b 1 1, a 1 2)
(a, b) S (b 1 3, a 1 3)
(a, b) S (2a 2 3, b 1 3)
(a, b) S (a 1 5, 2b)
(a, b) S (2a, b 2 6)
(a, b) S (a 1 4, 2b)
rx-axisT0,1
T0,1
rx-axis
T0,1
side in the original triangle. A dilation is not anisometry since distance is not preserved.
2. The length of each side of the image of thetriangle is times the length of its correspondingside in the original triangle.
Developing Skills3. (12, 28) 4. (�16, 8)5. (8, 0) 6. (�4, 36)7. (10, 10) 8. (5, 50)9. (�15, 25) 10. (0, 20)
11. (3, �1) 12. (2, 0)13. 14.15. (�4, 20) 16. (4, 6)17. 18. (15, 3)19. (6, �2) 20. (6, 12)21. D2 followed by rx-axis 22. D7 followed by ry-axis23. followed by rx-axis 24. followed by ry-axis
Applying Skills25. a. A�(4, �4), B�(12, �4), C�(12, 12), D�(4, 12)
b. Since A� and B� have the same y-coordinate,is a horizontal segment. Similarly, is
a horizontal segment. Since A� and D� havethe same x-coordinate, is a verticalsegment. Similarly, is a vertical segment.Vertical and horizontal lines are perpendicular. Thus, the sides of meet to form right angles, and is arectangle.
26. Dk(a, b) � (ka, kb), RO(ka, kb) � (�ka, �kb)D
�k(a, b) � (�ka, �kb)The two images are the same. Therefore, Dkfollowed by a point reflection in the origin givesthe same result as D–k.
27. a, b. C�
x
C
A
O 22 B
A�
y
B�
ArBrCrDrArBrCrDr
BrCrArDr
CrDrArBr
D14
D12
(3, 212)
(5, 312)(23, 221
2)
1k
274
c. As we can see from the graph, AB is not equalto A�B�. Therefore, distance is not preservedunder the dilation.
Hands-on Activitya. Yes b. Yes c. Yes
6-9 Transformations As Functions (pages254–255)
Writing About Mathematics1. No. For example, consider a reflection in y � x
and the y-axis. while . The two images are
not the same.2. Tyler is correct. Every opposite isometry changes
the order or orientation from clockwise tocounterclockwise or vice versa. Therefore, everypair (even number) of opposite isometries resultsin the original order or orientation. Every non-pair (odd number) of opposite isometries resultsin the reverse order or orientation.
Developing Skills3. A�(�3, 2) 4. A�(�3, 2)5. A�(�3, �2) 6. A�(�2, �3)7. A�(�3, 7) 8. A�(�3, 2)9. A�(2, 3) 10. A�(�6, 4)
11. A�(�2, �4)12. A reflection in the y-axis13. RO14. a. D�(0, �2), E�(2, �5), F�(1, 1)
b. Glide reflectionc. Opposite isometry
Applying Skills
15. A(a, b) and A�(�b, �a) are points. Let B(a, �a)and C(�b, b) be points on the line of reflection y � �x. The distance BA � �b � (�a)� � �b � a�and the distance BA� � ��b � a� � �b � a�.Therefore, B is equidistant from A and A�.Similarly, the distance CA � �a � (�b)� � �a � b�and the distance CA� � ��a � b� � �a � b�.Therefore, C is equidistant from A and A�. If twopoints are each equidistant from the endpoints ofa line segment, then they lie on the perpendicular bisector of the line segment. Therefore, ,which is a subset of the line y � �x, is theperpendicular bisector of . By the definitionof a line reflection, the image of is
.Ar(2b,2a)A(a,b)
AAr
CB
ry5x + ry-axis(a, b) 5 (b, 2a)ry-axis + ry5x(a, b) 5 (2b, a)
16.
The images are the same. Therefore,.
17.
The images are the same. Therefore,.
Review Exercises (pages 257–258)1. x � �2 2. y � �43. a. Graph
b. 16.5 sq units4. a. P�(5, 3)
b. Every point on the x-axis5. a. P�(�5, 3) 6. a. P�(3, 5)
b. (0, 0) b. (0, 0)7. a. P�(5, 0) 8. a. P�(7, 1)
b. No fixed point b. No fixed point9. a. P�(�3, �5) 10. a. P�(3, 5)
b. (0, 0) b. (0, 0)11. a. P�(�6, �2)
b. No fixed point12.
13.
14. S 15. Z16. Line and glide reflections 17. Dilation18. a. Graph
b. A�(�3, �2) , B�(�3, �7) , C�(2, �7)c. A(3, �2), B(3, �7), C(�2, �7)d. Reflection in the x-axis
19. a. Graphb. R�(4, 1), S�(1, 1), T�(1, �2)c. R(4, 7) , S(1, 7), T(1, 10)
ry52x + ry5x 5 RO
RO(a, b) 5 (2a, 2b)ry52x + ry5x(a, b) 5 ry52x(b, a) 5 (2a, 2b)
ry5x + ry52x 5 RO
RO(a, b) 5 (2a, 2b)ry5x + ry52x(a, b) 5 ry5x(2b, 2a) 5 (2a, 2b)
275
20. a. (3, 0), (3, 0), (�1, 0), (�1, 0)b. Two different points have the same image.
Hence, the correspondence is not one-to-one.21. a. M�(3, �9), A�(6, �6), T�(�6, �6)
b. M(3, �9), A(6, �6), T(�6, �6)c. Yes
Therefore, two transformations are the same.
Exploration (page 259)Answers will vary.
Cumulative Review (pages 259–261)Part I
1. 2 2. 1 3. 2 4. 4 5. 46. 3 7. 4 8. 3 9. 1 10. 2
Part II11. b � �2. Since A is a fixed point under a
reflection in y � x, A is on the line y � x, and so b � �2.
12. (2, �1)
Part III13. Statements Reasons
1. is perpendicular 1. Given.bisector of .
2. E is the midpoint of 2. Definition of and of . bisector.
3. and 3. Definition of midpoint.
4. �AEB � �BEC 4. Perpendicular lines � �CED � �AED intersect to form
congruent adjacentangles.
5. �AEB � �AED 5. SAS.� �CED � �CEB
6. �ABD � �ADB and 6. Corresponding �DBC � �BDC parts of congruent
triangles arecongruent.
7. �ABD � �DBC 7. Addition postulate.� �ADB � �BDC
8. �ABC 8. Partition postulate.� �ABD � �DBC�ADC� �ADB � �BDC
9. �ABC � �ADC 9. Substitutionpostulate.
BE > EDAE > EC
BEDAEC
BEDAEC
D3 + rx-axis(a, b) 5 D3(a, 2b) 5 (3a, 23b)rx-axis + D3(a, b) 5 rx-axis(3a, 3b) 5 (3a, 23b)
14. Let x � m�S, then 3x � 12 � m�R. Since theangles are supplementary:
Therefore, m�S � 42 and m�R � 138.Part IV15. Statements Reasons
1. �ABC is not scalene. 1. Assumption.2. �ABC has at least 2. Definition of a
one pair of congruent scalene triangle.sides.
3. �A � �B or 3. Isosceles triangle �B � �C or theorem.�A � �C
4. m�A � m�B or 4. Definition of m�B � m�C or congruent angles.m�A � m�C
5. m�A � m�B, 5. Givenm�B � m�C,m�A � m�C
6. �ABC is scalene. 6. Contradiction.
x 5 424x 5 168
x 1 3x 1 12 5 180
276
16. Statements Reasons1. T3,–3(�ABC) 1. Given.
� �A�B�C�2. A�B� � AB, 2. In a translation,
B�C� � BC distance ispreserved.
3. XY � A�B�, 3. Given.YZ � B�C�
4. XY � AB, 4. Transitive property YZ � BC of equality.
5. , 5. Definition of congruent segments.
6. �Y � �B 6. Given.7. �ABC � �XYZ 7. SAS.
YZ > BCXY > AB
Chapter 7. Geometric Inequalities
7-1 Basic Inequality Postulates (pages 266–267)
Writing About Mathematics1. No. The reflexive and symmetric properties do
not hold for inequality.2. No. By the trichotomy postulate, it is possible
that AB � BC.Developing Skills
3. a. Trueb. A whole is greater than any of its parts.
4. a. Falseb. The shortest distance between two points is
the length of the line segment joining thesetwo points.
5. a. Trueb. A whole is greater than any of its parts.
6. a. Trueb. Definition of a linear pair of angles
7. a. Falseb. Definition of straight angle
8. a. Trueb. Trichotomy postulate
9. a. Trueb. Transitive property of inequality
10. a. Trueb. Transitive property of inequality
11. a. Trueb. Substitution postulate
12. a. Trueb. Substitution postulate
Applying Skills13. Statements Reasons
1. 1. Given.2. �A � �CBA 2. Isosceles triangle
theorem.3. m�A � m�CBA 3. Definition of
congruent angles.4. m�CBD m�CBA 4. Given.5. m�CBD m�A 5. Substitution
postulate.
14. Statements Reasons1. 1. Given.2. PR PQ 2. A whole is greater
than any of its parts.3. PQ � RS 3. Given.4. PR RS 4. Substitution
postulate.
PQRS
AC > BC
15. Statements Reasons1. 1. Given.2. LM � NM 2. Definition of congruent
segments.3. 3. Given.4. KM LM 4. A whole is greater than
any of its parts.5. KM NM 5. Substitution postulate.
16. Statements Reasons1. KM KN, 1. Given.
KN NM2. KM NM 2. Transitive property of
inequality.3. NM � NL 3. Given.4. KM NL 4. Substitution postulate.
7-2 Inequality Postulates InvolvingAddition and Subtraction (page 269)
Writing About Mathematics1. No. It is possible for the differences to be
unequal in the same order or equal.2. Subtracting unequal quantities is the same as
adding the opposites of the unequal quantities.To find the opposites, we multiply the inequalityby �1, which changes the order of the inequality.Then, if unequal quantities are added to equalquantities, the sums are unequal in the sameorder.
Developing Skills3. If equal quantities (8 � 8) are added to unequal
quantities (10 7), then the sums are unequal inthe same order (18 15).
4. If equal quantities (11 � 11) are added tounequal quantities (4 � 14), then the sums areunequal in the same order (15 � 25).
5. If equal quantities (3 � 3) are subtracted fromunequal quantities (x + 3 12), then thedifferences are unequal in the same order (x 9).
6. If equal quantities (5 � 5) are added to unequalquantities (y � 5 � 5), then the sums are unequalin the same order (y � 10).
7. If unequal quantities (8 6) are added tounequal quantities in the same order (5 3),then the sums are unequal in the same order (13 9).
8. If equal quantities (2 � 2) are subtracted fromunequal quantities (7 � 12), then the differencesare unequal in the same order (5 � 10).
KLM
LM > NM
277
9. If equal quantities (1 � 1) are subtracted fromunequal quantities (y 8), then the differencesare unequal in the same order (y � 1 7).
10. If equal quantities (a � b) are subtracted fromunequal quantities (180 90), then thedifferences are unequal in the same order (180 � a 90 � b).
Applying Skills11. Statements Reasons
1. AB � AD, 1. Given.BC � DE
2. AB � BC � AD � DE 2. If equal quantities are added tounequal quantities,then the sums areunequal in thesame order.
3. AC � AB + BC, 3. Partition postulate.AE � AD � DE
4. AC � AE 4. Substitutionpostulate.
12. Statements Reasons1. AE BD, 1. Given.
AF � BF2. AE � AF � BD � BF 2. If equal quantities
are subtractedfrom unequalquantities, then thedifferences areunequal in thesame order.
3. AE � AF � FE, 3. Partition postulate.BD � BF � FD
4. FE FD 4. Substitutionpostulate.
13. Statements Reasons1. AE � BE 1. Given.2. 2. Definition of
congruentsegments.
3. �EAB � �EBA 3. Isosceles triangletheorem.
4. m�EAB � m�EBA 4. Definition ofcongruent angles.
5. m�DAC m�DBC 5. Given.6. m�DAC � m�EAB 6. If equal quantities
m�DBC � m�EBA are added tounequal quantities,then their sums areunequal in thesame order.
AE > BE
(Cont.)
Statements Reasons7. m�DAB � m�DAC 7. Partition
� m�EAB, postulate.m�CBA � m�DBC
� m�EBA8. m�DAB m�CBA 8. Substitution
postulate.
14. Yes. If equal weights are added to unequalweights (Blake weighs more than Caleb), thentheir new weights are unequal in the same order(Blake still weighs more than Caleb).
15. Not necessarily. If unequal weights aresubtracted from unequal weights (Blake weighsmore than Andre), then their new weights maybe unequal in the same order (Blake weighsmore than Andre), equal in different orders(Andre weighs more than Blake), or equal(Blake and Andre weigh the same).
7-3 Inequality Postulates InvolvingMultiplication and Division (page 272)
Writing About Mathematics1. No. It is only true for positive values of a.
Otherwise, if unequal quantities (1 � 2) aremultiplied by negative equal quantities, then theproducts are unequal in the opposite order (a 2a). If a � 0, then a � 2a.
2. No. It is possible that ac � 0 and bd � 0. Then a � b. It is also possible that ac � 0 and bd � 0 or ac � 0 and bd 0. Then ac � bc.
Developing Skills3. If unequal quantities (8 7) are multiplied by
positive equal quantities (3 � 3), then theproducts are unequal in the same order (24 21).
4. If unequal quantities (30 � 35) are divided bynegative equal quantities (�5 � �5), then thequotients are unequal in the opposite order (�7 � �6).
5. If unequal quantities (8 6) are divided bypositive equal quantities (2 � 2), then thequotients are unequal in the same order (4 3).
6. If unequal quantities (3x 15) are divided bypositive equal quantities (3 � 3), then thequotients are unequal in the same order (x 5).
7. If unequal quantities are multiplied bynegative equal quantities (�2 � � 2), then the products are unequal in the opposite order (8 �x).
A x2 . 24 B
278
8. If unequal quantities are multiplied bypositive equal quantities (6 � 6), then the products are unequal in the same order (y � 18).
9. Always true; if unequal quantities are multipliedby positive equal quantities, then the productsare unequal in the same order.
10. Always true; if equal quantities are added tounequal quantities, then the sums are unequal inthe same order.
11. Never true; let a � 5, b � 4, and c � 1. Then 1 � 5 � 1 � 4 or �4 � �3.
12. Always true; if equal quantities are subtractedfrom unequal quantities, then the differences areunequal in the same order.
13. Sometimes true; let a � 5, b � 3, and c � 2. Then5 � 3 3 � 2 or 2 1. Let a � 5, b � 4, and c � 2. Then 5 � 4 � 4 � 2 or 1 � 2.
14. Never true; let a � 5, b � 3, and c � 2. Then .
15. Always true; if unequal quantities are divided bypositive equal quantities, then the quotients areunequal in the same order.
16. Never true; let a � 3, b � 2, and c � 1. Then �3 � �2.
17. Always true; transitive property of inequality.Applying Skills18. Statements Reasons
1. BD � BE 1. Given.2. D is the midpoint 2. Given.
of , E is the midpoint of .
3. BA � 2BD, 3. Definition of BC � 2BE midpoint.
4. 2BD � 2BE 4. If unequal quantitiesare multiplied bypositive equalquantities, then theproducts are unequal inthe same order.
5. BA � BC 5. Substitution postulate.
19. Statements Reasons1. m�DBA 1. Given
m�CAB2. 2m�DBA 2. If unequal quantities
2m�CAB are multiplied bypositive equalquantities, then theproducts are unequal inthe same order.
BCBA
25 , 2
3
A y6 , 3 B
Statements Reasons3. m�CBA � 2m�DBA, 3. Given.
m�DAB � m�CAB4. m�CBA m�DAB 4. Substitution
postulate.
20. Statements Reasons1. AB AD 1. Given.2. 2. If unequal
quantities aremultiplied bypositive equalquantities, then theproducts areunequal in thesame order.
3. , 3. Given.4. AE AF 4. Substitution
postulate.
21. Statements Reasons1. m�CAB � m�CBA 1. Given.2. m�CAB � m�CAD � 2. Partition postulate.
m�DAB, m�CBA �m�CBE � m�EBA
3. m�CAD � m�DAB, 3. Definition of anglem�CBE � m�EBA bisector.
4. m�CAB � 2m�DAB, 4. Substitution m�CBA � 2m�EBA postulate.
5. 2m�DAB � 2m�EBA 5. Substitutionpostulate.
6. m�DAB � m�EBA 6. If unequalquantities aredivided by positiveequal quantities,then the quotientsare unequal in thesame order.
7-4 An Inequality Involving the Lengths ofthe Sides of a Triangle (pages 275–276)
Writing About Mathematics1. If s is not the longest side, then 12 is the longest
side. By the triangle inequality theorem, 12 � s +7, so s 5.
2. a. No. This may not be true for negative values.Example: �5 � �4 � �3, but �5 �3 �(�4) or �5 �7.
b. Yes. If a, b, and c are the lengths of the sides ofa triangle, then they are positive. By thetriangle inequality theorem, a � b � c.
AF 5 12ADAE 5 1
2AB
12AB . 1
2AD
279
Developing Skills3. Yes 4. No 5. Yes6. No 7. Yes 8. No9. Yes 10. Yes
11. 2 � s � 6 12. 19 � s � 4313. 0 � s � 13 14. 2.9 � s � 22.115. Since 3x = x + 2x, by the triangle inequality
theorem, x, 2x, and 3x cannot be the lengths ofthe sides of a triangle.
16. a 4; solve the inequalities given by the triangleinequality theorem:
17. Statements Reasons1. ABCD is a 1. Given.
quadrilateral.2. AD � AC � CD, 2. Triangle inequality
AC � AB � BC theorem.3. AC � CD 3. If equal quantities
� AB � BC � CD are added tounequal quantities,then the sums areunequal in thesame order.
4. AD � AB � BC � CD 4. Transitive propertyof inequality.
18. Statements Reasons1. AB � AD �BD 1. Triangle inequality
theorem.2. BC � BD � DC 2. Partition postulate.3. AD � DC 3. Given.4. BC � BD � AD 4. Substitution
postulate.5. AB � BC 5. Substitution
postulate.
19. Statements Reasons1. PZ � PQ � QZ 1. Triangle inequality
theorem.2. PY � PZ 2. If equal quantities
� PY � PQ � QZ are added tounequal quantities,then the sums areunequal in thesame order.
24 , a
a 2 2 , a 1 a 1 2
4 , a0 , a
a 1 2 , a 1 a 2 2a , a 1 2 1 a 2 2
(Cont.)
Statements Reasons3. YQ � XY � XQ 3. Triangle inequality
theorem.4. YQ � PY � PQ 4. Partition postulate.5. PY � PQ � XY � XQ 5. Substitution
postulate.6. PY � PQ � QZ 6. If equal quantities
� XY � XQ � QZ are added tounequal quantities,then the sums areunequal in thesame order.
7. XZ � XQ � QZ 7. Partition postulate.8. PY + PQ � QZ 8. Substitution
� XY � XZ postulate.9. PY � PZ � XY � XZ 9. Transitive property
of inequality.
Hands-On Activitya. 1, 6 in.; 2, 5 in.; 2, 6 in.; 3, 4 in.; 3, 5 in.; 3, 6 in.;
4, 4 in.; 4, 5 in.; 5, 5 in.; 5, 6 in.; 6, 6 in.b. {1, 6, 6}, {2, 5, 6}, {2, 6, 6}, {3, 4, 6}, {3, 5, 6}, {3, 6, 6},
{4, 4, 6}, {4, 5, 6}, {5, 5, 6}, {5, 6, 6}, {6, 6, 6}c. {1, 1, 6}, {1, 2, 6}, {1, 3, 6}, {1, 4, 6}, {1, 5, 6}, {2, 2, 6},
{2, 3, 6}, {2, 4, 6}, {3, 3, 6}d. In part b, the sum of the two shortest sides is
greater than 6. In part c, the sum is less than orequal to 6.
7-5 An Inequality Involving an ExteriorAngle of a Triangle (pages 280–281)
Writing About Mathematics1. Yes. By the exterior angle inequality theorem, the
exterior angles that are not adjacent to the rightangle have measures greater than the right angle,that is, greater than 90°.Therefore, they are obtuse.
2. Yes.The exterior angle adjacent to the right angleforms a linear pair with the right angle.Therefore,it also measures 90° and is a right angle.
Developing Skills3. a. �TRP
b. �T, �S4. a. True
b. Definition of median5. a. True
b. A whole is greater than any of its parts.6. a. True
b. Exterior angle inequality theorem7. a. True
b. A whole is greater than any of its parts.8. a. True
b. Exterior angle inequality theorem
280
9. a. Trueb. Exterior angle inequality theorem
10. a. Trueb. A whole is greater than any of its parts.
11. False12. a. True
b. A whole is greater than any of its parts.13. FalseApplying Skills14. Statements Reasons
1. Let N be the midpoint 1. Every line segment of . has one and only
one midpoint.2. Draw , extending 2. Two points
the ray through N to determine a line. A point G so that line segment can
. be extended to anylength.
3. Draw . 3. Two pointsdetermine a line.
4. Extend through C 4. A line segment canto point F. be extended to any
length.5. m�ACF � m�ACG 5. Partition postulate.
� m�FCG6. 6. Definition of
midpoint.7. �ANB � �GNC 7. Vertical angles are
congruent.8. �ANB � �CNG 8. SAS (steps 2, 6, 7)9. �A � �ACG 9. Corresponding
parts of congruenttriangles arecongruent.
10. m�ACF m�ACG 10. A whole is greaterthan any of itsparts.
11. m�ACF m�A 11. Substitutionpostulate.
12. �ACF � �BCD 12. Vertical angles arecongruent.
13. m�BCD m�A 13. Substitutionpostulate.
15. Statements Reasons1. �ABD � �DBE 1. Given.
� �ABE,�ABE � �EBC� �ABC
2. m�ABD � m�ABE, 2. A whole is greater m�ABE � m�ABC than any of its parts.
3. m�ABD � m�ABC 3. Transitive propertyof inequality.
AN > NC
BC
GC
BN > NG
BNh
AC
16. Statements Reasons1. DE � EF 1. Given.2. 2. Definition of
congruent segments.3. �EFD � �EDF 3. Isosceles triangle
theorem.4. m�EFD � m�EDF 4. Definition of
congruent angles.5. m�EFG m�EDF 5. Exterior angle
inequality theorem.6. m�EFG m�EFD 6. Substitution
postulate.
17. Statements Reasons1. �ABC with 1. Given.
m�C � 90.2. Extend through 2. A line segment may
C to point D, so that be extended to any is formed. length.
3. Exterior angle �BCD 3. Definition of an and interior angle exterior angle.�BCA form a linear pair.
4. m�BCD � m�MCA 4. If two angles form a� 180 linear pair, then
they aresupplementary.
5. m�BCD � 90 � 180 5. Substitutionpostulate.
6. m�BCD � 90 6. Subtractionpostulate.
7. 0 � m�A � 90 7. Exterior angleinequality theorem.
8. �A is acute. 8. Definition of anacute angle.
18. Statements Reasons1. m�RMP m�RTM, 1. Exterior angle
m�RTM m�SRT inequality theorem.2. m�RMP m�SRT 2. Transitive property
of inequality.
19. Statements Reasons1. FC � FD 1. Given.2. 2. Definition of
congruent segments.3. �DCF � �CDF 3. Isosceles triangle
theorem.4. �EDF � �BCF 4. If two angles are
congruent, thentheir supplementsare congruent.
FC > FD
ACD
AC
DE > EF
281
5. m�EDF � m�BCF 5. Definition ofcongruent angles.
6. m�ABF m�BCF 6. Exterior angleinequality theorem.
7. m�ABF m�EDF 7. Substitutionpostulate.
7-6 Inequalities Involving Sides and Anglesof a Triangles (pages 284–285)
Writing About Mathematics1. a. If the measures of two angles of a triangle are
equal, then the lengths of sides opposite theseangles are equal.
b. Yes2. a. If two angles of a triangle are congruent, then
sides opposite these angles are congruent.b. The converse statement and the
contrapositive statement both handle thesame relationship between the angles and thesides opposite the angle in a triangle.However, the converse deals with congruenceand the contrapositive deals with equality.
Developing Skills3. �B 4. 5.6. 7. �B 8. �C9. 10.
11. 12. RT RS ST
Applying Skills13. Statements Reasons
1. m�ABC m�CBD 1. Given.2. m�CBD m�CAB 2. Exterior angle
inequality theorem.3. m�ABC m�CAB 3. Transitive property
of inequality.4. AC BC 4. If the measures of
two angles of atriangle are unequal,then the lengths ofthe sides of oppositethese angles areunequal and thelonger side isopposite the largerangle.
14. a. The measure of the exterior angle adjacent to�C measures 90 degrees. By the exteriorangle inequality theorem, the nonadjacentinterior angles, �A and �B, have measuresless than 90 degrees. By definition, �A and�B are acute.
GH , FG , FHCD . BC . BDRS
ABACDF
b. �C is the largest angle, so hypotenuse ,which is opposite �C, is the longest side.
15. The exterior angle adjacent to the obtuse angleforms a linear pair with the obtuse angle. Sincethe measure of the obtuse angle is greater than90 degrees, the measure of the exterior angle isless than 90 degrees. By the exterior angleinequality theorem, the measures of the remoteinterior angles are also less than 90 degrees.Therefore, the two angles are acute.
Review Exercises (pages 286–287)1. A whole is greater than any of its parts.2. Transitive property of inequality3. Exterior angle inequality theorem4. If the measures of two angles of a triangle are
unequal, then the lengths of the sides oppositethese angles are unequal and the longer side liesopposite the larger angle.
5. If equal quantities are added to unequalquantities, then the sums are unequal in the sameorder.
6. Triangle inequality postulate7. If the measures of two angles of a triangle are
unequal, then the lengths of the sides oppositethese angles are unequal and the longer side liesopposite the larger angle.
8. A whole is greater than any of its parts.
9. Statements Reasons1. AE BD, EC DC 1. Given.2. AE + EC BD + DC 2. If unequal quantities
are added to unequalquantities in thesame order, the sumsare unequal in thesame order.
3. AC = AE + EC, 3. Partition postulate.BC = BD + DC
4. AC BC 4. Substitutionpostulate.
5. �B �A 5. If the lengths of twosides of a triangle areunequal, then themeasures of theangles opposite thesesides are unequaland the larger anglelies opposite thelonger side.
AB
282
10. Statements Reasons1. AD DC 1. Given.2. m�ACD m�CAD 2. If the lengths of two
sides of a triangle areunequal, then themeasures of theangles opposite thesesides are unequaland the larger anglelies opposite thelonger side.
3. �ABC � �CDA 3. Given.4. �ACD � �BAC 4. Corresponding parts
of congruent tri-angles are congruent.
5. m�ACD � m�BAC 5. Definition ofcongruent angles.
6. m�BAC m�CAD 6. SubstitutionPostulate.
7. does not 7. Definition of angle bisect �A. bisector.
11. Statements Reasons1. �ABC is isosceles 1. Given.
with .2. m�CAB � m�CBA 2. Base angles of an
isosceles triangle areequal in measure.
3. m�CBA m�DBA 3. A whole is greaterthan any of its parts.
4. m�CAB m�DBA 4. Substitutionpostulate.
5. DB DA 5. If the measures oftwo angles of a tri-angle are unequal,then the lengths ofthe sides of oppositethese angles are un-equal and the longerside is opposite thelarger angle.
12. Statements Reasons1. �RST is isosceles 1. Given.
with RS � ST.2. m�SRT � m�STR 2. Base angles of an
isosceles triangle areequal in measure.
3. �SRT and �SRP 3. Definition of anform a linear pair. exterior angle of a�STR and �STQ triangle.form a linear pair.
CA > CB
AC
(Cont.)
Statements Reasons4. m�SRP � m�STQ 4. If two angles are
congruent, then theirsupplements arecongruent.
13. ADB � 7 units; AEFGHIB � 7 units. The lengthof one side of a triangle is less than the sum ofthe lengths of the other two sides. Therefore,CB � AD � 1 and AC � CB � AC � AD � 1.We know that AC � 2 and AD � 4, so AC � CB � 7. Thus, ACB � ADB.a. ACB is the shortest path.b. ADB and AEFGHB are the longest paths.
Exploration (page 287)1. a. Proofs will vary.
Let ABC and DEF be two triangles with, , and m�B m�E.
Draw a point G on such that �ABG � �E. Then �ABG � �DEF by SAS.Corresponding parts of congruent trianglesare congruent, so and AG � DF.Since a whole is greater than any of its parts,AC AG. By the substitution postulate,AC DF.
b. Answers will vary.2. a. Proofs will vary.
Let ABC and DEF be two triangles with , , and AC DF. Draw a
point G on such AG � DF. Then �ABG � �DEF by SSS. Corresponding parts of congruent triangles are congruent, so�ABG � �E and m�ABG � m�E. Since awhole is greater than any of its parts,m�ABC m�ABG. By the substitutionpostulate, m�ABC m�E.
b. Answers will vary.
Cumulative Review (pages 288–289)Part I
1. 3 2. 4 3. 3 4. 15. 4 6. 3 7. 1 8. 19. 4 10. 2
Part II11. Yes. A statement and its contrapositive are
logically equivalent. Therefore, “If our meeting isnot cancelled, then the snow does not continue tofall” is true. By the law of detachment, since it istrue that “our meeting is not cancelled,” we canconclude that “snow does not continue to fall.”
ACBC > EFAB > DE
AG > DF
ACBC > EFAB > DE
283
12. A�(�2, �4), B�(�2, 0), C�(0, �1)T
�4,�5(0, 3) � (�4, �2), ry=x(�4, �2) � (�2, �4)T
�4,�5(4, 3) � (0, �2), ry=x(0, �2) � (�2, 0)T
�4,�5(3, 5) � (�1, 0), ry=x(�1, 0) � (0, �1)Part III13. Statements Reasons
1. AP � BP 1. Assumption.2. 2. Definition of con-
gruent segments.3. bisects . 3. Given.4. R is the midpoint of 4. Definition of
. bisector.5. 5. Definition of
midpoint.6. 6. Reflexive property.7. �APR � �BPR 7. SSS.8. �ARP � �BRP 8. Corresponding
parts of congruenttriangles arecongruent.
9. 9. If two lines intersectto form congruentadjacent angles,then they areperpendicular.
10. is not ⊥ to . 10. Given.11. AP � BP 11. Contradiction.
14. Statements Reasons1. bisectors �DAB, 1. Given.
bisects �DCB.2. �DAC � �BAC, 2. Definition of angle
�DCA � �BCA bisector.3. 3. Reflexive property.4. �DAC � �BAC 4. ASA.5. �B � �D 5. Corresponding parts
of congruenttriangles arecongruent.
Part IV15. m�PTR � m�QTS because they are vertical
angles, so x � y.m�PTR � m�RTQ � 180 since they aresupplements, so
x � 2x � y � 180
x � 2x � x � 180
4x � 180
x � 45
m�PTR � m�QTS � 45,m�RTQ � m�PTS � 2(45) � 45 � 135
AC > AC
CAhACh
ARBPR
PR ' ARB
PR > PR
AR > BRARB
ARBPR
AP > BP
16. a. m�BCA � m�DCB � 180, so
6x � 8 � 4x � 12 � 180
10x � 20 � 180
10x � 160
x � 16
m�BCA � 6(16) � 8 � 104,m�DCB � 4(16) � 12 � 76
284
b. m�A � m�B � m�BCA; we are given thatm�A � m�B. Since m�B � m�DCB andm�DCB � m�BCA, by the transitiveproperty of inequality, m�B � m�BCA.
c. BC � AC � AB; if the measures of two anglesof a triangle are unequal, then the lengths ofthe sides opposite these angles are unequal andthe longer side lies opposite the larger angle.
Chapter 8. Slopes and Equations of Lines
8-1 The Slope of a Line (page 295) Writing About Mathematics
1. “Delta y” means change of position with respectto the y-axis.
2. No. “No slope” is not the same as “zero slope.”“Zero slope” is a well-defined number. “Noslope” occurs when the formula for the slope of a line, , is undefined.
Developing Skills3. b. m � 1 c. Slant upward4. b. m � 3 c. Slant upward5. b. m � �2 c. Slant downward6. b. m � 1 c. Slant upward7. b. m � c. Slant downward8. b. No slope c. Vertical9. b. m � �1 c. Slant downward
10. b. m � 0 c. Horizontal11. b. m � 2 c. Slant upward12.
13. y
O1x
(�1, 3)
1
y
O1 x
(0, 1)
212
m 5y2
2 y1
x2 2 x1
14.
15.
16.
17. y
O1
x
(�4, 7)
�1
y
O1x
(�3, 2)
1
y
O1x
(�4, 5)
1
y
O1x
(2, 5)
1
18.
19.
20.
21.
22.
23. y
O1 x(�2, 0)
�1
y
O1x
(0, �2)
�1
y
O1 x
(�1, 0)
1
y
O1x
(�1, 5)
1
y
O1 x
(�2, 3)
�1
y
O1 x
(1, 3)
�1
285
Applying Skills24. a–c.
d. C(5, 10)e. , the altitude from C, is a vertical segment,
and by the reflexive property ofcongruence. Since and are horizontalsegments, AD � �5 � 2� � 3 and BD � �8 � 5� � 3. Thus, . Also,�ADC and �BDC are right angles sincevertical and horizontal lines areperpendicular. Therefore, �ADC � �BDCby SAS, and since corresponding parts of congruent triangles are congruent.
25. C(3, 2) and D(9, 2); C(3, �6) and D(9, �6)26. m �27. 7.5 feet
8-2 The Equation of a Line (pages 299–300)Writing About Mathematics
1. No. If the slope of is equal to the slope of ,
then and could be two distinct parallel
lines.2. a. Yes. The point-slope form of the equation of a
line requires that be set equal to the slope m. Since m does not exist, the equalitycannot be used.
b. A line with no slope is vertical. The equationof a vertical line is x � a, where a is the x-coordinate of any point on the line.
Developing Skills3. 4.5. 6.7. 8.9. 10.
11. 12.13. 14. x � 215. a. Yes. Slope of � slope of �
b. y 5 12x 1 32
12QRPQ
y 5 22x 1 4x 5 1y 5 5y 5 x 2 4y 5 23x 1 5y 5 21
2x 1 32y 5 24x 1 5y 5 2xy 5 24x 1 6y 5 x 2 1x 5 0
y 2 bx 2 a
CDg
ABg
CDAB
34
AC > BC
AD > BD
BDADCD > CD
CD
x
C
A
O1
1
B
y
D
16. a. No. Slope of � slope of
b. ; ;
Applying Skills17. a. y � 40x � 20 b. $140
c. The increase in cost of repair per hour of workd. The initial charge to repair a TV not including
any labor18. a. y � 18x � 8 b. $152
c. The price per cartridged. The shipping cost
19. To find the x-intercept, substitute y � 0:
Therefore, the x-intercept is (a, 0).To find the y-intercept, substitute x � 0:
Therefore, the y-intercept is (0, b).
8-3 Midpoint of a Line Segment (pages 306–307)
Writing About Mathematics
1.
✔
2.
✔
Developing Skills3. (3, 4) 4. (6, 6) 5. (5, 4)6. (2, 2) 7. (0, 0) 8. (4, 5.5)9. (0.5, 4) 10. (1, 8) 11. (�2, �3)
12. (3, 3.5) 13. (0.75, 2.75) 14. (0.5, 6)15. B(0, 5) 16. B(3, 15) 17. A(6, 3)18. A(�1, 2) 19. M(2, 6.5) 20. B(0, 0)Applying Skills21. a. B(9, 1), D(1, 7)
b. Midpoint of � � (5, 4)
Midpoint of � � (5, 4)A 9 1 12 , 7 1 1
2 BBDA 1 1 9
2 , 1 1 72 BAC
a 1 b 5 a 1 b2b 2 b 1 a 5? a 1 b
2 Ab 2 b 2 a2 B 5? 2 A a 1 b
2 Bb 2 b 2 a
2 5? a 1 b2
a 1 b 5 a 1 b2a 1 b 2 a 5? a 1 b
2 Aa 1 b 2 a2 B 5? 2 A a 1 b
2 Ba 1 b 2 a
2 5? a 1 b2
y 5 b
0a 1
yb 5 1
xa 1
yb 5 1
x 5 a
xa 1 0b 5 1
xa 1
yb 5 1
LNg
:y 5 35x 1 12
5
MNg
:y 5 23x 1 83LM
g:y 5 3
4x 1 94
MNLM
286
22. a. Q(x2, y1), S(x1, y2)
b. Midpoint of �
Midpoint of
�
23. a. M (2, 3) b. y � x � 1 c. N(2, 5)d. y � �x � 7 e. (4, 4) f. P(5, 4)g. y � 4 h. Yes i. Yes; (4, 4)
8-4 The Slopes of Perpendicular Lines(pages 310–312)
Writing About Mathematics1. The line x � 5 has no slope. Therefore, there can
be no reciprocal.2. Both are correct. The negative reciprocal of
.
Developing Skills3. a. m � 4 b. m �4. a. m � 1 b. m � �15. a. m � �1 b. m � 16. a. m � 2 b. m �
7. a. m � b. m �
8. a. m � b. m � �2
9. a. m � �2 b. m �
10. a. m � b. m � �211. a. No slope b. m � 012. a. m � 0 b. No slope
13. 14. y � �2x
15. y � 3x � 18 16. x � 217. No 18. y � �2x � 119. y � �x � 3 20. y � 2x � 121. y � 3x 22.
23. y � �3 24.Applying Skills25. a. y � �x � 1
b. Yes. It intersects M(1, 0), which is the midpoint of .
c. In �DEF, the altitude from D is also the median. Call this segment . By the definitionof altitude, , so �DGE � �DGF.By the definition of median, G is the midpointof , so . by the reflexive property of congruence. Therefore,�DGE � �DGF by SAS. Then since corresponding parts of congruenttriangles are congruent, which makes�DEF isosceles.
DE > DF
DG > DGEG > FGEF
DG ' EFDG
EF
y 5 72x 2 14
y 5 12x 1 74
y 5 715x 1 21
2
12
12
12
2323
2
212
14
213 5 2 5 3
A x2 1 x12 ,
y2 1 y12 B 5 A x1 1 x2
2 , y1 1 y2
2 BQS
A x1 1 x22 ,
y1 1 y22 BPR
1�1
3
26. The slopes of and are both . The slopesof and are both . Any two consecutive sides have slopes that are negative reciprocals.Therefore, any two consecutive sides areperpendicular and all angles are right angles. Bydefinition, ABCD is a rectangle.
27. a. 1 b. y � �x � 8c. No slope d. y � 3e. �0.5 f. y � 2x � 7g. The point (5, 3) makes the equations of the
perpendicular bisector true:
✔ ✔
✔
28. a. Altitude at A: y � x � 2;altitude at B: y � �0.5 x � 2;altitude at C: x � 2
b. (0, 2)29. a. The measure of the exterior angle adjacent to
�M is 90°. By the exterior angle inequalitytheorem, the nonadjacent interior angles, �Land �M, both measure less than 90°.Therefore, �L and �M are acute.
b. , ,c. �L, �N, �M
Hands-On ActivityIn a–c, answers will vary. Examples are given.
a. l1: y � �x � 7 b. l1: y �
l2: y � �x � 11 l2: y �
c. l1: y �
l2: y �
8-5 Coordinate Proof (pages 315–317)Writing About Mathematics
1. Yes. The slope of the segment connecting (a, 0)and (0, b) is . The slope of the segment
connecting (0, b) and (c, 0) is . If the segments are perpendicular, then the slopes are negative
reciprocals and would equal . Since
, the two sides of the triangle are
perpendicular, and the triangle is a right triangle.2. Yes. The point (0, b) lies on the y-axis and (a, 0)
and (c, 0) are on the x-axis. Therefore, the
5 ab 5 2b
c
b�a�1
2bc
b�a�1
2bc
2ba
213x 1 5
213x 1 20
3
32x 2 4
32x 2 4
LNLMMN
3 5 3y 5 3
3 5 33 5 33 5
? 2(5)273 5? 25 1 8
y 5 2x27y 5 2x 1 8
223BCAD
32DCAB
287
altitude from (0, b) is on the y-axis. Since a and chave the same sign, this altitude lies outside thetriangle. Thus, the triangle is obtuse.
Developing Skills
3. Midpoint of � (2, 2)Midpoint of � (2, 2)The two segments have the same midpoint.Therefore, the two segments bisect each other.
4. Midpoint of �
Midpoint of �The two segments have the same midpoint.Therefore, the two segments bisect each other.Slope of �
Slope of � 7The slopes are negative reciprocals.Therefore, thetwo segments are perpendicular to each other.
5. a. K � (3, 2)b. Slope of � �3
Slope of �
The slopes are negative reciprocals. Therefore,is perpendicular to .
c. Since the altitude of the triangle is also itsmedian, the triangle is isosceles.
6. a. Slope of � �3Slope of � 3The slopes are negative reciprocals. Therefore,the two sides are perpendicular.
b. �C c. d. (5, 5)e. y � 2x � 5 f. y � 2x � 5g. Yes. The altitude and the median are the same
line segment.
In 7–8, part a, answers will be graphs.7. b. Midpoint of � (�2, 4)
Midpoint of � (6, 4)Midpoint of � (4, 0)
c. Slope of � 2Slope of �
Slope of � 0d. : ; : ; : x � 4e. (4, 1)
8. a. Midpoint of � (6, �1)Midpoint of � (6, �1)The two segments have the same midpoint.Therefore, the segments bisect each other.
b. Slope of �
Slope of � �2The slopes are negative reciprocals.Therefore, the two segments areperpendicular to each other.
BD
12AC
BDAC
ACy 5 232x 2 5BCy 5 21
2x 1 3ABAC
223BC
ABACBCAB
AB
BCAC
LNMK
13LN
MK
BD
217AC
A512, 11
2 BBD
A512, 11
2 BAC
CDAB
Applying Skills9. Since and are vertical segments,
SP � �b � 0� � �b� and RQ � �b � 0� � �b�.Therefore, . is a horizontal segment.Since horizontal and vertical segments areperpendicular, and .�SPQ and �RQP are both right angles, so�SPQ � �RQP. By the reflexive property,
. Then �SPQ � �RQP by SAS.Corresponding parts of congruent triangles arecongruent, so .
10. (1) The midpoint of is .
The midpoint of is .
Therefore, and intersect each other attheir midpoints so, by definition, bisect eachother.
(2) The slope of � 1 and the slope of � �1, so the slopes are negative
reciprocals. Therefore, and areperpendicular to each other.
11. Let A � (0, 0), B � (2a, b), and C � (a, b). Sinceis a horizontal segment, the altitude from C is
a vertical line intersecting in the point D(a, 0). Then, AD � �0 � a� � �a� and DB � �a � 2a� � �a�, so . Since and
are vertical and horizontal segments,respectively, they are perpendicular to eachother. Thus, �ADC and �CDB are both right angles and �ADC � �CDB. Since bythe reflexive property, �ADC � �BDC by SAS.Therefore, because corresponding parts of congruent triangles are congruent. Bydefinition �ABC is isosceles.
12. Let A � (�a, 0), B � (a, 0), and C � (0, b).Under the translation , the images of thevertices of the triangle are:
By Exercise 11, �A�B�C� is an isosceles triangle.Since distance is preserved under translation, theimage of a triangle under a translation iscongruent to the given triangle by SSS.Therefore, �ABC is also an isosceles triangle.
13. a. E(a, b); F(c, d)b. Slope of �
Slope of � 2b 2 2d2a 2 2c 5 b 2 d
a 2 cBC
b 2 da 2 cEF
C(0, b) S Cr(a, b)B(a, b) S Br(2a, b)
A(2a, 0) S Ar(0, 0)
Ta,0
AC > CB
CD > CD
ABCDAD > DB
ABAB
FHEGFH
EG
FHEG
A 0 1 a2 , a 1 0
2 B 5 A a2, a2 BFH
A 0 1 a2 , 0 1 a
2 B 5 A a2, a2 BEG
SQ > PR
PQ > PQ
RQ ' PQSP ' PQ
PQSP > RQ
RQSP
288
14. a. We are given a segment with endpoints A(�a, 0) and B(a, 0), and points P(0, b) andQ(0, c). Consider �AOP and �BOP, with O the origin. AO � �0 � (�a)� � a and BO � �0 � a� � a, so . Since and
are vertical and horizontal segments,respectively, they are perpendicular to eachother. Thus, �AOP and �BOP are both rightangles, and �AOP � �BOP. By the reflexive property, , and �AOP � �BOP bySAS. Therefore, because corre-sponding parts of congruent triangles arecongruent, and P is equidistant from the endpoints of . Similarly, and�AOQ � �BOQ by SAS. Therefore,
and Q is equidistant from theendpoints of .
b. If two points are each equidistant from theendpoints of a line segment, then the pointsdetermine the perpendicular bisector of the
line segment. Therefore, by part a, is the
perpendicular bisector of .
8-6 Concurrence of the Altitudes of aTriangle (pages 321–322)
Writing About Mathematics1. Yes. Using the formula for the coordinates of the
orthocenter from the proof, the orthocenter is . Since A is to the left of the origin, a is
negative. Since C is to the right of the origin, c ispositive. Thus, ac is a negative number, and so �ac is a positive number. Since b is positive,is a positive number. Therefore, the y-coordinateof the orthocenter is positive, and so theorthocenter is located above the origin.
2. No. Using the formula for the coordinates of theorthocenter from the proof, the orthocenter is
. Since both A and C are located to the right of the origin, both a and c are positive. Thus,ac is positive, and so �ac is negative. Since b is positive, is negative. Therefore, the y-coordinate of the orthocenter is negative, andso the orthocenter is located below the origin.
Developing Skills
3. a. Slope of �
Slope of EF 5 234
43DE
2acb
A0, 2acb B
2acb
A0, 2acb B
AB
PQg
ABAQ > BQ
QO > QOAB
AP > BPPO > PO
ABPOAO > BO
The slopes are negative reciprocals. Therefore,and are perpendicular, and so �DEF
is a right angle.b. Since �DEF is a right angle, is the
altitude from D to and is the altitudefrom F to . The two altitudes intersect at E.Therefore, E is the orthocenter.
4. (0, 1) 5. (0, 8) 6. (2, �2.5)7. (1, 6) 8. (3.5, 3.2) 9. (12.5, �5.5)
Applying Skills10. a. y � 2x � 6 b. y � �x � 6
c. (0, 6) d. x � 0e. Point P has coordinates (0, 3) and so it lies on
the line x � 0.11. a. y � �x b. y � x �
c. d. y �
e. Substitute x � 1 and y � 1 into the equationfrom part d:
✔
The coordinates of P make the equation true.Therefore, P lies on the line.
Hands-On Activity(1) b. (0, 6) (2) b. (6, �2) (3) b. (0, �4)
Review Exercises (pages 323–324)1. a. (3, �2) b. m � c. y �
2. a. (2, �1) b. m � � c. y �
3. a. (3, �3) b. m � �1 c. y � �x4. a. m � �0.5 b. y � �0.5x � 4
c. (2, 3) d. y � 2x � 15. a. Slope of � 2
Slope of �
Therefore, the two segments areperpendicular, and the triangle is a righttriangle.
b. c. x � 1 d. y � �3x � 7
e. The equation of the line that contains is. This line is perpendicular to the
median from S. Therefore, the median is alsothe altitude.
6. (6, 7)
y 5 13x 2 11
3
RT
A212, 21
2 B
212ST
RS
23x5 1 15
35
223x22
3
1 5 1
1 5? 77
1 5? 237(1) 1 10
7
237x 1 10
7A2212, 21
2 B103
13
DEEFEF
DE
EFDE
289
7. a. Midpoint of � (3, 3)Midpoint of � (2, 5)Midpoint of � (0, 3)
b. Slope of � 1Slope of � 0Slope of � �2
c. Slope of the altitude from F � �1Slope of the altitude from D � undefinedSlope of the altitude from E �
d. Perpendicular bisector of : x � 2Perpendicular bisector of : y � �x � 6Perpendicular bisector of : y � 0.5x � 3
e. (2, 4)8. a. Slope of � 1
Slope of � �1The two sides are perpendicular and form aright angle. Therefore, the triangle is a righttriangle.
b. The slope of is �1. Therefore, isperpendicular to since the slope of is1, and �ANM � �CNM since they are bothright angles. because N is themidpoint of . by the reflexiveproperty of congruence. Therefore, �ANM ��CNM by SAS. Since corresponding parts ofcongruent triangles are congruent,and AM � CM. Since M is the midpoint of
, AM � MB. By the transitive property ofequality, AM � BM � CM, and M is equi-distant from the vertices of A, B, and C.
9. a. Midpoint of � (�2, 0)
b. Since is a vertical line and is a horizontal line, the two lines are perpendicular. Since A is the midpoint of ,
is a perpendicular bisector. Since E is onthe perpendicular bisector of , it is equidistant from points D and F. Therefore,DE � FE, and .
10. Slope of � 3
Slope of �
Therefore, the two sides are perpendicular, and�D is a right angle.
213DC
g
DFg
DE > FE
DFAE
DF
AEg
DFg
DF
AB
AM > CM
NM > NMACAN > NC
ACACNMNM
CBAC
DFDEEF
12
DFEFDE
DFEFDE
Exploration (page 324)a.
Area of the enclosing rectangle � 8(8) � 64Area of Triangle 1 � (1)(8) � 4
Area of Triangle 2 � (5)(7) � 17.5
Area of Triangle 3 � (3)(8) � 12Area of enclosed triangle � 64 � (4 � 17.5 � 12)
� 30.5 sq unitsb.
Area of the enclosing rectangle � 8(10) � 80Area of Triangle 1 � (8)(5) � 20
Area of Triangle 2 � (5)(5) � 12.5
Area of Triangle 3 � (3)(10) � 15Area of enclosed triangle � 80 � (20 � 12.5 � 15)
� 32.5 sq unitsc.
Area of the enclosing rectangle � 6(11) � 66Area of Triangle 1 � (6)(11) � 33
Area of Triangle 2 � (2)(10) � 1012
12
x
A
O�1
By
C
1
12
12
12
x
A
O1�1
By
C
12
12
12
x
C
A O11
By
290
Area of Triangle 3 � (1)(4) � 2Area of remaining rectangle � 2Area of enclosed triangle � 66 � (33 � 10 � 2 � 2) � 19 sq units
Yes. The area of the quadrilateral is 19.
Cumulative Review (pages 325–327)Part I
1. 3 2. 1 3. 2 4. 15. 2 6. 4 7. 2 8. 19. 2 10. 3
Part II11. a.
b. See graph; B� � (5, �7), C� � (9, 0)c. a � 4, b � �2
12. Yes. Lines and have slopes of �2 and ,
respectively. Since these slopes are negative
reciprocals, the lines are perpendicular and �C is
a right angle. Therefore, �ABC is a right triangle.Part III13. Statements Reasons
1. bisects �RTS. 1. Given.2. �RTQ � �STQ 2. Definition of angle
bisector.3. 3. Given.4. �TQR and �SQT are 4. Definition of
right angles. perpendicular lines.5. �TQR � �SQT 5. Right angles are
congruent.6. 6. Reflexive property.7. �TQR � �TQS 7. ASA.8. 8. Corresponding
sides of congruenttriangles arecongruent.
9. �RST is isosceles. 9. Definition ofisosceles triangle.
TR > TS
TQ > TQ
TQ ' RS
TQh
12BC
gACg
x
AO
�1
B
yC
�1C�
A�
B�
12
14. Let e � “Evanston is the capital of Illinois.”Let c � “Chicago is the capital of Illinois.”Let s � “Springfield is the capital of Illinois.”Then in symbols, the given statements are:
�e → �c
s ∨ c
�e
Since �e is true, �c is true by the Law ofDetachment.Since �c is true, c is false.Since c is false and s ∨ c is true, s is true by theLaw of Disjunctive Inference.Therefore, Springfield is the capital of Illinois.
Part IV15. Substitute the first equation for y in the second
equation, and solve for x:
x 5 1
7x 5 7
x 1 6x 2 2 5 5
x 1 2(3x 2 1) 5 5
291
Substitute x � 1 into either equation to find y:
Therefore, the coordinates of the intersectionpoint are (1, 2).
16. Statements Reasons1. and bisect 1. Given.
each other at G.2. G is the midpoint of 2. Definition of
and of . bisector.3. and 3. Definition of
midpoint.4. �CGD � �EGF 4. Vertical angles are
congruent.5. �CGD � �EGF 5. SAS.6. 6. Corresponding
sides of congruenttriangles arecongruent.
CD > EF
DG > GECG > GF
DGECGF
DGECGF
y 5 2
y 5 3(1) 2 1
y 5 3x 2 1
Chapter 9. Parallel Lines
9-1 Proving Lines Parallel (page 334)Writing About Mathematics
1. Corresponding angles2. Yes. This is the contrapositive of Theorem 9.1a: If
two coplanar lines cut by a transversal are notparallel, then the alternate interior angles formedare not congruent.
Developing Skills3. If two coplanar lines are cut by a transversal so
that the alternate interior angles formed arecongruent, then the two lines are parallel.
4. If two coplanar lines are cut by a transversal sothat the corresponding angles are congruent,then the two lines are parallel.
5. If two coplanar lines are cut by a transversal sothat the interior angles on the same side of thetransversal are supplementary, then the lines areparallel.
6. Vertical angles (�2 and �4) are congruent. If twocoplanar lines are cut by a transversal so that theinterior angles on the same side of the transversalare supplementary, then the lines are parallel.
7. Vertical angles (�2 and �4) are congruent. Iftwo coplanar lines are cut by a transversal so thatthe corresponding angles are congruent, then thelines are parallel.
8. Vertical angles (�7 and �5) are congruent. Iftwo coplanar lines are cut by a transversal so thatthe interior angles on the same side of thetransversal are supplementary, then the lines areparallel.
Applying Skills
9. Given: intersects and ; �1 � �5.
Prove:
Statements Reasons
1. is not parallel 1. Assumption.
to .2. �1 � �3 2. Vertical angles are
congruent.3. �3 is not congruent 3. If two coplanar
to �5. lines cut by atransversal are notparallel, then thealternate interiorangles formed arenot congruent.
4. �1 is not congruent 4. Transitive property.to �5.
5. �1 � �5 5. Given.
6. 6. Contradiction.ABg
7CDg
CDg
ABg
ABg
7CDg
CDg
ABg
EFg
10. Given: and
Prove:Statements Reasons
1. and 1. Given.
2. �1 and �2 are right 2. Definition of angles. perpendicular lines.
3. �1 � �2 3. Right angles arecongruent.
4. 4. If two coplanar lines are cut by atransversal so thatthe alternate interiorangles are congruent,then the two lines areparallel.
11. m�A � m�B � 3x � (180 � 3x) � 180. Sinceinterior angles on the same side of the transversal are supplementary, .
12. We are given that , so �BCD is a right angle and m�BCD � 90. Therefore,m�BCD � m�ADC � 180. Since interior same angles on the same side of the transversal aresupplementary, .
13. a. Statements Reasons1. and bisect 1. Given.
each other at E.2. E is the midpoint 2. Definition of
of and of . bisector.3. , 3. Definition of
midpoint.4. �CEA � �DEB 4. Vertical angles are
congruent.5. �CEA � �DEB 5. SAS.
b. Statements Reasons1. �CEA � �DEB 1. Part a.2. �ECA � �EDB 2. Corresponding parts
of congruent anglesare congruent.
c. Statements Reasons1. �ECA � �EDB 1. Part b.2. 2. If two coplanar lines
are cut by a trans-versal so that thealternate interiorangles formed arecongruent, then thetwo lines areparallel.
CA y DB
CE > EDAE > EB
CDAB
CDAB
AD y BCCD
DC ' BCAD y BCAB
ABg
y CDg
CDg
' EFg
ABg
' EFg
ABg
7CDg
CDg
' EFg
ABg
' EFg
292
14. Given: intersects and ; �1 � �2.
Prove:
Statements Reasons
1. intersects 1. Given.
and ;�1 � �2.
2. �1 � �3 2.Vertical angles arecongruent.
3. �2 � �3 3. Transitive property.
4. 4. If two coplanar lines are cut by a transversalso that thecorresponding anglesare congruent, then thetwo lines are parallel.
9-2 Properties of Parallel Lines (pages 340–341)
Writing About Mathematics1. a. Yes. The inverse of Theorem 9.1a states: If two
coplanar lines are cut by a transversal so thatthe alternate interior angles formed are notcongruent, then the lines are not parallel. Thecontrapositive of this statement, which has thesame truth value, is: If two coplanar lines cutby a transversal are parallel, then the alternateinterior angles formed are congruent. This isthe same as Theorem 9.1b.
b. Yes. The inverse of Theorem 9.6 states: If atransversal is not perpendicular to one of twoparallel lines, then it is not perpendicular tothe other. Alternate interior angles of parallellines are congruent, so if the transversal doesnot form a right angle with one of the parallellines, it will not form a right angle with theother.
2. The measures of the angles formed by theparallel lines and the transversal are all 90degrees. Alternate interior angles of parallel linesare congruent, so their measures are equal. If theangles are also supplementary, then they mustboth be right angles.
Developing Skills3. 80 4. 150 5. 1206. 75 7. 115 8. 509. 42 10. 80 11. 65
12. 60 13. 44 and 136 14. 21 and 21
ABg
y CDg
CDg
ABg
EFg
ABg
y CDg
CDg
ABg
EFg
15. a. m�1 � 70, m�2 � 110, m�3 � 70, m�4 � 50,m�5 � 130, m�6 � 50, m�7 � 50, m�8 � 60,m�9 � 70
b. 120 c. Yes d. 120e. 110 f. 130 g. 180
16. m�A � m�C � 75; m�ABC � m�D � 105Applying Skills17. Statements Reasons
1. and 1. Given.
transversal 2. �3 � �5 2. If two parallel lines
are cut by atransversal, then thealternate interiorangles formed arecongruent.
3. �4 is the 3. If two angles form supplement of �3. a linear pair, then they
are supplementary.4. �4 is the 4. If two angles are
supplement of �5. congruent, then theirsupplements arecongruent.
18. Statements Reasons
1. , 1. Given.2. �AEF � �EFD 2. If two parallel lines are
cut by a transversal,then the alternateinterior angles formedare congruent.
3. �AEF is a right 3. Perpendicular lines angle. intersect to form right
angles.4. �EFD is a right 4. Definition of
angle. congruent angles.
5. 5. Perpendicular lines intersect to form rightangles.
19. When two parallel lines are cut by a transversal,alternate interior angles are congruent. Theseangles are also congruent to their respectivevertical angles. Since these vertical angles arealternate exterior angles, alternate exteriorangles are congruent.
EFg
' CDg
EFg
' ABg
ABg
y CDg
EFg
ABg
y CDg
293
20. Statements Reasons
1. bisects exterior 1. Given.�BCD
2. �BCE � �DCE 2. Definition of anglebisector.
3. 3. Given.4. �DCE � �A 4. If two parallel lines
are cut by a trans-versal, then corre-sponding angles arecongruent.
5. �BCE � �B 5. If two parallel lines arecut by a transversal,then the alternateinterior angles formedare congruent.
6. �A � �B 6. Substitution postulate.
21. a. Statements Reasons1. �CAB � �DCA 1. Given.
2. 2. If two coplanar lines are cut by atransversal so thatthe alternate inte-rior angles formedare congruent, thenthe two lines areparallel.
b. Statements Reasons
1. 1. Part a.2. �CAB � �DCA, 2. Given.
�DCA � �ECB3. �CAB � �ECB 3. Transitive property.4. �ECB � �ABC 4. If two parallel lines
are cut by a trans-versal, then thealternate interiorangles formed arecongruent.
5. �CAB � �ABC 5. Transitive property.6. m�ABC � 6. Angles forming a
m�CBG � 180 linear pair aresupplementary.
7. m�CAB � 7. Substitution m�CBG � 180 postulate.
8. �CAB is the sup- 8. Definition of sup-plement of �CBG. plementary angles.
AB y DCEg
AB y DCEg
CEh
y AB
CEh
22. If two parallel lines are cut by a transversal, thenthe two interior angles on the same side of thetransversal are supplementary. Therefore, since
and �P is a right angle, its supplement,�S, is also a right angle. Similarly, since and �P and �S are right angles, their respectivesupplements, �Q and �R, are also right angles.
23. If two parallel lines are cut by a transversal, thenthe two interior angles on the same side of thetransversal are supplementary. Therefore, since
and �K is an acute angle measuring less than 90°, its supplement, �N, must measuremore than 90° and be obtuse. Similarly, since
and �K is acute, its supplement, �L,must be obtuse. Since �N is an obtuse anglemeasuring more than 90°, its supplement, �M,must measure less than 90° and be acute.
9-3 Parallel Lines in the Coordinate Plane(pages 345–347)
Writing About Mathematics1. They are the same line.2. Vertical lines are parallel, but they do not have
the same slope since vertical lines do not haveslope.
Developing Skills3. Perpendicular 4. Parallel 5. Parallel6. Neither parallel nor perpendicular7. Parallel 8. Perpendicular9. y � �3x � 4 10.
11. 12. y � �3
Applying Skills
13. a. b. c.d. 3 e. 3 f. y � 3x � 1g. (2, 5)
14. a. 1 b. �1
c. �1; since and , .d. 1; since and , .e. y � x � 2 f. y � �x g. (�1, 1)
15. a. Slope of , slope of
Slope of , slope of
Therefore, and .
b. The slopes of the adjacent sides, and , are not negative reciprocals. Therefore, no sidesare perpendicular and PQRS does not have aright angle.
232
14
QR y SPPQ y RS
SP 5 232QR 5 23
2
RS 5 14PQ 5 1
4
AB y DCDA ' DCDA ' ABBC y DADA ' ABBC ' AB
y 5 212x 1 621
2212
y 5 12x 1 3
y 5 13x 1 4
LM y NK
KL y MN
QR y SPPQ y RS
294
16. a. Slope of , slope of
Slope of � �2, slope of � 3Only one pair of sides has equal slopes.Therefore, KLMN has only one pair ofparallel sides.
b. The slopes of the adjacent sides, and 3, are negative reciprocals, so the respective sidesare perpendicular and form right angles.Therefore, �K and �N are right angles.
Hands-On Activity 1a. If two coplanar lines are each perpendicular to
the same line, then they are parallel.b. (1)
(2)(3)
Hands-On Activity 21. Answers will vary.
a. A(a, b); B(c, d); C(e, f)
b. Midpoint of
Midpoint of
c. Slope of
Slope of
The slope of the midsegment is equal to theslope of the third side of the triangle.Therefore, they are parallel.
2. Results will vary.
9-4 The Sum of the Measures of theAngles of a Triangle (pages 351–352)
Writing About Mathematics1. Yes. The sum of the angles of a triangle is 180
degrees. Therefore, if one angle has a measuregreater than 90 degrees, then the sum of theother two angles must be less than 90 degrees,and they must both be acute.
2. No. By definition, exactly two angles can besupplementary. Since a triangle has three angles,and no angle of a triangle can measure 0 degrees,the sum of the measure of any two angles mustbe less than 180 degrees.
Developing Skills3. Yes 4. No 5. No6. Yes 7. 40 8. 509. 54 10. 50 11. 80
12. 45 13. 52 14. 35
5 d 2 bc 2 aDE 5
AB 5 d 2 bc 2 a
BC 5 E A c 1 e2 ,
d 1 f2 B
AC 5 D A a 1 ec ,
b 1 f2 B
y 5 219x 1 4
y 5 12x 2 148y 5 1
4x 1 194
213
NKLM
MN 5 213KL 5 21
3
c 1 e2 2 a 1 e
2
d 1 f2 2
b 1 f2
15. 20 16. 140 17. 9018. 54 19. 12020. m�ACD � 60; m�ACB � 12021. m�ACD � 90; m�ACB � 9022. m�ACD � 60; m�B � 2023. m�B � 95; m�ACD � 135; m�ACB � 45Applying Skills24. 46°, 67°, 67° 25. 70°, 55°, 55° 26. 20°, 80°, 80°27. m�N � 58; measure of exterior angle � 12228. a. m�A � 72; m�B � 67; m�C � 41 b.29. m�A � m�C � 30; m�B � 120; measure of
exterior angle � 6030. Let �ABC be a triangle with �BCA the exterior
angle at C. Then m�A � m�B � m�C � 180since the sum of the measures of the angles of atriangle is 180. Also, �BCA and �C form a linearpair, so m�BCA � m�C � 180. By thesubstitution postulate, m�A � m�B � m�C �m�BCA � m�C. By the subtraction postulate,m�A � m�B � m�BCA, or the measure of anexterior angle of a triangle is equal to the sum ofthe measures of the nonadjacent interior angles.
31. a. Graph b. Graphc. is a vertical line, so BC � |�1 � 2| � 3.
is a horizontal line, so CD � |�1 � 2| � 3.Since BC � CD, and �BDC isisosceles. Horizontal and vertical lines areperpendicular, so �BCD is a right angle.
d. m�BDC � 45; each acute angle of anisosceles right triangle measures 45°.
e. is a horizontal line and, therefore,perpendicular to . By the partition postulate, m�DBA � m�DBC � m�ABC.Since m�DBC � 45 and m�ABC � 90,m�DBA � 135.
32. In hexagon ABCDEF, draw . This divides the hexagon into two quadrilaterals, ABCDand ADEF. The sum of the measures of theangles in each quadrilateral is 360°. Therefore,the sum of the measures of the angles in ABCDEF � 360 � 360 � 720°.
33. bisects �ABC, so �ABD � �CBD.bisects �ADC, so �ADB � �CDB.by the reflexive property of congruence, so�ABD � �CBD by SAS. Corresponding partsof congruent triangles are congruent, so �A � �C.
BD > BDDBh
BDh
AD
BCABg
BC > CD
CDBC
BC
295
9-5 Proving Triangles Congruent by Angle,Angle, Side (pages 356–357)
Writing About Mathematics1. No, it simply means that SSA is insufficient to
prove the triangles congruent. The triangles mayor may not be congruent.
2. Yes. and are perpendicular, so �C is a right angle and measures 90°. Since the sum ofthe measures of the angles of a triangle is 180°,the measures of �CAB and �CBA must be 90°,which makes them complementary.
Developing Skills3. Yes, by AAS. 4. Yes, by AAS.5. No. AAA is insufficient to prove congruence.6. Yes, by AAS.7. No. SSA is insufficient to prove congruence.8. Yes, by SAS.
Applying Skills
9. Let �ABC � �DEF, so �A � �D and. Draw , the altitude from �A in
�ABC, and , the altitude from �F in �DEF,to form right triangles �ABG and �DFH. Tworight triangles are congruent if the hypotenuseand an acute angle of one right triangle arecongruent to the hypotenuse and an acute angleof the other right triangle, so �ACG � �DFH.Corresponding parts of congruent triangles are congruent, so the altitudes, and , arecongruent.
10. Let �ABC � �DEF, so �A � �D, ,and . Draw , the median from �A in �ABC, and , the median from �F in�DEF, to form triangles �ABG and �DFH.Since G is the midpoint of and H is the midpoint of , AG � and DH � .Halves of congruent segments are congruent, so
, and �ACG � �DFH by SAS.Corresponding parts of congruent triangles arecongruent, so the altitudes, and , arecongruent.
11. Let �ABC � �DEF, so �A � �D, ,and �C � �F. Draw , the bisector of �A in �ABC, and , the bisector of �F in �DEF, to form triangles �ABG and �DFH. By the definition of angle bisector, m�ACG �
and m�DFH = . Halves of congruent angles are congruent, so �ACG � �DFH, and
12m/F
12m/C
FHCG
AC > DF
FHCG
AG > DH
12DE1
2ABDEAB
FHCGAB > DE
AC > DF
FHCG
FHCGAC > DF
BCAC
�ACG � �DFH by AAS. Corresponding partsof congruent triangles are congruent, so thealtitudes, and , are congruent.
12. Statements Reasons
1. �A � �C and is 1. Given.the bisector of �ABC.
2. �ABD � �CBD 2. Definition of anglebisector.
3. 3. Reflexive property.4. �ABD � �CBD 4. AAS.5. �ADB � �CDB 5. Corresponding parts
of congruent tri-angles are congruent.
6. bisects �ADC. 6. Definition of angle bisector.
13. Statements Reasons1. , , 1. Given.
and .2. 2. If a transversal is
perpendicular to oneof two parallel lines,then it is perpendic-ular to the other.
3. �B and �C are right 3. Definition of angles. perpendicular lines.
4. �B � �C 4. Right angles arecongruent.
5. �AEB � �DEC 5. Vertical angles arecongruent.
6. �ABE � �DCE 6. AAS.7. , 7. Corresponding parts
of congruent anglesare congruent.
8. and bisect 8. Definition of each other. bisector.
14. a. Under the translation T9,0, A(�6, 0) → D(3, 0),B(�1, 0) → E(8, 0), and C(�5, 2) → F(4, 2).Distance is preserved under translation, so�ABC � �DEF by SSS.
b. Answers will vary. Under rx=1, A(�6, 0) → (8, 0),B(�1, 0) → (3, 0), and C(�5, 2) → (7, 2).Under rx=5.5, (8, 0) → D(3, 0), (3, 0) → E(8, 0),and (7, 2) → F(4, 2).Distance is preserved under reflection, so�ABC � �DEF by SSS.
15. Let �ABC and �DEF be right triangles with �Band �E the right angles, , and AC > DF
BECAED
AE > DEBE > CE
CD ' BECAB ' BEC
AB > CDAB y CD
DBh
BD > BD
BDh
FHCG
296
�A � �D. Since right angles are congruent,�B � �E. Therefore, �ABC � �DEF by AAS.
16. Let P be a point on the bisector of �ABC.By the definition of angle bisector, �ABP ��CBP. By the reflexive property of congruence,
. Draw and . Then �BAP and �BCP are right angles andcongruent. Therefore, �ABP � �CBP by AAS.Corresponding parts of congruent triangles are congruent, so and PA � PC. By defi-nition, P is equidistant from and from .
17. Through a point E on side , draw parallelto . If two parallel lines are cut by a transversal, then the corresponding angles arecongruent. Therefore, �A � �BED and �C � �BDE. By the reflexive property ofcongruence, �B � �B. If AAA were sufficientto establish congruence, then �EBD would becongruent to �ABC. But AB � DB and CB � DB, so �EBD is not congruent to �ABC.
9-6 The Converse of the Isosceles TriangleTheorem (pages 360–362)
Writing About Mathematics1. Yes. The contrapositive of Theorem 7.3, which is
also true, states: If the measures of the anglesopposite two sides of a triangle are equal, thenthe lengths of the sides opposite these angles areequal.
2. Yes. The sum of the measures of the angles of atriangle is 180°, so if two angles measure 45° and90°, then the third must measure 45°. Since twoangles of the triangle have equal measures, theyare congruent, so the sides opposite these anglesmust be congruent. By definition, the righttriangle is isosceles.
Developing Skills3. a. 70 b. Isosceles4. a. 30 b. Isosceles5. a. 65 b. Isosceles6. a. 60 b. Not isosceles7. x � 4 8. PR � RQ � 219. m�R � 72; m�N � 36
10. AB � AC � 36; BC � 4611. x � 10 � 2x � 2x � 30 � 180, so x � 40. The
three angles then measure 50°, 80°, and 50°. Sincetwo of the angles of the triangle measure 50°, thetriangle has two congruent angles and, therefore,
ACDEBC
BCh
BAh
PA > PC
PC ' BCh
PA ' BAh
BP > BP
BPh
two congruent sides. By definition, the triangle isisosceles.
12. x � 35 � 2x � 10 � 3x � 15 � 180, so x � 25. Thethree angles all measure 60°. Therefore, all of theangles are congruent and the triangle isequiangular. If a triangle is equiangular, then it isequilateral.
13. 3x � 18 � 4x � 9 � 10x � 180, so x � 9. Themeasures of the three angles are 45°, 45°, and 90°.Since the triangle has a right angle, it is a righttriangle. Since two of the angles of the trianglemeasure 40°, the triangle has two congruentangles and, therefore, two congruent sides. Bydefinition, the triangle is isosceles.
14. 120 15. 360Applying Skills16. Statements Reasons
1. �ABP � �PCD 1. Given.2. �ABP and �PBC are 2. If two angles form a
supplements. �PCD linear pair, then and �PCB are they are supple-supplements. mentary.
3. �PBC � �PCB 3. If two angles arecongruent, thentheir supplementsare congruent.
4. 4. If two angles of a triangle are congru-ent, then the sidesopposite these an-gles are congruent.
5. �BPC is isosceles. 5. Definition ofisosceles triangle.
17. Statements Reasons1. �PAB � �PBA 1. Given.2. 2. If two angles of a
triangle are congru-ent, then the sidesopposite these an-gles are congruent.
3. PA � PB 3. Definition of con-gruent segments.
4. P is on the perpen- 4. A point is on the dicular bisector of . perpendicular
bisector of a linesegment if and onlyif it is equidistantfrom the endpointsof the line segment.
AB
PA > PB
PB > PC
297
18. Statements Reasons
1. bisects �DBC. 1. Given.2. m�DBC � 2m�DBE 2. Definition of angle
bisector.
3. 3. Given.4. �A � �DBE 4. If two parallel lines
are cut by a trans-versal, then thecorresponding an-gles are congruent.
5. m�A � m�DBE 5. Definition ofcongruent angles.
6. m�DBC 6. Exterior angle � m�A � m�C theorem.
7. 2m�A 7. Substitution � m�A � m�C postulate.
8. m�A � m�C 8. Subtractionpostulate.
9. �A � �C 9. Definition ofcongruent angles.
10. 10. If two angles of a triangle arecongruent, then thesides oppositethese angles arecongruent.
19. Statements Reasons1. �PBC � �PCB, 1. Given.
�APB � �DPC2. , 2. If two angles of a
triangle are congru-ent, then the sidesopposite these an-gles are congruent.
3. �ABP and �PBC 3. If two angles form a are supplements. linear pair, then �DCP and �PCB are they are supple-supplements. mentary.
4. �ABP � �DCP 4. If two angles arecongruent, thentheir supplementsare congruent.
5. �ABP � �DCP 5. AAS.6. 6. Corresponding parts
of congruent trian-gles are congruent.
20. In �ABC, �A � �B. Draw , the altitude from �C. By definition, , so �ADCand �BDC are right angles and congruent.
CD ' ABCD
AP > DP
AB > CDPB > PC
AB > CB
BEh
y AC
BEh
by the reflexive property of congruence. �ACD � �BCD by AAS.Therefore, because correspondingparts of congruent triangles are congruent.
9-7 Proving Right Triangles Congruent byHypotenuse, Leg (pages 365–367)
Writing About Mathematics1. In two right triangles, the hypotenuse and one
leg, both legs, or one leg and its adjacent acuteangle must be congruent for the triangles to beproved congruent.
2. PD � PQ. The distance from a point to a line isthe length of the perpendicular from the point tothe line. Therefore, �PDQ is a right angle and
can be considered a leg of right triangle.Then, is the hypotenuse of the triangle, and the hypotenuse is longer than either leg.
Developing Skills3. a. m�BCA � 80
b. m�PAN � 20; m�PBN � 30; m�APB � 130c. m�PCL � 40; m�PBL � 30; m�BPC � 110d. m�PAM � 20; m�PCM � 40;
m�APC � 120e. No. �APB and �BPL form a linear pair so
are supplementary. Since m�APB � 130,m�BPL � 50. Similarly, �APC and �CPLform a linear pair so are supplementary.Since m�APC � 120, m�CPL � 60.m�BPL � m�CPL, so they are not congruent.
Therefore, does not bisect �CPB.4. a. m�A � 45; m�B � 45
b. m�PAB � 22.5; m�PBA � 22.5;m�APB � 135
c. m�PBC � 22.5; m�PCB � 45;m�BPC � 112.5
d. m�PAC � 22.5; m�PCA � 45;m�APC � 112.5
e. Yes. Extend through P to a point M.Since angles forming a linear pair are supple-mentary and m�APC � m�BPC � 112.5, wehave that m�APM � m�BPM � 67.5. Thus,
�APM � �BPM and bisects �APB.5. a. m�A � 20; m�B � 20
b. m�PAB � 10; m�PBA � 10; m�APB � 160c. m�PBC � 10; m�PCB � 70; m�BPC � 100d. m�PAC � 10; m�PCA � 70; m�APC � 100
CPh
CPh
ALh
PQPD
CA > CB
CD > CD
298
e. Yes. Extend through P to a point M.Since angles forming a linear pair are supple-mentary and m�APC � m�BPC � 100, wehave that m�APM � m�BPM � 80. Thus,
�APM � �BPM and bisects �APB.6. m�TRS � 70; m�RST � 60; m�SPT � 1257. a–c. Graph
d. No. Since �A is not congruent to �B,�PAB and �PBA (halves of these angles) are also not congruent. If the measures of two angles of a triangle are unequal, thelengths of the sides opposite these angles are unequal. Therefore, in �APB, sincem�PAB � m�PBA, AP � BP. Similarly,AP � CP and BP � CP.
e. Equilateral
Applying Skills8. Statements Reasons
1. at D, 1. Given.
at F.2. �PDB and �PFB 2. Definition of
are right angles. perpendicular lines.3. PD � PF 3. Given.4. 4. Definition of con-
gruent segments.5. 5. Reflexive property.6. �ABP � �PFB 6. HL.7. �ABP � �CBP 7. Corresponding parts
of congruent trianglesare congruent.
9. When �ADC is an isosceles right triangle,� and �A and �C are congruent
complementary angles. Since ⊥ ,�BDC and �C are also complementary. If twoangles are complements of the same angle,then they are congruent. Thus, �BDC ��A � �C, and so � by the converse of the Isosceles Triangle Theorem. By HL,we can conclude that �ABD � �DBC. Thus,�ABD � �DBC when �ADC is an isoscelesright triangle.
10. If and are parallel, then �PAB and�PBA are supplementary. By the partitionpostulate, and
. Since a whole isgreater than any of its parts, m�PAB � m�Aand m�PBA � m�B. But in �ABC, the sum of
m/B 5 m/CBP 1 m/PBAm/A 5 m/CAP 1 m/PAB
BMAL
BCDB
ACDBDCAD
BP > BP
PD > PF
PF ' BCh
PD ' BAh
CPh
CPh
the measures of �A, �B, and �C is 180°. Thus,m�A � m�B � 180, and so:
Since �PAB and �PBA are supplementary,m�PAB � m�PBA � 180, and we arrive at
, a contradiction. Thus,the assumption is false, and and are notparallel.
11. Statements Reasons1. , 1. Given.2. �ABD and �CDB 2. Definition of
are right angles. perpendicular lines.3. 3. Given.4. 4. Reflexive property.5. �ABD � �CDB 5. HL.6. �A � �C, 6. Corresponding
�ADB � �CBD parts of congruenttriangles arecongruent.
7. 7. If two coplanar lines are cut by atransversal so thatthe alternate inte-rior angles formedare congruent, thetwo lines areparallel.
12. a. Statements Reasons1. In �QRS, the 1. Given.
bisector of �QRSis perpendicular to
at P.2. �QRP � �SRP 2. Definition of angle
bisector.3. 3. Reflexive property.4. �QPR and �PSR 4. Definition of
are right angles. perpendicular lines.5. �QPR � �PSR 5. Right angles are
congruent.6. �QPR � �PSR 6. ASA.7. 7. Corresponding
parts of congruenttriangles arecongruent.
8. �QRS is isosceles. 8. Definition ofisosceles triangle.
QR > SR
RP > RP
QS
AD y CB
BD > BDAD > CB
BD ' DCAB ' BD
BMAL180 , m/A1m/B , 180
m/PAB1m/PBA , m/A1m/B , 180
299
b. Statements Reasons1. �QPR � �PSR 1. Part a.2. 2. Corresponding parts
of congruent trian-gles are congruent.
3. P is the midpoint 3. Definition of of . midpoint.
13. Given: Isosceles triangle �ABC with vertex �B,D the midpoint of , , and
.Prove:Proof: By the isosceles triangle theorem, since
in �ABC, �A � �C. Since D is themidpoint of , . Since and
, �CED and �AFD are right anglesand congruent. Therefore, �CED � �AFD byAAS. Then because correspondingparts of congruent triangles are congruent.
14. a. In ABCD, �A and �C are right angles, so �A� �C. Since AB � CD, . By thereflexive property of congruence, ,so �ABD � �CDB by HL. Sincecorresponding parts of congruent triangles are congruent, and AD � BC.
b. From part a, �ABD � �CDB. Correspondingparts of congruent triangles are congruent, so�ABD � �CDB.
c. From parts a and b, �ABD � �CDB. Sincecorresponding parts of congruent triangles are congruent, �ADB � �CBD and m�ADB � m�CBD. The acute angles of aright triangle are complementary, so m�CDB � m�CBD � 90. By the substitutionpostulate, m�CDB � m�ADB � 90.Since m�CDB � m�ADB � m�ADC,m�ADC � 90, and �ADC is a right angle.
15. In ABCD, �ABC and �BCD are right anglesand therefore congruent. Since AC � BD,
. By the reflexive property ofcongruence, , and �ABC � �DCB byHL. Corresponding parts of congruent trianglesare congruent, so and AB � CD.AB > CD
BC > BCAC > BD
AD > BC
BD > BDAB > CD
DE > DF
DF ' ABDE ' BCAD > CDAC
AB > BC
DE > DFDF ' AB
DE ' BCAC
QS
QP > SP
16. Since and �ABP and �ADP are right angles and congruent. We are
given that PB � PD, so . by the reflexive property of congruence. Therefore,�ABP � �ADP by HL. Corresponding parts ofcongruent triangles are congruent, so �BAP ��DAP or �CAS � �EAS. By the definition of
angle bisector, bisects �CAE.
9-8 Interior and Exterior Angles ofPolygons (pages 371–372)
Writing About Mathematics1. Yes. A diagonal may be drawn from every vertex
of a polygon to every other vertex except foritself and the two adjacent vertices. Therefore,each vertex of an n-sided polygon is an endpointof (n � 3) diagonals.
2. Yes. Each vertex is an endpoint of (n – 3)diagonals, so n vertices are the endpoints of n(n � 3) diagonals. A diagonal has two endpoints, so there are diagonals.
Developing Skills3. a. 180 b. 900 c. 1,2604. a. 360 b. 1,080 c. 540 d. 1,4405. a. 360 b. 360 c. 360 d. 3606. a. 90 b. 907. a. 72 b. 1088. a. 60 b. 1209. a. 45 b. 135
10. a. 40 b. 14011. a. 30 b. 15012. a. 18 b. 16213. a. 10 b. 17014. a. b.15. a. 12 b. 8 c. 6 d. 316. a. 4 b. 6 c. 9 d. 1817. a. 3 b. 4 c. 5 d. 7
e. 10 f. 17 g. 12 h. 22Applying Skills18. 8 sides 19. 9 sides20. No. If two angles were greater than 180 degrees,
then the sum of the interior angles would alreadyexceed 360 degrees, which is the sum of themeasures of the angles of a quadrilateral.
21. a. From �A in ABCDE, draw diagonals andto form �ABC and �AED. Since
ABCDE is regular, it is both equilateral and equiangular. Therefore,
, �B � �E, and �ABC � �AED by SAS.AB > BC > AE > ED
ADAC
1713784
7
n2(n 2 3)
APS›
AP > APPB > PD
PD ' ADE›,
PB ' ABC›
300
b. Since in �ABC and in�ABD, the triangles are isosceles.
c. Since �ABC � �AED, and correspondingparts of congruent triangles are congruent,
and �DAC is isosceles.
22. a. From �L in LNMRST, draw diagonals ,, and . Since LMNRST is regular, it is
both equilateral and equiangular. Therefore,, �M � �T, and
�LMN � �LTS by SAS.
b. �LMN � �LTS and corresponding parts of congruent triangles are congruent, so
. Because LMNRST is regular,
. By the reflexive property ofcongruence, so �LNR � �LSR by SSS.
c. m�M � m�T � 120;m�MLN � m�MNL � m�TLS
� m�TSL � m�NLR � m�SLR � 30;m�LSR � m�LNR � 90;m�LRS � m�LRN � 60
23. a. Slope of � slope of � �1Slope of � slope of � 1The slopes of adjacent sides are negativereciprocals, therefore , ,
, and �A, �B, �C, and �D areright angles.
b. The x-axis is horizontal and the y-axis isvertical, so they intersect to form right angles, which are congruent, so �AOB � �BOC � �COD � �DOA.Also, AO � CO � BO � DO � 2, so
and �AOB � �BOC � �COD � �DOAby SAS.
c. Quadrilateral ABCD is equiangular because it contains four right angles. It is equilateralbecause each of its sides is the hypotenuse of a congruent right triangle. Therefore, it isregular.
Hands-On Activitya. No b. Yesc. The angle bisectors of a polygon are concurrent
if and only if the polygon is regular.
Review Exercises (pages 375–376)1. 15 2. 30 3. 24 4. 555. a. 18 b. 72 c. 72 d. 108 e. 72
AO > CO > BO > DO
CD ' DABC ' CDAB ' BC
DABCCDAB
LR > LRNR > RSLN > LS
LM > MN > LT > TS
LSLRLN
AC > AD
AE > EDAB > BC
6. 30 7. 6 8. 369. 42, 48, 90 10. 120 11. 80
12. 40 13. 154 14. 6015. 28 16. 20, 80, 80 17. 9018. 30 19. 12 20. 1,26021. In �ABC, �C is a right angle so m�C � 90.
Therefore, m�B � 90 since �B arecomplementary so �B must be acute. If themeasures of two angles of a triangle are unequal,the lengths of the sides opposite these angles areunequal and the longer side lies opposite thelarger angle. Therefore, AB � AC.
22. Since and bisect each other,and . Vertical angles are
congruent, so �AEC � �BED. Therefore,�AEC � �BED by SAS. Corresponding parts of congruent triangles are congruent, so �EAC � �EBD. If two coplanar lines are cut bya transversal so that the alternate interior anglesformed are congruent, then the two lines are parallel. Therefore, .
23. In �BPA, , so by the isosceles triangle theorem �PBC � �PCB. �PBA and �PBCform a linear pair, so they are supplementary.Also, �PCD and �PCB form a linear pair, sothey are supplementary. If two angles arecongruent, then their supplements are congruent, so �PBA � �PCD. It is given that�APB � �DPC, so �ABP � �DCP by ASA.Corresponding parts of congruent triangles are congruent, so .
24. In �BPC, �PBC � �PCB. If two angles of atriangle are congruent, then the sides opposite these angles are congruent, so . It is given that �PBC � �PCB. �PBA and �PBCform a linear pair, so they are supplementary.Also, �PCD and �PCB form a linear pair, sothey are supplementary. If two angles arecongruent, then their supplements are congruent,so �PBA � �PCD. Since , �ABP ��DCP by SAS. Corresponding parts of congruent triangles are congruent, so .
25. No. The sum of the measures of the interior an-gles of a pentagon is 540 degrees, and angles A, B,C, and D have measures which already sum to 540.Since m�E � 0, Herbie’s pentagon is not possible.
Exploration (page 376)Results will vary.
PA > PD
AB > DC
PB > PC
PA > PD
BP > CP
AC y BD
CE > EDAE > EBCEDAEB
301
Cumulative Review (pages 376–378)Part I
1. 4 2. 2 3. 1 4. 3 5. 26. 3 7. 1 8. 1 9. 1 10. 2
Part II
11. ; m�C � 45 and m�B � 90. If the measures of two angles of a triangle are unequal, thelengths of the sides opposite these angles areunequal and the longer side lies opposite the larger angle. Therefore, is the longest side.
12. Since a point on the perpendicular bisector of asegment is equidistant from the endpoints of the segment, AP � BP. Therefore, and�ABP is isosceles.
Part III
13. Since ABCD is equilateral,. By the reflexive
property of congruence, , so �ABC � �ADC by SSS. Corresponding parts of congruent triangles are congruent, so �BAC � �DAC and �BCA � �DCA. By thedefinition of angle bisector, bisects �DAB and �DCB.
14. We are given that . If two parallel lines are cut by a transversal, then the alternateinterior angles formed are congruent. Therefore,�ECB � �EDA and m�ECB � m�EDA.By the exterior angle theorem,
m�DEB � m�EBC � m�ECB.
By the substitution postulate,
m�DEB � m�EBC � m�EDA.Part IV15.
3x � 36, 4x � 48, 8x � 96The smallest exterior angle is the supplement of the largest interior angle, so its measure is 180 � 96 � 84.
16. The midpoint of
The slope of
The perpendicular bisector has a slope of �1 andgoes through (3, 2):
The equation of the perpendicular bisector is y � �x � 5.
b 5 52 5 23 1 by 5 2x 1 b
AB 56 2 (22)7 2 (21) 5 1
AB 5 A21 1 72 , 22 1 6
2 B 5 (3, 2)
x 5 1215x 5 180
3x 1 4x 1 8x 5 180
ADg
y CBg
AC
AC > ACAB > BC > CD > DE
AB > BP
AC
AC
10-2 The Parallelogram (pages 383–385)Writing About Mathematics
1. No. If opposite sides are supplementary, thequadrilateral is not necessarily a parallelogram.For example, consider a quadrilateral with anglemeasures 30°, 40°, 150°, and 140°. Oppositeangles are supplementary but not congruent, andso the quadrilateral is not a parallelogram.
2. No. Consider parallelogram ABCD with thediagonals intersecting at the point E. The fourtriangles formed are �ABE, �BCE, �CDE, and �ADE. Since and , in orderfor �ABE to be congruent to �BCE, wouldneed to be congruent to adjacent side .However, this is not necessarily true.
Developing Skills3. a. 70, 110, 110 b. 65, 115, 115
c. 90, 90, 90 d. 50, 130, 130e. 25, 155, 155 f. 12, 168, 168
4. x � 50; m�A � 80; m�B � 1005. x � 34; m�A � 78; m�B � 1026. x � 70; m�A � m�C � 110; m�B � m�D � 707. x � 22; m�A � m�C � 66; m�B � m�D � 1148. x � 15; m�A � m�C � 75; m�B � m�D � 1059. x � 120; m�A � m�C � 60; m�B � m�D � 120
10. x � 5; AB � CD � 2711. x � 2; y � 2; AB = CD � 12; BC � DA � 812. x � 3; AC � 1213. y � 2; DB � 9
Applying Skills
14. In parallelogram ABCD, diagonal divides the parallelogram into two congruent triangleswith �ABC � �CDA. Since corresponding parts of congruent triangles are congruent,and . Therefore, the opposite sides of a parallelogram are congruent.
15. In a parallelogram ABCD, diagonal divides the parallelogram into two congruent triangleswith �ABC � �CDA. Since corresponding partsof congruent triangles are congruent, �B � �D.Similarly, diagonal divides the parallelogram into two congruent triangles with �ABD ��CDB, and �A � �C. Therefore, the oppositeangles of a parallelogram are congruent.
BD
AC
AD > BCAB > CD
AC
BCAB
BE > EDAE > EC
302
16. Statement Reason1. Parallelograms EBFD 1. Given.
and ABCD2. 2. Opposite sides of a
parallelogram arecongruent.
3. �E � �F, 3. Opposite angles of �BAD � �DCB a parallelogram are
congruent.4. �BAD and �EAD 4. If two angles form a
are supplements, linear pair, then �DCB and �FCB are they are supple-supplements. mentary.
5. �EAD � �FCB 5. If two angles arecongruent, thentheir supplementsare congruent.
6. �EAD � �FCB 6. AAS.
17. In a parallelogram, consecutive angles aresupplementary. Thus, the corners adjacent to thecorner forming a right angle are also right angles.In a parallelogram, opposite angles arecongruent. Thus, the corner opposite the cornerforming a right angle is also a right angle.Therefore, Petrina’s floor has four right angles.
18. Suppose that ABCD is a parallelogram. Thenopposite angles �A and �C are congruent andm�A � m�C. However, we are given that m�Aand m�C are not equal, which is a contradiction.Therefore, the assumption is false, and ABCD isnot a parallelogram.
19.
20. Since , , and �ABC � �PQR by SAS. Since a diagonaldivides a parallelogram into two congruenttriangles, �ABC � �CDA and �PQR � �RSP.By the transitive property of congruence,
/B > /Q,QR > BCPQ > AB
B C
A D
Q R
P S
ED > BF
Chapter 10. Quadrilaterals
�ABC � �CDA � �PQR � �RSP. Sincecorresponding parts of congruent triangles are congruent, ,
, and �D � �B � �S � �Q. Since �C and �A are supplementary to �B, and �Rand �P are supplementary to �Q, �B � �Q,�C � �A � �R � �P because the supplementsof congruent angles are congruent. Thus, all sidesand angles of one parallelogram are congruent tothe corresponding sides and angles of the otherparallelogram, and so the parallelograms arecongruent.
10-3 Proving That a Quadrilateral Is aParallelogram (pages 387–388)
Writing About Mathematics
1. because opposite sides of aparallelogram (EBFD) are congruent.
2. because opposite angles in aparallelogram (EBFD) are congruent.
Developing Skills3. A pair of opposite sides is both parallel and
congruent.4. Two pairs of opposite sides are congruent.5. Both pairs of opposite angles are congruent.6. Both pairs of opposite sides are parallel.7. The diagonals bisect each other.
8. Since , �D is the supplement of �A, and �B is the supplement of �C. Since thesupplements of congruent angles are congruentand �A � �C, . If both pairs ofopposite angles are congruent, then thequadrilateral is a parallelogram. Therefore,ABCD is a parallelogram.
9. Since is supplement of , . is a
transversal that cuts and . Thus, �S is the supplement of �R. Since the supplements ofcongruent angles are congruent, . Ifboth pairs of opposite angles are congruent, thenthe quadrilateral is a parallelogram. Therefore,ABCD is a parallelogram.
10. Since �GDF � �DFE, �EDF � �GFD, andby the reflexive property, �EDF �
�GFD by ASA. Therefore, andsince corresponding parts of
congruent triangles are congruent. If oppositesides of a quadrilateral are congruent, then thequadrilateral is a parallelogram. Therefore,DEFG is a parallelogram.
GD > FEGF > DE
DF > DF
/Q > /S
QRg
SPg
SRPSg
y QRg
/Q/P
/B > /D
ABg
y CDg
/DEB > /BFD
DE > FB
SR > PQDC > AB >DA > BC > SP > QR
303
11. In a parallelogram, opposite sides are congruent.Thus, . Since E and F are the midpointsof and , respectively, and
. Thus, since congruenthalves of congruent segments are congruent.
Since ABCD is a parallelogram, , and so
and are also parallel. Therefore, AEFD is a parallelogram because one pair of sides is bothcongruent and parallel.
12. EFGH is a parallelogram, so and. Segments of parallel lines are
parallel, so . Since F is the midpoint of, . By the transitive property of
congruence, . If one pair of opposite sides of a quadrilateral is both congruent andparallel, the quadrilateral is a parallelogram.Therefore, FJGH is a parallelogram.
13. a. Since P, Q, R, and S are midpoints of , ,, and , respectively, ,
, , and . SinceABCD is a parallelogram, opposite sides arecongruent, and so and .Since congruent halves of congruent segmentsare congruent, , ,
, and . Since ABCD is a parallelogram, opposite angles are congruent,and so �A � �C and �D � �B. Therefore,�APS � �CRQ and �BQP � �DSRby SAS.
b. In part a we showed that �APS � �CRQ and�BQP � �DSR. Since corresponding partsof congruent triangles are congruent,
and . Therefore, PQRS is a parallelogram since both pairs of oppositesides are congruent.
14. Yes. Since sum of the measures of the angles in aquadrilateral is 360°, the fourth angle also has tomeasure 90°. Therefore, the quadrilateral is aparallelogram since both pairs of opposite anglesare congruent.
Applying Skills
15. Since the diagonals and bisect eachother, and . Since vertical angles are congruent, �AEB � �DEC.Therefore, �ABE � �CDE by SAS. Therefore,
and �CDE � �EBA because corresponding parts of congruent triangles are AB > DC
BE > EDAE > ECDBAC
SP > RQSR > PQ
DS > QBSA > CQRD > PBAP > RC
BC > DAAB > DC
DS > SACR > RDBQ > QCAP > PBDACD
BCAB
FJ > GHEF > FJEJ
FJ y GHEF > GH
EF y GH
AEg
DFg
ABg
y DEg
AE > DFAE > EBDF > FEDEAB
AB > DE
congruent. is a transversal that cuts and
with congruent interior angles �CDE and �EBA. If two lines are cut by a transversal andthe alternate interior angles formed arecongruent, then the two lines are parallel. Thus,
, and so ABCD is a parallelogram since a pair of sides is both congruent and parallel.
16. If the diagonals of a quadrilateral bisect eachother, then the quadrilateral is a parallelogram.Therefore, a quadrilateral drawn by joining theendpoints of two line segments that bisect eachother is a parallelogram.
17. Slope of
Slope of Slope of
Slope of Both pairs of opposite sides are parallel.Therefore, ABCD is a parallelogram.
18. Suppose that the diagonal paths bisect eachother. Then the quadrilateral is a parallelogram.However, we are given that the quadrilateral isnot a parallelogram, a contradiction. Therefore,the assumption is false, and the diagonal paths donot bisect each other.
19. Yes. Since �ABC � �A�B�C�, �C � �C�,, and .
Thus, opposite angles �C and �C� are congruentin the quadrilateral, ,and . Since A coin-cides with B� and B coincides with A�,
andby
the partition postulate. Substitutingand
, we arrive at . Thus, m�CAC� � m�CBC� by the
transitive property, and �CAC� � �CBC�.Therefore, both pairs of opposite angles arecongruent in the quadrilateral, and so thequadrilateral is a parallelogram.
20. a. In a parallelogram, opposite sides arecongruent. Thus, . Since M and Nare midpoints, and .Since congruent halves of congruent segmentsare congruent, and .
Since ABCD is a parallelogram, and
. Therefore, AMND and MBCN are MBg
y NCg
AMg
y DNg
MB > CNAM > DN
CN > NDAM > MBAB > CD
1 m/CABm/CBCr 5 m/CrBrArm/CrArBr
m/CAB 5m/CBA 5 m/CrBrAr
m/CACr 5 m/CAB 1 m/CrBrArm/CBCr 5 m/CBA 1 m/CrArBr
m/CAB 5 m/CrArBrm/CBA 5 m/CrBrAr
/CAB > /CrArBr/CBA > /CrBrAr
AD 5 1
CD 5 212
BC 5 1
AB 5 212
ABg
y CDg
DCg
ABg
DBg
304
parallelograms because, in both quadrilaterals,one pair of sides is both congruent andparallel.
b. Since opposite angles of parallelogramsABCD, AMND, and MBCN are congruent,�A � �C, �D � �B, , �D ��AMN, �NMB � �C, and �MNC � �B. Bythe transitive property, �A � �NMB ��DNM � �C and �D � �MNC � �AMN� �B. Thus, corresponding angles arecongruent.
Since opposite sides of a parallelogramsAMND and MBCN are congruent,
and . By transitivity,. Since M and N are
midpoints, and . Thus,corresponding sides are congruent.
Therefore, the two parallelograms arecongruent since corresponding sides andangles are congruent.
10-4 The Rectangle (pages 392–393)Writing About Mathematics
1. Yes. If the parallelogram were a rectangle, thenall of its angles would be right angles, but thiscontradicts the given condition.
2. Yes. By Exercise 16 on page 388, thequadrilateral is a parallelogram. If the diagonalsof a parallelogram are congruent, theparallelogram is a rectangle. The given segmentsform the diagonals of this quadrilateral. Since thesegments are congruent, the quadrilateral is alsoa rectangle.
Developing Skills
3. Since diagonals of a rectangle are congruent andalso bisect each other, . Therefore,
is isosceles.
4. x � 8; AC � BD � 38; AE � BE � 145. y � 10; AE � DE � 22; AC � BD � 446. a � 4; BE � ED � 13; BD � AC � 267. x � 12; AE � 12; BD � AC � 24; BE � 128. ; m�AEB � 110;
m�AED � 709. x � 40; m�AEB � m�DEC � 120;
m�CAB � 30; m�CAD � 6010. y � 40; m�AED � 50; m�AEB � 130;
m�CAB � 25; m�CAD � 65Applying Skills
11. Sides and are horizontal segments sincethe endpoints of the segments have the same
DCAB
m/ACB 5 m/CAD 5 55
nAEBAE > EB
DN > NCAB > MBAD > NM > CB
NM > CBAD > NM
/A > /DNM
y-coordinates. Similarly, and are verticalsegments. Thus, , , ,and . Since perpendicular lines meet toform right angles, �A, �B, �C, and �D are rightangles.
12. Since rectangle ABCD is a parallelogram,opposite sides and are congruent. Since �DAB and �ABC are right angles, �DAB ��ABC. Since by the reflexive property, �DAB � �CBA by SAS. Therefore,
because corresponding sides of congruent triangles are congruent.
13. Since opposites sides of a parallelogram arecongruent, . Since the diagonals arecongruent, . Thus, since by the reflexive property, �DAB � �CBA by SSS.Consecutive angles of a parallelogram aresupplementary, and so �DAB and �CBA aresupplementary. However, they are alsocorresponding angles of congruent triangles�DAB and �CBA. Thus, �DAB � �CBA, andthese two angles are both congruent andsupplementary. Since congruent andsupplementary angles are right angles, �DABand �CBA are both right angles. Therefore,parallelogram ABCD is also a rectangle.
14. Since M is the midpoint of , . Sinceopposite sides of a rectangle are congruent,
. Since the angles of a rectangle are allright angles, �S � �R. Thus,by SAS. Since corresponding sides of congruenttriangles are congruent, .
15. a. Slope of Slope of
Slope of Slope of The slopes of opposite sides are equal.Therefore, the opposite sides are parallel, andthe quadrilateral is a parallelogram. The slopes of adjacent sides and are negative reciprocals. Therefore, these twosides are perpendicular and form a right angle.Since a parallelogram with one right angle is arectangle, ABCD is a rectangle.
b.
c. 90° clockwise rotation about the origin.d. Under a rotation, angle measure is preserved.
Thus, , , , and, and the angles of areArBrCrDr/D > /Dr
/C > /Cr/B > /Br/A > /Ar
A 32, 2 B
ABAD
BC 5 2AD 5 2
DC 5 212AB 5 21
2
PM > QM
nQRM > nPSMSP > RQ
SM > MRRS
AB > ABDB > ACDA > CB
AC > BD
AB > AB
CBDA
CD ' DABC ' CDAB ' BCDA ' AB
DABC
305
all right angles. Therefore, is arectangle.
16. a. Slope of Slope of
Slope of Slope of The slopes of opposite sides are equal.Therefore, the opposite sides are parallel andthe quadrilateral is a parallelogram.
b. Suppose that PQRS were a rectangle. Then allof its angles would be right angles. However,since the slopes of adjacent sides and are not negative reciprocals, and are not perpendicular, and �Q is not a right angle,a contradiction. Therefore, the assumption isfalse, and PQRS is not a rectangle.
c. P�(�2, �5), Q�(�6, �2), R�(�2, 2), S�(2, �1)d. Since reflections and translations preserve
distance, , , ,and . Thus, corresponding sides arecongruent. Since reflections and translationspreserve angle measure, ,
, , and . Thus,corresponding angles are congruent, and
.
17. Suppose that ABCD were a parallelogram.Since �A is a right angle, ABCD is a rectangle.However, we are given that ABCD is not arectangle, a contradiction. Therefore, theassumption is false, and ABCD is not aparallelogram.
18. Since she can only measure distance, she can first find the midpoint of the segment formed by the stakes she has already placed in theground. Through this midpoint, she can drawanother line. She can then draw two other points on each side of the midpoint at an equaldistance away. The two remaining stakes can be placed on these two points to form arectangle.
19. The diagonals of a rectangle are congruent andbisect each other. Archie should locate themidpoint of each string by folding it in half. Then,with one end of each string at a different vertex,he should cross the strings at their midpoints. Thetwo remaining ends of the stings are located atthe other two vertices.
PQRS > PrQrRrSr
/S > /Sr/R > /Rr/Q > /Qr/P > /Pr
SP > PrSrRS > RrSrQR > QrRrPQ > PrQr
PQQRPQQR
SR 5 243PS 5 1
PQ 5 243QR 5 1
ArBrCrDr
10-5 The Rhombus (pages 396–398)Writing About Mathematics
1. Yes. In rhombus ABCD with diagonals andintersecting at E, the sides are all congruent,
and the diagonals bisect each other and areperpendicular. Thus, � ,
, , and �AED, �DEC,�AEB, and �BEC are all right angles (sinceperpendicular lines meet to form right angles).Thus, �AED, �CED, �CEB, and �AEB are allright triangles. Since right angles are congruent,�AED � �CED � �CEB � �AEB by SAS.
2. Yes. The diagonals of a rhombus separate therhombus into four congruent right triangles.
Thus, each triangle has a base of length and a
height of . The area of each triangle is .
Therefore, the area of the rhombus is
.
Developing Skills3. AB, BC, CD, DA4. AE and EC; DE and EB5. AC, DB 6.7. No, since the diagonals are not congruent.8. No 9. C, D, A, B, E
10. Yes11. a. The diagonal bisects �P into two 60°
angles and �R into two 60° angles. Since thesum of the measures of the angles of a triangleis 180°, �S and �Q both measure 60°. Thus,diagonal separates the rhombus into two equilateral triangles �PRS and �PRQ.
b. 24 cm12. Yes13, 14. No 15. Yes
Applying Skills
16. In rhombus ABCD with diagonals and intersecting at E, the sides are all congruent, andthe diagonals bisect each other and are perpendicular. Thus, � ,
, , and �AED , �DEC,�AEB, and �BEC are all right angles (sinceperpendicular lines meet to form right angles). Since right angles are congruent,�AED � �CED � �CEB � �AEB by SAS.Since corresponding sides of congruent triangles
DE > EBAE > ECAD > BCAB > DC
DBAC
30°20°
120°40° 90°
PR
PR
/AEB, /BEC, /CED, AED
(4) A d1d28 B 5
d1d22
d1d28
d22
d12
DE > EBAE > ECAD > BCAB > DC
DBAC
306
are congruent, , ,, and .
Therefore, bisects �DAB and �DCB, andbisects �CDA and �CBA.
17. In parallelogram ABCD with diagonals andintersecting at E, since the
diagonals bisect each other. Since the diagonalsare perpendicular, �AED and �AEB are bothright angles. Thus, �AED � �AEB. Since
by the reflexive property,�AED � �AEB by SAS. Therefore,because corresponding parts of congruenttriangles are congruent, and ABCD is a rhombus.
18. a. Slope of Slope of
Slope of Slope of The slopes of opposite sides are equal.Therefore, the opposite sides are parallel,and ABCD is a parallelogram.
b. Slope of � 1 Slope of The slopes are negative reciprocals. Therefore,the two diagonals are perpendicular to eachother.
c. Yes. If the diagonals of a parallelogram areperpendicular, then the parallelogram is arhombus. Since ABCD is a parallelogram withperpendicular diagonals, ABCD is a rhombus.
19. �KLM is an equilateral triangle. Each angle of an equilateral triangle measures 60°. Thus,�KLM is a 60° angle.
20. In rhombus ABCD with diagonals and intersecting at E, the sides are all congruent, andthe diagonals bisect each other and are perpendicular. Thus, � ,
, , and �AED, �DEC,�AEB, and �BEC are all right angles (sinceperpendicular lines meet to form right angles).Thus, �AED, �CED, �CEB, and �AEB are allright triangles. Since right angles are congruent,�AED � �CED � �CEB � �AEB by SAS.
21. Since the diagonals bisect each other, thequadrilateral is a parallelogram. Since thediagonals are perpendicular to each other, theparallelogram is a rhombus.
22. In parallelogram ABCD, opposite angles arecongruent, so . If bisects�DAB and �DCB, then and/DAC > /CAB
AC/DAB > /DCB
DE > EBAE > ECAD > BCAB > DC
DBAC
DB 5 21AC
BC 5 5CD 5 15
DA 5 5AB 5 15
AD > ABAE > AE
DE > EBDBAC
DBAC
/ABE > /CBE/ADE > /CDE/DCE > BCE/DAE > /EAB
. Since congruent halves ofcongruent angles are congruent. �DAC hascongruent base angles, and so �DAC is isosceles.Since �DAC is isosceles, , and the parallelogram has two congruent consecutivesides. Therefore, ABCD is a rhombus.
23. Since a rhombus is a parallelogram, the diagonalsof ABCD bisect each other. Since the diagonalsof a rhombus are perpendicular to each other,the diagonals of ABCD are the perpendicular bisectors of each other. Therefore, is theperpendicular bisector of . Thus, P is on theperpendicular bisector of . If a point is on the perpendicular bisector of a segment, then it isequidistant from the endpoints of the segment.Therefore, AP � CP, and so .
24. a. Slope of
Slope of
The slopes of opposite sides and arethe same. Therefore, and are parallel.Since and are both vertical lines, they are parallel. Therefore, oppositesides of quadrilateral ABCD are parallel, andthe quadrilateral is a parallelogram.
b. (0, 0)c. Slope of
Slope of The slopes are negative reciprocals.Therefore, the diagonals are perpendicular toeach other.
d. Yes. If the diagonals of a parallelogram areperpendicular, then the parallelogram is arhombus. Since ABCD is a parallelogram withperpendicular diagonals, ABCD is a rhombus.
25. a, b. In Exercise 20 on page 388, we proved thatAMND and MBCN are both parallelograms. Since , AMND isa rhombus since it is a parallelogram withtwo congruent consecutive sides. Since M isthe midpoint of , . By thetransitive property, . Sinceopposite sides are congruent inparallelogram ABCD, . Thus,
by transitivity, and MBCN isalso a rhombus.
c. Since opposite angles of parallelogramsABCD, AMND, and MBCN are congruent,
MB > BCAD > BC
MB > ADAM > MBAB
AD > AM
DB 5 22
AC 5 12
CBADABDC
ABDC
DC 5 234
AB 5 234
AP > CP
ACAC
DB
AD > DC
/DCA > /ACB
307
�A � �C, �D � �B, ,�D � �AMN, �NMB � �C, and �MNC � �B. By the transitive property,�A � �NMB � �DNM � �C and �D � �MNC � �AMN � �B. Thus,corresponding angles are congruent.
Since the sides of rhombuses AMND andMBCN are all congruent,
and .Thus, corresponding sides are congruent.
Therefore, the two parallelograms arecongruent since corresponding sides andangles are congruent.
Hands-On Activity1–2. Answers will vary.3–4. The two endpoints of the diagonal constructed
in step 1 will be two opposite vertices of therhombus. Since any point on the perpendicularbisector of a segment is equidistant from theendpoints of the segment, the third vertex willbe equidistant from the two opposite vertices.Thus, the two consecutive sides formed will becongruent. The fourth vertex will be on theperpendicular bisector at the same distanceaway from the segment as the third vertex.Since the diagonals formed bisect each other,the resulting quadrilateral is a parallelogram.Since the parallelogram has two congruentconsecutive sides, the parallelogram is arhombus.
5. Answers will vary.
10-6 The Square (pages 401–402)Writing About Mathematics
1. Yes. Since a square is a rhombus, its diagonalsseparate it into four congruent right triangles andbisect the angles of the square. Since each angleof a square measures 90°, the resulting angles areall 45°. Thus, the base angles of the trianglesformed are congruent, and the triangles arecongruent isosceles right triangles.
2. Yes. If a quadrilateral is equilateral, thenconsecutive sides are congruent. If aquadrilateral is equiangular, then it is a rectangle.If a quadrilateral is a rectangle with twocongruent consecutive sides, then it is a square.Therefore, an equilateral and equiangularquadrilateral is a square. Conversely, since asquare is a rhombus, it is equilateral, and since allof its angles are right angles, it is equiangular.
NM > MB > BC > CNNM > DNAD > AM >
/A > /DNM
Developing Skills3. Since a square is a rhombus, its diagonals are the
perpendicular bisectors of each other.
4. x � 3; AC � BD � 11; AM � MB �
5. a � b � 5; AB � BC � CD � DA � 106. x � 54; y � 18 7. Yes8. Yes9. No 10. Yes
11. See the counterexample for Exercise 9.Applying Skills12. A rhombus is a parallelogram. Thus, ABCD is a
parallelogram. Since a parallelogram with oneright angle is a rectangle, ABCD is a rectangle. Arectangle with two congruent consecutive sides isa square. Therefore, since ABCD is also a rhombus, , and so ABCD is a square.
13. Since a square is a rhombus, its diagonals are theperpendicular bisectors of each other.
14. See the answer to Exercise 1.
15. Since the diagonals bisect each other, thequadrilateral formed is a parallelogram. Sincethe diagonals are congruent, the parallelogram isa rectangle. Since the diagonals areperpendicular to each other, the parallelogram isa rhombus. Thus, the parallelogram is a rhombusand a rectangle. In particular, it is a rhombus witha right angle. Therefore, since a rhombus with aright angle is a square, the quadrilateral formedis a square.
16. Let ABCD be a square with A(0, 0), B(2a, 0),C(2a, 2a), and D(0, 2a). Then P, the midpoint of
, is at (0, a); Q, the midpoint of , is at (2a, a); R, the midpoint of , is at (a, 2a); and S the midpoint of , is at (0, a). The slope of
� the slope of � 1, so . The slopeof � the slope of � �1, so .Therefore, PQRS is a parallelogram. Since isvertical and is horizontal, the diagonals areperpendicular to each other and PQRS is arhombus. Since the slopes of and areQRPQ
SQPR
QR y SPSPQRPQ y SRSR PQ
DACD
BCAB
AB > BC
512
308
negative reciprocals, �PQR is a right angle andPQRS is a square.
17. a. Midpoint of � (4, 1)Midpoint of � (4, 1)The midpoints are the same. Therefore, thediagonals bisect each other.
b. is a horizontal segment. is vertical segment. Since vertical and horizontal linesare perpendicular, the two segments areperpendicular to each other.
c. Since the diagonals bisect each other, PQRS isa parallelogram. Since PR � SQ � 6, thediagonals are congruent, and so parallelogramPQRS is a rectangle. Thus, PQRS has a rightangle. Since the diagonals are perpendicular toeach other, parallelogram PQRS is also arhombus. A rhombus with a right angle is asquare. Therefore, PQRS is a square.
d. 90° rotation about the origin.18. a. Midpoint of � (1, 2)
Midpoint of � (1, 2)The midpoints are the same. Therefore, thediagonals bisect each other.
b. is a horizontal segment. is a vertical segment. Since vertical and horizontal linesare perpendicular, the two segments areperpendicular to each other.
c. Since the diagonals bisect each other, ABCDis a parallelogram. Since AC � DB � 8, thediagonals are congruent, and so parallelogramABCD is a rectangle. Thus, ABCD has a rightangle. Since the diagonals are perpendicular toeach other, parallelogram ABCD is also arhombus. A rhombus with a right angle is asquare. Therefore, ABCD is a square.
d.
10-7 The Trapezoid (pages 408–409)Writing About Mathematics
1. No. If ABCD is a trapezoid with , then �A and �D are supplementary. Thus, if �A is aright angle, �D must also be a right angle.
2. No. Suppose that trapezoid ABCD has threeobtuse angles �A, �B, and �C. Then, �A and�B cannot be supplementary, and �B and �Ccannot be supplementary (since each angle measures more than 90°). Thus, cannot beparallel to , and cannot be parallel to .DCABCB
AD
ABg
y CDg
T23,1
DBAC
BDAC
SQPR
QSPR
The trapezoid has no parallel sides, acontradiction.
Developing Skills3. a. b.
c.4. x � 6 5. y � 3; AD � 66. x � 20; m�A � m�B � 75; m�D � m�C � 1057. x � 10; ;
8. AD � DC � BC � 11, AB � 229. True. An isosceles trapezoid is a trapezoid in
which nonparallel sides are congruent.10. False. Two nonparallel sides can be congruent to
one of the bases.
11. False. Base angles are congruent only in anisosceles trapezoid.
12. True. The diagonals of a trapezoid are congruentif and only if the trapezoid is isosceles.
13. True. The trapezoid is a quadrilateral. The sum ofthe measures of the angles of a quadrilateral is360 degrees.
14. True. The trapezoid consists of two parallel linescut by two transversals. The angles formed by thetwo sides and the two parallel bases form twopairs of supplementary angles.
Applying Skills
15. Since , �T is the supplement of �Q, and�S is the supplement of �R. Since thesupplements of congruent angles are congruentand �Q � �R, �T � �R.
16. Draw P on such that . Then QPST is a parallelogram since both pairs of sides areparallel. In a parallelogram, consecutive anglesare supplementary. Thus, �Q is the supplementof �QPS. Since �QPS and �SPR form a linearpair, �SPR is the supplement of �QPS. Thesupplements of congruent angles are congruent.Thus, �Q � �SPR since �QPS is congruent toitself. By the transitive property, �SPR � �R,and so �SPR is an isosceles triangle by theconverse of the isosceles triangle theorem. In particular, . But in parallelogram QPST,opposite sides are congruent. Thus, . Bythe transitive property, .QT > SR
QT > PSPS > SR
QT y PSQR
QR y ST
m/BAD 5 m/ABC 5 120m/ADC 5 m/BCD 5 60
m/DAB 5 70m/ABC 5 70m/BCD 5 110
309
17. a. Draw and . Then, �DEBand �CFA are both right angles. Since
, �DEB � �CFA by HL. Thus,�CAB � �DBA because corresponding partsof congruent triangles are congruent. Since
and (by the reflexiveproperty), �ACB � �BDA by SAS.Therefore, because correspondingparts of congruent triangles are congruent.
b. Because we cannot assume that the two baseangles are congruent.
18. Since translations preserve distance, we canassume that B, C, and D are to the right of theorigin and that C is to the right of D.
AB � �0 � b� � b
DC � �d � c� � c � d
Thus, .
M � and
Thus, , and
MN = .
19. a. Slope of Slope of Slope of Slope of
b. Since the slopes of only two sides are equal,only two sides are parallel. Therefore, ABCDis a trapezoid.
c. E(3,5), F(0, �1) d. Slope of e. Since the slope of the median is the same as
the slope of the bases, it is parallel to them.
20. Suppose that the diagonals of the trapezoidbisected each other. A trapezoid is aquadrilateral. Since a quadrilateral withdiagonals that bisect each other is aparallelogram, the trapezoid is a parallelogram.Therefore, it has two pairs of parallel sides, acontradiction.
21. A trapezoid is isosceles if and only if the diag-onals are congruent. Therefore, if the diagonalsare not congruent, the trapezoid is not isosceles.
22. If a quadrilateral does not have pair ofconsecutive angles that are supplementary, thenno pair of opposite sides is parallel. Therefore,the quadrilateral is not a trapezoid.
EF 5 2
DA 5 22.5CD 5 2BC 5 21AB 5 2
12(AB1CD)
MN 5 Pd2 2 c 1 b2 P 5 1
2(b 1 c 2 d)
N 5 A c 1 b2 , e2 BA d
2, e2 B
12(AB 1 CD) 5 1
2(b 1 c 2 d)
AD > BC
AB > ABDB > AC
AC > DB
CF ' ABDE ' AB
10-8 Areas of Polygons (pages 410–412)Writing About Mathematics
1. Yes. The area of ABCD is and thearea of PQRS is . Since AB � PQ andBC � QR, by substitution, the area of PQRS is
. Therefore, the two areas are equal.2. No. The area of a parallelogram is equal to the
product of the length of a base and the height,where the height is the distance from the base toits parallel side. If we are only given that the sidesof the parallelogram are congruent, the height ofeach parallelogram may be different, and so eachparallelogram can have a different area.
Developing Skills3. 40 sq units4. a. Graph
b. and because alternate interior angles formed by atransversal cutting two parallel lines are congruent. Since by the reflexiveproperty, �ABC � �DBC by ASA.
c. ABCD is a parallelogram since both pairs ofopposite sides are parallel. From Example 1,the area of the parallelogram is (AB)(CD) � bh. The area of congruentfigures are congruent, so
(area of �ABC) � (area of �DBC).Therefore, 2(area of �ABC) � bh, or
(area of �ABC) � .
5. 24 sq units6. a. Answers will vary.
b. Since the lines perpendicular to the same lineare parallel, . Since segments ofparallel lines are parallel, . Thus,FECD is parallelogram, and CE � DF sinceopposite sides of a parallelogram arecongruent.
c. Area of ABCD � area of �AFD � area ofFECD � area of �CEB
Let AF � x and EB � y.Then .
Area of �AFD
7. 16 sq units5 1
2h(b1 1 b2)
5 hS12(b1 2 b2) 1 b2T
5 hS12(x 1 y) 1 b2T
5 12xh 1 b2h 1 12yh
x 1 y 5 b1 2 b2
FE y DCCE y DF
12bh
BC > BC
/ACB > /CBD/ABC > /BCD
(AB)(BC)
(PQ)(QR)(AB)(BC)
310
8. a. Answers will vary.b. By Exercise 20 on page 398, the diagonals of a
rhombus separate the rhombus into fourcongruent right triangles. Therefore,�ABE � �CBE � �CDE � �ADE.
c. The diagonals of a rhombus separate therhombus into four congruent right triangles.Thus, each diagonal is a height of one of thetriangles, and the area of �ABE is
.
d. The area of the rhombus is the sum of eachtriangle formed by the intersecting diagonals.Since the triangles are all congruent, the area
of the rhombus is .
9. 12 sq unitsApplying Skills
10. a. Midpoint of Midpoint of
The midpoints are the same. Therefore, thetwo diagonals of ABCD bisect each other.
is a horizontal segment. is avertical segment. Horizontal and vertical linesare perpendicular. Therefore, the twodiagonals of ABCD are perpendicular.
If the diagonals of a quadrilateral bisecteach other, then the quadrilateral is a paral-lelogram. Thus, ABCD is a parallelogram. Ifthe diagonals of a parallelogram areperpendicular, then the parallelogram is arhombus. Thus, ABCD is a rhombus.
b. 2411. a. Midpoint of � (1, 2)
Midpoint of � (1, 2)The midpoints are the same. Thus, the
diagonals bisect each other, and ABCD is a parallelogram. Since the diagonals arevertical and horizontal segments, they areperpendicular to each other. Thus,parallelogram ABCD is a rhombus. Since AC � DB � 6, the diagonals are congruent,and parallelogram ABCD is a rectangle. Inparticular, ABCD has a right angle. Arhombus with a right angle is a square.Therefore, ABCD is a square.
b. 18 sq unitsc. A�(2, 2), B�(�1, �1), C�(�4, 2), D�(�1, 5)d. 18
DBAC
DBAC
DB 5 (2, 1)AC 5 (2, 1)
(4) A d1d28 B 5
d1d22
12 A
d12 ?
d22 B 5
d1d28
e. The diagonals of AEA�F are congruent andare the perpendicular bisectors of each other.Therefore, AEA�F is a square.
f. 8 sq units12. a. 189 sq units b. 9 units13. The two triangles have the same base . Also,
, and so SM � TN where and are the altitudes from S and T to . Let h � SM � TN. Thus, the areas of both trianglesare .
14. a. 16 sq units b. 16 sq units c. 32 sq units15. The length of the median is equal to one-half
the sum of the bases. Thus, EF = .Therefore, area of the trapezoid is
or bysubstitution.
16. a. 24 sq unitsb. Area of sq units
Area of sq unitsArea of sq units
c. 16.5 sq units17. 35 sq unitsHands-On Activity
1. BD = d,
2. Area of �ACD � area of �ACB �
3. Area of ABCD �4. Area of ABCD � s2
5. 6.
7. Length of a hypotenuse � (length of a leg)
Review Exercises (pages 414–415)1. No. If a parallelogram has one right angle, then
the parallelogram is a rectangle by the definitionof a rectangle. But in a rectangle, all the anglesare right angles.
2. x � 24; 72°, 108°, 72°, 108° 3.4. If a transversal is perpendicular to one of two
parallel lines, then it is perpendicular to the other. Thus, since transversal isperpendicular to , it is also perpendicular to
. If two lines are each perpendicular to thesame line, they are parallel. Thus, since
and , . Sincesegments of parallel lines are parallel, .Thus, MBND is a parallelogram. Since oppositesides of a parallelogram are congruent,
DN y MBDM y NBNB ' DCDM ' DC
DCAB
DM
m/APC 5 135
!2
d 5 s 3 !2s 5 d!2
d2
2
5 d2
4
12 3 d 3 d
2
AE 5 d2, EC 5 d
2, BE 5 d2, ED 5 d
2
nSMP 5 2nQLR 5 3
2
nPNQ 5 4
(DH)(EF)12(DH)(AB 1 CD)
12(AB 1 CD)
12(AB)(h)
ABTNSMAB y DC
AB
311
and . Since and , �AMD and �CNB are both right
angles, and so �AMD and �CNB are both righttriangles. Therefore, �AMD � �CNB by HL.
5. ANCM is a parallelogram since and(segments of parallel lines are
parallel). Thus, since opposite sides of aparallelogram are congruent, ,
, , and ,and so DC � AB and NC � AM. By the partition postulate, DC � DN � NCand AB � AM � MB. Thus,DN = DC � NC � AB � AM � MB, and so
. Therefore, bySSS.
6. a.b.
7. If �PTQ is isosceles with �PTQ the vertexangle, then . Thus, since doubles ofcongruent segments are congruent, ,and PQRS is a parallelogram with congruentdiagonals. Therefore, since a parallelogram withcongruent diagonals is a rectangle, PQRS is arectangle.
8. Since a rectangle is a parallelogram, its oppositesides are congruent. Thus, . Since P isthe midpoint of , . Since �C and �Bare both right angles, �C � �B. Thus,
by SAS. Therefore, sincecorresponding parts of congruent triangles arecongruent, .
9. a. Yes. Every equilateral triangle is equiangular.b. No. An equilateral quadrilateral does not
have to be equiangular. For example, arhombus is equilateral but consecutive anglesare not necessarily congruent.
10. Since ABEF and BCDE are rhombuses, theirdiagonals are perpendicular, so �EGB and�EHB are right angles. Opposite sides of a
rhombus are parallel, so . If two parallel lines are cut by a transversal, then the alternateinterior angles formed are congruent. Therefore,�ABE � �DEB. Since the sides of a rhombus
are congruent, and , and so
. By the reflexive property, ,so �ABE � �DEB by SAS. Correspondingparts congruent triangles are congruent, so�GEB � �HBE and m�GEB � m�HBE.
BE > BEAB > DE
BE > DEAB > BE
FEg
y ABg
AP > DP
nABP > nDCP
BP > PCBCAB > DC
PR > SQPT > TQ
/EBA, /EBC, /EDC, /EDA/EAD, /ECB, /ECD
nAND > nCMBDN > MB
NC > AMAN > MCDC > ABAD > BC
AM y CNAN y CM
NB ' DCDM ' ABAD > BCDM > NB
Since �EGB and �EHB are right triangles,their acute angles are complementary.Then, m�GBE � m�GEB � 90 and m�HBE � m�HEB � 90. By the substitutionpostulate, m�GBE � m�HBE � 90 andm�GEB � m�HEB � 90. Therefore, �GEHand �GBH are right angles and BHEG is arectangle.
11. Since a square is a rhombus, the result fromExercise 10 applies. Thus, BHEG is a rectangle,and BHEG has a right angle. Since ABEF is a square, . Thus, BHEG is also a rhombus. A rhombus with a right angle is asquare. Therefore, BHEG is a square.
12. a. Graphb. Parallelogram; slope of � slope of � 0;
slope of � slope of � 2.5. The slopes of opposite sides are equal. Therefore,opposite sides are parallel, and ABCD is aparallelogram.
c. 30 sq units13. a. Midpoint of � midpoint of � (2, 2)
Since the midpoints are the same, the twodiagonals bisect each other. If the diagonals ofa quadrilateral bisect each other, thequadrilateral is a parallelogram. Therefore,DEFG is a parallelogram.
b. The diagonals are perpendicular since theyare horizontal and vertical segments. If thediagonals of a parallelogram areperpendicular to each other, then theparallelogram is a rhombus. Therefore, DEFGis a rhombus.
c. Slope of and slope of . The slopes are not negative reciprocals. Thus, thetwo consecutive sides are not perpendicular,and �FGD is not a right angle. The angles in asquare are all right angles. Therefore, DEFG isnot a square.
d. Since DEFG is a rhombus, the area of the
quadrilateral is by Exercise 8 on page 411.
14. a. Midpoint of
� (b � d, a)
Midpoint of
� (b � d, a)
Since the midpoints are the same, the twodiagonals bisect each other. If the diagonals ofa quadrilateral bisect each other, the
DB 5 A 2b 1 2d2 , 0 1 2a
2 B
AC 5 A 2b 1 2d 1 02 , 2a 1 0
2 B
(DF)(GE)2
GF 5 23GD 5 22
3
FDGE
BCDADCAB
EG > GB
312
quadrilateral is a parallelogram. Therefore,ABCD is a parallelogram.
b. P(b, 0), Q(2b � d, a), R(b � 2d, 2a), S(d, a)c. Midpoint of
� (b � d, a)
Midpoint of
� (b � d, a)
Since the midpoints are the same, the twodiagonals bisect each other. If the diagonals ofa quadrilateral bisect each other, thequadrilateral is a parallelogram. Therefore,PQRS is a parallelogram.
15. a. 12 � xy b. 16 � 2(x � y)c. Solving for x in the second equation gives:
Substituting this value of x into the firstequation and solving for y gives:
Therefore, y � 6 or y � 2. When y � 6,x � 2, and when y � 2, x � 6. Thus, thedimensions of the rectangle are 6 centimetersby 12 centimeters. (The length and width canbe either of these two dimensions.)
16. In �ABD and �PQS, �A � �P, , and. Thus, �ABD � �PQS by SAS. Since
corresponding parts are congruent, .Since we are given that and ,�BCD � �QRS by SSS. Thus, �C � �Rbecause corresponding parts are congruent. Bythe partition postulate:
But since corresponding parts are congruent incongruent triangles:
m/BDC 5 m/QSR
m/DBC 5 m/SQR
m/ADB 5 m/PSQ
m/ABD 5 m/PQS
m/S 5 m/PSQ 1 m/QSR
m/Q 5 m/PQS 1 m/SQR
m/D 5 m/ADB 1 m/BDC
m/B 5 m/ABD 1 m/DBC
CD > RSBC > QRBD > QS
DA > SPAB > PQ
0 5 (y 2 6)(y 2 2)
0 5 2y2 1 8y 2 12
12 5 8y 2 y2
12 5 (8 2 y)y
12 5 xy
x 5 82y
SQ 5 A d 1 2b 1 d2 , a 1 a
2 B
PR 5 A b 1 b 1 2d2 , 0 1 2a
2 B
By the substitution postulate:
Thus, in ABCD and PQRS, all correspondingangles and sides are congruent, and so the twoquadrilaterals are congruent.
Exploration (page 416)a.
m/D 5 m/PSQ 1 m/QSR 5 m/S
m/B 5 m/PQS 1 m/SQR 5 m/Q
313
b. Rectangle, square, isosceles trapezoidc. Circumcenter: square, rectangle, isosceles
trapezoidNo circumcenter: rhombus, parallelogram(without a right angle), non-isosceles trapezoid
d. See graph of part a.e. Answers will vary. For example: a circumcircle is
a circle such that all of the vertices of a polygonare on the circle and the center is thecircumcenter of the polygon.
Cumulative Review (pages 417–418)Part I
1. 4 2. 2 3. 1 4. 1 5. 16. 4 7. 3 8. 2 9. 1 10. 2
Part II11. Since the sum of the angles of a triangle is 180°:
Therefore, the measures of the angles of thetriangle are 3(16) � 48°, 4(16) � 5 � 69°, and5(16) � 17 � 63°.
12. Statements Reasons
1. 1. Given.2. Through G, draw 2. Through a given
. point not on a given line, thereexists one and onlyone line parallel tothe given line.
3. 3. If two of three lines in the same planeare each parallel tothe third line, thenthey are parallel toeach other.
4. m�EGJ � m�FGH 4. Partition postulate.� m�z
HJg
y CDg
HJg
y ABg
ABg
y CDg
x 5 1612x 5 192
12x 2 12 5 180
3x 1 4x 1 5 1 5x 2 17 5 180
(Cont.)
Statements Reasons5. �x � �EGJ, 5. If two coplanar
�y � �FGH lines are cut by atransversal, thenthe alternateinterior anglesformed arecongruent.
6. m�x � m�EGJ, 6. Definition ofm�y � m�FGH congruent angles.
7. m�x � m�y � m�z 7. Substitutionpostulate.
Part III
13. Since the triangles are isosceles with bases and , and
by the isosceles triangletheorem. Since �CAB � �FDE,�CAB � �FDE � �CBA � �FED by thetransitive property. Since ,�ABC � �DEF by ASA.
14. since radii of a circle arecongruent. Since , by the transitive property and �OCD is equilateral.
DC > CO > DOAO > DCAO > DO > CO > BO
AB > DE
/FDE > /FED/CAB > /CBADEAB
314
Since and segments of parallel lines areparallel, . Therefore, AOCD contains a pair of opposite sides that are both congruentand parallel, and AOCD is a parallelogram. In aparallelogram, opposite sides are congruent.Thus, , and �AOC and �BOD areboth equilateral.
Part IV
15. Slope of Slope of
Slope of Slope of The slopes of only two pairs of sides are equal.Therefore, KLMN has only one pair of parallelsides, and so it is a trapezoid.
16. Under :
Under :
Cs(2, 4) S Cr(22, 4)
Bs(0, 4) S Br(0, 4)
As(2, 2) S Ar(22, 2)
ry-axis
C(4, 2) S Cs(2, 4)
B(4, 0) S Bs(0, 4)
A(2, 2) S As(2, 2)
ry5x
NK 5 1NM 5 212
LM 5 1KL 5 215
AD > OC
AO y DCAB y CD
Chapter 11.The Geometry of Three Dimensions
11-1 Points, Lines, and Planes (pages 422–423)
Writing About Mathematics1. Yes. If two lines intersect, then they determine a
plane. If two lines are parallel, then they lie in thesame plane. Since skew lines are neither parallelnor intersecting, skew lines can be defined as twolines that do not lie in the same plane.
2. Yes. Two intersecting lines determine a plane, soA, B, C, and D all lie in the same plane. Sincethey are distinct points in the same plane, theycan be the vertices of a quadrilateral.
Developing Skills
3. Since , the lines are coplanar and have no points in common, so A, B, C, and D are fourpoints that lie in the same plane. Since AB � CD,ABCD is not a parallelogram, so is notparallel to . A quadrilateral in which two and only two sides are parallel is a trapezoid.Therefore, ABCD is a trapezoid.
BCAD
CDg
ABg
y
4. Since , the lines are coplanar and have no points in common, so A, B, C, and D are fourpoints that lie in the same plane. A parallelogramis a quadrilateral in which two pairs of opposite
sides are parallel. Since , ABCD is a
parallelogram.
5. Two intersecting lines determine a plane, so and are coplanar. Therefore, A, B, C, and D are four points that lie in the same plane. Sincethe diagonals are perpendicular bisectors of eachother, �AEB and �AED are congruent right angles and . By the reflexive propertyof congruence, . Then �AEB ��AED by SAS, and . A quadrilateral with diagonals that are perpendicular is arhombus, and a rhombus with two congruentconsecutive sides is a square. Therefore, ABCD is a square.
AB > ADAE > AE
BE > DE
BEDAEC
ADg
y BCg
y CDg
ABg
In 6–9, answers will vary. Examples are given.6. and , and 7. and and 8. and , and 9. The skew lines
10. a. p ↔ (q ∧ r)b. p states that two lines in space are parallel. By
definition, parallel lines in space are coplanar,so q is true.
Applying Skills11. He should choose a base with three legs.
Consider the bottom of each leg of the camerabase. There is one and only one plane containingthree non-collinear points, so if the base hasthree legs, the bottom of each leg will touch theground at the same time and the base will besturdy. Four points can be in different planes, sothe bottom of each leg is not guaranteed to touchthe ground at the same time and it can wobble.
12. Two intersecting lines determine a plane, so ifeach pair of strings intersect, then they all lie inthe same plane and the stakes to which they areconnected lie in the same plane.
13. The four triangles are equilateral, so, ,, and . By the
transitive property of congruence,and
triangles are congruent by SSS.
11-2 Perpendicular Lines and Planes (pages 432–433)
Writing About Mathematics1. Yes. Since the four dihedral angles have equal
measures, they each measure 90 degrees and areright dihedral angles. Perpendicular planes aretwo planes that intersect to form a right dihedralangle.
2. a. Yes. If a line is perpendicular to a plane, it isperpendicular to each line in the planethrough the point of intersection of the lineand the plane.
b. CubeDeveloping Skills
3. False. At a given point on a given line, any planeperpendicular to the line at that point containsinfinitely many lines perpendicular to the line atthe point.
4. True. If a line not in a plane intersects the plane,then it intersects the plane in exactly one point.
AB > BC > AC > AD > CD > BD
AB > BD > ADAC > CD > ADBC > CD > BDAB > BC > AC
CGAEBFAEHGAEDC,AEHEDHADAE
315
5. False. If a line is perpendicular to a plane at B, and C and D are two other points on the
plane, then is not perpendicular to .6. False. The line and the plane can intersect the
given line in the same point.7. False; p ⊥ r and q ⊥ r, but p is not perpendicular
to q.
8. True. If a line is perpendicular to a plane, thenevery plane containing the line is perpendicularto the given plane.
9. False; p ⊥ r at A and q ⊥ r at A.
10. False. Let plane p be perpendicular to at A,
and let intersect . Through a given point on a plane, there is only one line perpendicular toa given plane. Therefore, p is not perpendicular
to .11. False. Let the two perpendicular planes, p and q,
intersect in line . Let line be perpendic-
ular to plane p in q. Then cannot be perpen-dicular to plane q because it lies in the plane.
Applying Skills
12. , so �RPA and �SPA are right anglesand congruent. P is the midpoint of , so
. by the reflexive property of congruence, so �RPA � �SPA by SAS.Corresponding parts of congruent triangles are congruent, so .
13. From steps 1 and 2, and .by the reflexive property of
congruence, so �RAB � �SAB by SSS.Corresponding parts of congruent triangles are congruent, so �RAB � �SAB.
14. From step 1, . From step 4, .by the reflexive property of
congruence, so �RPQ � �SPQ by SSS.Corresponding parts of congruent triangles arecongruent, so �RPQ � �SPQ.
PQ > PQRQ > SQRP > SP
AB > ABBR > BSAR > AS
AR > AS
AP > APRP > SPRS
l ' AP
BCg
BCg
ABg
ACg
ACg
ABg
ABg
r
q
A
p
r
q
p
CDg
ABg
ABg
15. We are given that ⊥ plane q at M, themidpoint of . A line is perpendicular to aplane if and only if it is perpendicular to eachline in the plane through the intersection of theline and the plane. Since R is a point in p,
and �AMR and �BMR are rightangles and congruent. by thedefinition of midpoint and by the reflexive property of congruence. Therefore,�AMR � �BMR by SAS. Corresponding partsof congruent triangles are congruent, so
and AR � BR.
16. We are given that M is the midpoint of , so. We are also given that
and . and by the reflexive property of congruence. Therefore,�AMS � �BMS and �AMR � �BMR by SSS. Corresponding parts of congruent trianglesare congruent, so �AMS � �BMS and �AMR � �BMR.If two lines intersect to form congruent adjacentangles, then they are perpendicular, so and . If a line is perpendicular to each of two intersecting lines at their point ofintersection, then the line is perpendicular to the plane determined by these lines. Therefore, is perpendicular to the plane determined by M, R,and S.
17. We are given equilateral �ABC in plane p and
perpendicular to plane p. A line is perpendicular to a plane if and only if it isperpendicular to each line in the plane throughthe intersection of the line and the plane.Therefore, and , so �BADand �CAD are right angles and congruent. Since�ABC is equilateral, . by the reflexive property of congruence. Therefore,�ABD � �ACD by SAS. Corresponding partsof congruent triangles are congruent, so
.
18. We are given that and intersect at A and
determine plane p, and is perpendicular to plane p at A. A line is perpendicular to a plane ifand only if it is perpendicular to each line in theplane through the intersection of the line and the
plane. Therefore, and , so �BAD and �CAD are right angles andcongruent. We are given that AB � AC, so
, and by the reflexive AD > ADAB > AC
ADg
' ACADg
' AB
ADg
ACg
ABg
BD > CD
AD > ADAB > AC
AD ' ACAD ' AB
ADg
AB
AB ' MRAB ' MS
MS > MSMR > MRSA > SBRA > RBAM > BM
AB
AR > BR
MR > MRAM > BM
AB ' MR
ABAB
316
property of congruence. Therefore,�ABD � �ACD by SAS.
19. We are given �QRS in plane p with perpendicular to plane p. A line is perpendicularto a plane if and only if it is perpendicular toeach line in the plane through the intersection of the line and the plane. Therefore,and , so �QST and �RST are rightangles and congruent. We are given that �QTS � �RTS, and by the reflexiveproperty of congruence. Therefore, �QST ��RST by ASA. Corresponding parts of con-gruent triangles are congruent, so .
20. If a line is perpendicular to two intersecting linesat their point of intersection, then the line isperpendicular to the plane determined by theselines. Therefore, the pole is perpendicular to theground.
21. A line is perpendicular to a plane if and only if itis perpendicular to each line in the plane throughthe intersection of the line and the plane.Therefore, the pole and the ground between thepole and each wire form right angles. The pole iscongruent to itself. The wires are of equal lengthsand are, therefore, congruent. The two trianglesformed by the pole, a wire, and the groundbetween the pole and wire are congruent by HL.Corresponding parts of congruent triangles arecongruent, so the other legs of the triangles arecongruent, and the wires are fastened to theground at equal distances from the pole.
11-3 Parallel Lines and Planes (pages 439–440)
Writing About Mathematics1. No. For example, consider the floor and any two
adjacent walls of a room. The walls are bothperpendicular to the floor but intersect in thecorner of the room.
2. Yes. Draw a line l perpendicular to the givenplane. If two planes are parallel, then a lineperpendicular to one plane is perpendicular tothe other plane. Thus, since l is perpendicular tothe given plane and since both planes are parallelto the given plane, both planes are perpendicularto l. Two planes perpendicular to the same lineare parallel. Therefore, the two planes are alsoparallel.
TQ > TR
ST > ST
ST ' RSST ' QS
ST
Developing SkillsIn 3–9, there may be more than one answer.
3. a. Plane EFGH ⊥ plane BCFE , , and
.
b. and
, but .
4. a. is parallel to planes ABEH and CDGF,
and .
b. is parallel to planes DCFG and BCFE,
but .
5. a. is perpendicular to and , and
.
b. is perpendicular to and , but
and intersect.
6. a. and do not intersect and are parallel.
b. and do not intersect but are skew
lines.7. a. Plane ABEH is perpendicular to planes
AHGD and BEFC, and .b. Plane ABEH is perpendicular to AHGD and
EFGH, but .
8. a. Plane EFGH is parallel to and , and
.
b. Plane EFGH is parallel to and , but
.
9. a. and is a skew line of both lines.
b. , is a skew line of , but
intersects .Applying Skills
10. We are given that p || q. �ABC determines aplane. If a plane intersects two parallel planes,then the intersection is two parallel lines.Therefore, . If two coplanar parallel linesare cut by a transversal, then the correspondingangles are congruent. Therefore, �ABC ��ADE and �ACB � �AED. Since �ABC isisosceles, �ABC � �ACB. By the transitiveproperty of congruence, �ADE � �AED. If twoangles of a triangle are congruent, then the sidesopposite these angles are congruent. Therefore,
and �ADE is isosceles.AD > AE
DE y BC
BEg
FEg
AHg
FEg
AHg
y BEg
DFg
AHg
y BEg
ABg
' ADg
ADg
ABg
ABg
y IJg
IJg
ABg
AHGD ' EFGH
AHGD y BEFC
DCg
AHg
BEg
AHg
DFg
DCg
DFg
DCg
ADg
AHg
y BEg
BEg
AHg
ABg
DCFG ' BCFE
AHg
ABEH y CDGF
IJg
AHg
y DCFGAHg
y DCFG
Plane DCFG ' plane BCFE
AHg
' EFGH
AHg
y BCFE
317
11. Consider the two triangles and .Because , and , so
. Since we are givenand by the reflexive
property, �PQA � �PQB by HL. Hence,.
12. Yes. Two lines perpendicular to the same planeare parallel. Thus, each pair of posts is parallel toeach other.
13. No. Parallel planes are everywhere equidistant.He has to make the poles the same length.
11-4 Surface Area of a Prism (pages 444–445)
Writing About Mathematics1. a. Equilateral triangle. An equilateral triangle is
equiangular. The base cannot change shapebecause the angles must always measure 60degrees.
b. Rhombus. The base can change shape becausewhile opposite angles must be congruent,these angle measures can change.
c. Yes. The lateral sides are rectangles and theyare perpendicular to the bases.
2. a. No. The rectangular faces are notperpendicular to the bases, so the height of theprism will be less than the length of thealtitude of one of the rectangular faces.
b. Yes. The faces that are parallelograms areperpendicular to the bases, so the height ofthe altitude will be equal to the length of thealtitude of one of the faces that areparallelograms.
Developing Skills3. 158 cm2
4. 1,500 in.2 or 125 ft2
5. 292 ft2
6. 9,605.62 cm2 or 96.0562 m2
7. a. Rectangleb. 9.00 cm 14.5 cm, 40.0 cm 14.5 cm,
41.0 cm 14.5 cmc. 1,665 cm2
8. a. Rectanglesb. 5 cm 12 cm, 5 cm 12 cm, 6 cm 12 cmc. 216 cm2
9. 6 faces. There are 2 bases, and each base has 4sides, so there are 4 lateral faces.
10. No. A parallelepiped must have parallelogramsas bases.
11. 162 in.2
AQ > BQ
PQ > PQPA > PBm/PQA 5 m/PQB 5 90
PQ ' QBPQ ' QAPQ ' pnPQBnPQA
Applying Skills12. We are given that the bases of a parallelepiped
are ABCD and EFGH with ABCD � EFGH.A parallelepiped is a prism, and the lateral edges of a prism are congruent and parallel.Therefore, or AE � BF � CG � DH, and .
13. a. We are given a prism with congruent bases�ABC and �DEF. The bases of a prism areparallel, so a line perpendicular to one of theplanes is perpendicular to the other.Therefore, and ;
and ; and and. If a plane contains a line
perpendicular to another plane, then theplanes are perpendicular. Therefore, thelateral faces are all perpendicular to the bases.By definition, the prism is a right prism andthe lateral faces are rectangles.
b. The lateral faces are congruent polygonswhen �ABC and �DEF are equilateral. Then
. Also,since parallel planes are
everywhere equidistant. The sides are allrectangles, so all angles are right angles.
14. 4 ft15. Let the bases of the parallelepiped be ABCD and
DEFG. A parallelepiped is a prism, and thelateral edges of a prism are parallel, so
. ABCD and DEFG arecongruent parallelograms, so and
. The lateral faces of the parallelepipedare parallelograms, so and .Similarly, .
16. Rhombuses. The lateral edges of a prism are con-gruent. Since the lateral faces are squares, thesides are all congruent to the lateral edges.Therefore, the edges of the bases are congruent.The bases of a parallelepiped are parallelograms,and a parallelogram with all sides congruent is arhombus.
17. By Exercise 16, a parallelepiped with squarelateral faces has bases that are rhombuses. Sinceone of the bases has a right angle, that base is asquare. A parallelepiped is a prism, so the basesare congruent. Therefore, both bases are squaresand the parallelepiped is rectangular. Since theedges are all congruent, all of the faces of theparallelepiped are congruent, and it is a cube.
BC y AD y EF y DGCD y FGAB y DE
DE y FGAB y CD
AD y BE y CF y DG
AD > BE > CFAB > BC > AC > DE > EF > FD
BE ' EFBE ' BCAD ' DEAD ' AB
AD ' DFAD ' AC
AE y BF y CG y DHAE > BF > CG > DH
318
18. a. 512 ft2 b. 8 ftHands-On Activity
1–3. Demonstrated with manipulatives.
11-5 Volume of a Prism (page 448)Writing About Mathematics
1. No. The bases need only have equal area.Polygons with equal area can be different shapes,and therefore, not congruent.
2. Yes. The height of the prism is equal to the heightof each of the lateral sides if the sides are allperpendicular to the base. By definition, such aprism is a right prism. The lateral sides of a rightprism are rectangles.
Developing Skills3. 72 ft3 4. 27.2 ft3 5. 157.5 in.3
6. 1,080 cm3 7. 50,008 cm3
Applying Skills8. 25 cm 9. 18 cm
10. AB : CD � 1 : 2;
11. Let the bases of the prism be �ABC and�A�B�C�. Consider lateral side ABB�A and aplane p parallel to the bases and intersecting thelateral sides. Then plane p will intersect lateral
side ABB�A in a line that is parallel to
and . Since ABB�A� is a parallelogram, we have �A � �B� and �A� � �B�. Since
, corresponding angles are congruent,so and . By thetransitive property, and
. Thus, both pairs of oppositeangles are congruent in , so is
a parallelogram. Similarly, since , we have and .Thus, is also a parallelogram because
and by thetransitive property. Since opposite sides of a parallelogram are congruent, and
. Therefore, plane p intersects lateral side in a segment that is congruent tothe base edges of the lateral side. Since was an arbitrary side, the result is true for theother two sides. Therefore, the figure formed byplane p is congruent to the bases of the prism.
ABBrArABBrAr
AB > AsBsArBr > AsBs
/Br > /ArAsBs/Ar > /AsBsBrAsBsBrAr
/B > /AsBsBr/A > /ArAsBsAsBsg
y ABg
ABBsAsABBsAs/AsBsB > /A
/AAsBs > /B/Br > /AsBsB/Ar > /AAsBs
AsBsg
y ArBrg
ArBrg
ABg
AsBsg
12 5 AB
CD
12CD 5 AB
12(AB)(CD) 5 (AB)2
12(AB)(CD) 5 (PQ)2
12. Let the sides of a right prism be s1, s2, . . . , snand the height be h. All of the lateral sides of a right prism are rectangles. The lateral area �s1h � s2h � . . . � snh � (s1 � s2 � . . . � sn)h.Since the perimeter of base � s1 � s2 � . . . � sn,the lateral area is equal to the perimeter of thebase times the height of the prism.
11-6 Pyramids (pages 452–453)Writing About Mathematics
1. They are both correct. In an equilateral triangle,the perpendicular bisectors intersect at the samepoint at which the medians intersect.
2. No. The lateral faces of a pyramid are congruentif the pyramid is regular.
Developing Skills3. 240 cm2 4. 12 ft2 5. 4,608 cm2
6. 576 cm3 7. 46 in.3 8. 2 ft3
9. cm3 10. 96 in.2
Applying Skills11. a. 49.5945 cm2 b. 148.7835 cm2
c. 198.378 cm2 d. 144.48531 cm3
12. 216 ft3
13. Let the base of the regular pyramid be �ABC,and the vertex, D. The lateral faces of a regular pyramid are isosceles triangles, so and
. By the transitive property ofcongruence, , so the lateral edges are congruent.
14. We are given a pyramid with square base ABCDand vertex F with . Since ABCD is a square, its diagonals are congruent and bisect each other at a point E, so
. By the reflexiveproperty of congruence, . Therefore,�AEF � �CEF by SSS, so �AEF � �CEF. Iftwo lines intersect to form congruent adjacent angles, then they are perpendicular, so
. Since ABCD is a square, .Since and we are given ,�ABF � �CBF by SSS. Thus, we also have�AEF � �BEF by SSS. Since correspondingangles are congruent, �AEF � �BEF, and �FEB is a right angle. Thus, is alsoperpendicular to . A line perpendicular to two lines in a given plane is perpendicular to the plane. Thus, is the altitude of the pyramid and intersects the base in its center E. Since aregular pyramid is a pyramid whose base is aregular polygon and whose altitude is
FE
DBFE
AF > FCFB > FBAB > BCAC ' EF
EF > EFAE > CE > DE > EB
AF > CF
AD > BD > CDBD > CD
AD > BD
2,25813
319
perpendicular to the base at its center, thepyramid is regular.
15. Let the base of the regular pyramid be �ABC,and the vertex, D. The lateral faces of a regularpyramid are congruent isosceles triangles, so �ABD � �BCD � �ACD with �
and �DAB � �DBC � �DCA. Draw thealtitudes such that , , and
. This forms congruent right angles�DEA, �DFB, �DGA. Therefore, �DEA ��DFB � �DGA by AAS, so the altitudes
, , and are congruent.16. Let the sides of the base of a regular pyramid
be s1, s2, . . . , sn . Then the perimeter p � s1 � s2 � . . . � sn. The lateral faces of aregular pyramid are triangles, so the lateral
area � �
� .
17. a. When both perimeter and slant height aredoubled, the lateral area increases by amultiple of 4. When tripled, the lateral areaincreases by a multiple of 9.
b. When both the sides of the base and theheight are doubled, the volume increases by amultiple of 8. When tripled, the volumeincreases by a multiple of 27.
11-7 Cylinders (pages 455–456)Writing About Mathematics
1. No. When the radius is doubled and the height is halved, the volume is .Therefore, the volume is doubled.
2. Yes. When the radius is doubled and the height is halved, the lateral surface area is
. Therefore, the lateral surface area remains unchanged.
Developing Skills3. a. 4,080p cm2 b. 6,392p cm2 c. 69,360p cm3
4. a. 96p in.2 b. 128p in.2 c. 192p in.35. a. 864p in.2 or 6p ft2
b. 1,512p in.2 or 10.5p ft2
c. 7,776p in.3 or 4.5p ft3
6. a. 15,000p cm2 or 1.5p m2
b. 35,000p cm2 or 3.5p m2
c. 750,000p cm3 or 0.75p m3
7. 6.2 cm 8. 6.4 cm9. 9 in. 10. 16.3 cm
Applying Skills11. 27.7 in.3 12. 12,117 gal
2p(2r) A 12h B 5 2prh
p(2r)2 A 12h B 5 2pr2h
12phs
12hs(s1 1 s21c1 sn)
12s1hs 1 12s2h21c1 12snhs
DGDFDE
DG ' ACDF ' BCDE ' AB
CDAD > BD
13. a. 47.7 cm3 b. 158.3 cm3
c. Volume of the lateral surface of the vase(without the base) � 158.3 cm3
But this is the height of the vase not includingthe base. The height of the vase is 14.65 � 0.75 � 15.4 centimeters.
d. cm2
14. 2.2 in., 5.5 in.
11-8 Cones (pages 458–459)Writing About Mathematics
1. No. For the lateral areas to be equal, the pyramidand the cone must have equal slant heights.
2. No. The volume of the larger cone is 4 times thevolume of the smaller since .
Developing Skills
3. a. b. c.4. a. b. c.5. a. b. c.6. a. b. c.7. 4.5 in. 8. 3.5 cm 9. 6 ft
Applying Skills10. 377 ft3 11. 4 gal 12. 112.32p in.3
13. Let the cone cut by the plane create �ABC withvertex B. since they are both the slantheight, hs, of the cone. Let hc, the altitude of thetriangle, intersect at D. Then �BDA and�BDC are right angles. The altitude of anisosceles triangle is also the median, so
. Since hc � r, and�ADC and �BDC are isosceles right triangles.The acute angle of an isosceles right trianglemeasures 45°, so m�ABD � m�CBD �m�ABC � 90 and �ABC is a right triangle.
11-9 Spheres (pages 463–464)Writing About Mathematics
1. Yes. The surface area is .2. No. The volume of the cone is one-fourth the
volume of the sphere since the volume of the cone is and the volume of the sphere is .4
3pr313pr3
4pr2 5 4p A 12d B 2
5 pd2
AD > BD > CDAD > CD
AC
AB > BC
128p cm3144p cm280p cm21,344p cm31,176p cm2600p cm2100p cm390p cm265p cm212p cm324p cm215p cm2
13p(2r)2hc 5 4
3pr2hc
2p(4.5)(15.4) < 435.4
< 14.65 cmh 5 158.3pf (4.5)2
2 (4.1)2g
158.3 5 phf(4.5)2 2 (4.1)2g
158.3 5 p(4.5)2h 2 p(4.1)2h
320
Developing Skills3. S � 225p in.2, V � 563p in.34. S � 697p cm2, V � 3,067p cm3
5. S � 16p ft2, V � 11p ft3
6. S � 1,989p cm2, V � 14,786p cm3
7. 5 ft 8. 2.6 cm 9. 6 in.10. in.3
Applying Skills11. 7 cups 12. 13.1 oz13. 6,3361,600p mi2 14. 4,665,600p mi2
15. LA � SA � 1,024p cm2 16. LA � SA � 4pr2
Hands-On Activitya. 9; each face of the prism has four symmetry
planes perpendicular to that face: two thatcontain the midpoints of opposite sides and twothat contain the opposite vertices.
b. 2; one symmetry plane is the plane parallel to thebases at the midpoint of the altitude of the prism;the other is the plane containing the altitudes ofthe bases.
c. 4; two symmetry planes are the planesperpendicular to the base containing thediagonals of the base, the other two are theplanes perpendicular to the base containing theperpendicular bisectors of the sides of the base.
d. 9; each face of the cube has four symmetry planesperpendicular to that face: two that contain themidpoints of opposite sides and two that containthe opposite vertices.
e. Infinitely many; a symmetry plane is any planecontaining a radius of the sphere.
f. Infinitely many; one symmetry plane is the planeparallel to the bases at the midpoint of thealtitude of the cylinder; the others are any planescontaining a radius of the base.
Review Exercises (pages 468–469)1. Yes. If a line is perpendicular to each of two
intersecting lines at their point of intersection,then the line is perpendicular to the planedetermined by these lines.
2. No. Through a given point on a line, there can beonly one plane perpendicular to the given line.
3. Only if and coincide. Through a given point on a plane, there is only one line to thegiven plane.
4. Yes. Two lines perpendicular to the same planeare coplanar.
5. Yes. Two planes are perpendicular if and only ifone plane contains a line perpendicular to theother.
RSg
RTg
45713p
6. Yes. Two planes are perpendicular if and only ifone plane contains a line perpendicular to theother.
7. Yes. If a line is perpendicular to a plane, then anyline perpendicular to the given line at its point ofintersection with the given plane is in the plane.
8. Yes. If a line is perpendicular to a plane, thenevery plane containing the line is perpendicularto the given plane.
9. Yes. If a plane intersects two parallel planes, thenthe intersection is two parallel lines. Therefore,
intersects if planes p and q are not parallel.
10. Yes. Two planes are perpendicular to the sameline if and only if they are parallel.
11. No. The lateral edges of a prism are parallel.12. Yes. The lateral edges of a prism are congruent.13. Yes. The volume of a prism is equal to the area of
the base times the height of the prism.14. No. The volume of the pyramid is equal to one-
third the area of the base times the height of thepyramid.
15. Yes. If two planes are equidistant from the centerof a sphere and intersect the sphere, then theintersections are congruent circles.
16. No. A great circle of a sphere is the intersectionof the sphere and a plane through the center ofthe sphere. The intersection of plane p and thesphere is not a great circle and the intersection ofplane q and the circle is a great circle. Therefore,the circles are not congruent.
17. L � 384 cm2, S � 512 cm2, V � 768 cm3
18. L � 180 in.2, S � 288 in.2, V � 270 in.319. L � 128 ft2, S � 254 ft2, V � 252 ft3
20. L � 60 in.2, S � 96 in.2, V � 48 in.321. L � 204 ft2, S � 283 ft2, V � 314 ft3
22. L � 396 cm2, S � 704 cm2, V � 1,385 cm3
23. 2.8 m24. cm25. a. 117.8 in.3 b. 7.2 in.3 c. 16 scoops
Exploration (pages 470–471)b.
No. No. No.vertices faces edges
Tetrahedron 4 4 6
Cube 8 6 12
Octahedron 6 8 12
Dodecahedron 20 12 30
Icosahedron 12 20 30
7!2
CDg
ABg
321
Vertices + Faces � 2 � Edgesc. Results will vary.
Cumulative Review (pages 471–473)Part I
1. 1 2. 2 3. 4 4. 35. 3 6. 4 7. 4 8. 49. 4 10. 3
Part II11. Statements Reasons
1. Isosceles �ABC 1. Given.and �DEF with
2. 2. Definition of isosceles triangle.
3. �B � �E 3. Given.4. �ABC � �DEF 4. SAS.
12. Statements Reasons1. Altitude bisects 1. Given.
�C in �ABC.2. �ADC � �BDC 2. Definition of altitude.3. �ACD � �BCD 3. Definition of angle
bisector.4. �A � �B 4. If two angles of one
triangle arecongruent to twoangles of anothertriangle, then thethird angles arecongruent.
5. 5. If two angles of a triangle arecongruent, then thesides opposite theseangles are congruent.
6. �ABC is isosceles. 6. Definition ofisosceles triangle.
Part III13. Statements Reasons
1. BCDE is a 1. Given.parallelogram.
2. 2. Definition of parallelogram.
3. 3. Opposite sides of a parallelogram arecongruent.
4. B is the midpoint of 4. Given..ABC
BC > DE
BC y ED
AC > BC
CD
BC > EFAB > DE
(Cont.)
Statements Reasons5. ABDE is a parallel- 5. If one pair of
ogram. opposite sides of aquadrilateral is bothcongruent andparallel, then thequadrilateral is aparallelogram.
14. is perpendicular to plane p at B, themidpoint of . By the definition of midpoint,
. Let D be a point on plane p. If a lineis perpendicular to a plane, then any lineperpendicular to the given line at its point ofintersection with the given plane is in the line.Therefore, and �ABD and �CBDare right angles and congruent. by the reflexive property of congruence, so
BD > BDAD ' ABC
AB > CBABC
ABC
322
�ABD � �CBD by SAS. Corresponding partsof congruent triangles are congruent, so
and AD � CD. Therefore, any point on p is equidistant from A and C.
Part IV
15. y � �3x � 1; the median is the line throughB(�1, 4) and the midpoint of . The midpoint
is at or (1, �2).
m � � �3
y � �3x � b
�2 � �3(1) � b
b � 1
16. A�(�3, �1), B�(1, 5); R90° (1, 3) � (�3, 1),rx-axis (�3, 1) � (�3, �1); R90° (5, �1) � (1, �5),rx-axis (1, �5) � (1, 5)
4 2 (22)21 2 1
A23 1 52 , 22 1 22
2 BAC
AD > CD
Chapter 12. Ratio, Proportion, and Similarity
12-1 Ratio and Proportion (pages 479–480)Writing About Mathematics
1. Yes. Consider the proportion . If the means are interchanged with the extremes, and theextremes are interchanged with the means, the resulting proportion is . This proportion is equivalent to the original.
2. Yes. If the extremes are perfect squares, then theproportional equation can be written as
or with a and b integers. Thus, ,an integer.
Writing About Mathematics3. Yes 4. Yes 5. No6. Yes 7. No 8. Yes9. , 10. ,
11. , 12. 20
13. 14. 6 15. 40.516. 2 17. 18.19. �6 or 5Applying Skills20. AB � 12, BC � 21 21. 8 cm, 40 cm22. 18 cm, 30 cm 23. 20 in., 24 in., 28 in.24. If the measures of the sides of the triangle are in
the ratio 2 : 3 : 7, then the measures of the sides ofthe triangle are 2x, 3x, and 7x for some x. In atriangle, the length of any side of the trianglemust be less than the sum of the lengths of the
6106306!3
315 5 2
1032 5 15
10
186 5 12
41812 5 6
4305 5 6
1306 5 5
1
x 5 aba2b2 5 x2
a2
x 5 xb2
ba 5 d
c
ab 5 c
d
other two sides. Since 7x � 2x � 3x � 5x, themeasures of the sides of the triangle cannot be inthe ratio 2 : 3 : 7.
25. Length � 30, width � 48 26. 40°, 140°
12-2 Proportions Involving Line Segments(pages 484–485)
Writing About Mathematics1. In any segment, the midpoint divides the
segment into two congruent parts. Thus, the ratioof the lengths of the two parts formed by themidpoint is always equal to 1. Therefore, themidpoints of any two line segments divide thetwo segments proportionally.
2. No. Let x � AB � BC � CD � DE � EF.Then BF � BC � CD � DE � EF � 4x, and AB : BF � x : 4x � 1 : 4.
Developing Skills3. 18 4. 8.5 5. 126. 5 7. 1 : 2 8. 3 : 19. 76 10. 138 11. 6 cm
12. 21 13. 12 14. 3015. AB � 6, BC � 1516. a. 2 : 9 b. 7 : 3 c. 4 : 7 d. 6 : 717. Since AB : BC � 2 : 3, AB � 2x and BC � 3x for
some x. Since DE : EF � 2 : 3, DE � 2y and EF � 3y for some y. Thus, AC � AB � BC � 5x
and DF � DE � EF � 5y. and ABAC 5 2x
5x 5 25
. Therefore, and
AB : AC � DE : DF.Applying Skills18. a. A line segment joining the midpoints of two
sides of a triangle is parallel to the third sideand its length is one-half the length of the third side. Thus, , joining the midpoints ofsides and , is parallel to and
. Since is a segment of and, . Since M is the midpoint of
, , so CM � NL. Thus,
and LMCN is a parallelogram because a pair of opposite sides is bothcongruent and parallel.
b. 18 units19. By Exercise 18, PMQC is a parallelogram. Since
�C is a right angle, PMQC is a parallelogramwith a right angle. Therefore, PMQC is arectangle.
20. a. A line segment joining the midpoints of twosides of a triangle is parallel to the third sideand its length is one-half the length of the third side. Thus, since joins the midpoints
of sides and , and .
Since is a segment on , is also parallel to . In particular, . Since
Q is the midpoint , , so AQ � MP by the transitive property. We aregiven PM � MD. Thus, AQ � DM, and
. Therefore, QADM is a paral-lelogram since it has a pair of opposite sidesthat are congruent and parallel. Exercises 18and 19 show that PMQC is a rectangle. Thus,�CQM is a right angle. Since �AQM and�CQM together form a linear pair, �AQM isalso a right angle. Thus, QADM is a parallelogram with a right angle, so it is arectangle.
b. Since Q is the midpoint of , .Since both QADM and PMQC are rectangles,�AQM and �CQM are both right angles, so�AQM � �CQM. Since by thereflexive property, �AQM � �CQM by SAS.Therefore, since correspondingparts of congruent segments are congruent.
AM > CM
QM > QM
AQ > QCAC
AQ > DM
12AC 5 AQAC
AQ y DMACDMPM
hDM
12AC 5 MPMP y ACCBAB
MP
CM 5 NL
12AC 5 CMAC
CM y NLAC y NLACCM1
2AC 5 NLACBCAB
NL
ABAC 5 DE
DFDEDF 5
2y5y 5 2
5
323
c. Since P is the midpoint of , .Since PMQC is a rectangle, �CPM is a rightangle. Since �CPM and �BPM together forma linear pair, �BPM is also a right angle, so �CPM � �BPM. Since by thereflexive property, �CPM � �BPM by SAS.Therefore, because correspondingparts of congruent segments are congruent. Bypart b, . By the transitive property,
, so M is equidistant fromthe vertices of �ABC.
21. Given: and with
Prove:
Statements Reasons1. 1. Given.2. 2. The product of
the means equalsthe product of the extremes.
3. (BC)(DF) � (BC)(EF) 3. Subtraction � (AC)(EF) � (EF)(BC) postulate.
4. 4. Distributive � property.
5. 5. Substitutionpostulate.
6. 6. If the products of two pairs of fac-tors are equal,one pair of factorscan be the meansand the other theextremes of aproportion.
22. Let M be the midpoint of , N be the midpointof , P be the midpoint of , and Q be themidpoint of . Then, diagonal divides the quadrilateral into two triangles �ACD and�ACB. A line segment joining the midpoints oftwo sides of a triangle is parallel to the third sideand its length is one-half the length of the third side. Thus, since joins the midpoints of thesides of �ACD and joins the midpoints ofthe sides of �ACB, we have , ,
, and . Two lines parallelto the same line are parallel. Thus, .
Since and , we have 12AC 5 MN1
2AC 5 QP
QP y MN
12AC 5 MN1
2AC 5 QP
MN y ACQP y ACMN
QP
ACDACDBC
AB
BCAB 5 EF
DE
(BC)(DE) 5 (EF)(AB)(EF)(AC2BC)
(BC)(DF2EF)
(AC)(EF)(BC)(DF) 5
BCAC 5 EF
DF
BCAB 5 EF
DE
BCAC 5 EF
DFDEFABC
AM > CM > BMCM > AM
BM > CM
MP > MP
CP > PBCB
QP � MN, and and are congruent.Therefore, MNPQ is a parallelogram since it has a pair of opposite sides that are congruentand parallel.
12-3 Similar Polygons (pages 488–489)Writing About Mathematics
1. Yes. Two polygons are similar if and only if allpairs of corresponding angles are congruent andthe ratios of the lengths of all pairs ofcorresponding sides are equal. Given any twosquares, all pairs of corresponding angles arecongruent since all right angles are congruent.The ratios of the lengths of any two pairs ofcorresponding sides are equal because the sidesof a square are all congruent. Therefore, any twosquares are similar.
2. No. For example, a square is not similar to anequilateral triangle since it does not have thesame number of sides.
Developing Skills3. 14. Yes. Two polygons are congruent if and only if all
pairs of corresponding parts are congruent. Thus,given any pair of corresponding sides of twocongruent polygons, the ratio of the lengths ofthe sides is equal to 1. Therefore, the twopolygons are similar.
5. No. Similar polygons are not congruent unlessthe ratio of the lengths of all pairs ofcorresponding sides is equal to 1. In that case,corresponding sides have equal lengths and arecongruent, and corresponding angles arecongruent.
6. 1 7. 27, 338. 3, , 4 9. 3a, 3b, 3c
Applying Skills10. Since the first triangle is equilateral, each side
has the same length, a. Since the second triangleis equilateral, each side has the same length, b.The ratio of the length of any side in the firsttriangle to the length of any side in the second triangle is . The angles of both triangles are all congruent since they all measure 60°. Therefore,the triangles are similar.
11. In any two regular polygons with the same num-ber of sides, the measure of any interior angle is
. Thus, all pairs of corresponding angles are congruent. If the length of one side of one ofthe regular polygons is a, then all of its sides have
180(n 2 2)n
ab
412
MNQP
324
the same measure. Similarly, if the length of oneside of the other regular polygon is b, then all ofits sides have the same measure. Thus, the ratio of the lengths of any pair of corresponding sidesis a : b, so the two regular polygons are similar.
12. a. A line segment joining the midpoints of twosides of a triangle is parallel to the third sideand its length is one-half the length of the third side. Thus, since joins the midpointsof the sides of �ABC, and
. Since M and N are midpoints of the sides of �ABC, we have and
, and the ratios of the lengths ofall pairs of corresponding sides are equal.Since , �CMN � �CAB and �CNM � �CBA because correspondingangles are congruent. Since �C � �C, allpairs of corresponding angles are congruent.Therefore, �ABC � �MNC.
b. 2 : 113. a. A line segment joining the midpoints of two
sides of a triangle is parallel to the third sideand its length is one-half the length of the third side. Thus, since joins the midpointsof the sides of �MNC, and
. Since M and N are midpoints of
the sides of �MNC, we have and
, and the ratios of the lengths of all pairs of corresponding sides are equal.Since , �CPQ � �CMN and �CQP� �CNM because corresponding angles arecongruent. Since �C � �C, all pairs ofcorresponding angles are congruent.Therefore, �PQC � �MNC. In Exercise 12awe showed that �MNC � �ABC. Therefore,since similarity is transitive, �PQC � �ABC.
b. 4 : 114. The angles of rectangles ABCD and EFGH are
all congruent since all right angles are congruent.Thus, all pairs of corresponding angles arecongruent. Since opposite sides are congruent ina rectangle, AB � DC, DA � BC, EF � GH, andHE � FG. Since , by substitution. Therefore, the ratios of the lengthsof all pairs of corresponding sides are equal, andthe two rectangles are similar.
DCHG 5 DA
HEABEF 5 BC
FG
PQ y MN
12NC 5 QC
12MC 5 PC
12MN 5 PQ
PQ y MNPQ
MN y AB
12CB 5 CN
12AC 5 MC
12AB 5 MN
MN y ABCM
15. Since and opposite sides of a
parallelogram are congruent, by substitution. Thus, the ratios of the lengths of allpairs of corresponding sides are equal. Since �K � �P, �M ��R because opposite sides of aparallelogram are congruent. In parallelogramKLMN, �N and �L are both the supplement of�K. In parallelogram PQRS, �S and �Q areboth the supplement of �P. Since thesupplements of two congruent angles arecongruent, �N � �L � �S � �Q. Therefore, allpairs of corresponding angles are congruent, andthe two parallelograms are similar.
12-4 Proving Triangles Similar (pages 494–495)
Writing About Mathematics1. Yes. The two right angles are congruent since all
right angles are congruent. Thus, if one acuteangle in one right triangle is congruent to anacute angle in the other triangle, the twotriangles are similar by AA�.
2. No. If the ratios of the lengths of two pairs ofcorresponding sides are equal, correspondingangles are not necessarily congruent. Thus, thetwo triangles do not need to be similar.
Developing Skills3. A line segment joining the midpoints of two sides
of a triangle is parallel to the third side and itslength is one-half the length of the third side.Thus, since joins the midpoints of the sides of �ABC, and . Since D and E
are midpoints of the sides of �ABC,
and . Thus, the ratios of the lengths of all pairs of corresponding sides are equal. Since
, �CDE � �CAB and �CED � �CBAbecause corresponding angles are congruent.Since �C � �C, all pairs of corresponding an-gles are congruent. Therefore, �ABC � �DEC.
4. 5 5. 6 6. 97. 12 8. 6 9. 15
10. 6 11. 10 12. 1613. 30 14. 9 15. DE � 4, DA � 3Applying Skills
16. In the proof of Theorem 12.5, we showed that CB � C�E and �A�B�C� � �DEC�. If twopolygons are similar, then their corresponding sides are in proportion, so .DE
ArBr 5 CrECrBr
DE y AB
12CB 5 CE
12AC 5 CD
12AB 5 DEDE y AB
DE
MNRS 5 NK
SP
KLPQ 5 LM
QR
325
Substituting CB � C�E into the proportion gives
. We are given that so by
the transitive property, . Therefore,
(DE)(A�B�) � (AB)(A�B�) or DE � AB.
17. Choose point D on such that B�D � BC.Choose point E on such that .Corresponding angles of parallel lines arecongruent, so and
. Therefore,by AA�. If two polygons are similar,
then their corresponding sides are in propor-tion, so . Substituting B�D � BC
into the proportion gives . We are
given that so by the transitivity
property, , and EB� � AB.Therefore, by SAS and
. Then by the transitiveproperty of similarity, .
18. Since �ABC is an isosceles right triangle,
and �A � �B. Since is the angle
bisector of �ACB, �ACD � �DCB.by the reflexive property of congruence. Then�ACD � �BCD by ASA. Thus, the angles�ADC and �BDC are congruent. If the anglesof a linear pair are congruent, then they are bothright angles. In particular, �CDA is a right angle.Therefore, �CDA � �C and �A � �A, so�ABC � �ACD by AA�.
19. Since and alternate interior angles in parallel lines are congruent, �DCA � �BAC.Since vertical angles are congruent,�DEC � �AEB. Therefore, by AA�,�ABE � �CDE.
20. We are given that �DAE � �BCE. Sincevertical angles are congruent, �AED � �CEB,so �ADE � �CBE by AA�.
21. Since perpendicular lines meet to form rightangles, �DFA and �BEA are both right angles.Thus, �DFA � �BEA since right angles arecongruent. Since ABCD is a parallelogram,opposite angles are congruent, so �D � �B.Therefore, �ABE � �ADF by AA�.
22. Since and are vertical and horizontal segments, respectively, they are perpendicular, so�EDC is a right angle. Similarly, �ABC is a rightangle. Thus, �EDC � �ABC. In �DEC, DE � 2
DCDE
AB y CD
CD > CD
CDh
AC > CB
nArBrCr , nABCnEBrD , nABC
nEBrD > nABC
EBrArBr 5 AB
ArBr
ABrArBr 5 BC
BrCr
EBrArBr 5 BC
BrCr
EBrArBr 5 BrD
BrCr
nEBrDnArBrCr , /CrArBr > /DEBr
/ArCrBr > /EDBr
ArCr y DEArBrBrCr
DEArBr 5 AB
ArBr
BCBrCr 5 AB
ArBrDE
ArBr 5 CBCrBr
and DC � 1. In �ABC, AB � 2 and CB � 4.Thus, , and the ratios of two pairs of corresponding sides are equal. Therefore,�ABC � �CDE by SAS�.
23. SP � 4 and SR � 2. Since :
, , and
. Therefore, the ratios of all
three pairs of corresponding sides are equal, so�PQS � QRS by SSS�.
24. Since and are vertical and horizontalsegments, respectively, , so �BOA is a right angle. Similarly, �DCA is a right angle.Thus, �BOA � �DCA. In �OAB, BO � 6 and OA � 4. In �CDA, CA � 3 and DC � 2. Thus,
. Therefore, since the ratios of two pairs of corresponding sides are equal and thecorresponding angles included between them arecongruent, �OAB � �CDA by SAS�.
25. Let the base of the pyramid be �ABC and let Pbe the vertex of the pyramid. If a plane p parallelto the base cuts the lateral faces of the pyramid,triangle �A�B�C� is formed (with A� on , B�
on , and C� on .) Consider the lateral face�ABP. Since p is parallel to the base, .A line is parallel to one side of a triangle andintersects the other two sides if and only if thepoints of intersection divide the sides propor-tionally. Thus, . Two line segments are divided proportionally if and only if the ratio ofthe length of a part of one segment to the lengthof the whole is equal to the ratio of thecorresponding lengths of the other segment.Thus, . Since �P � �P, �PA�B� ��PAB by SAS�. Therefore,
since corresponding sides are in proportion. Asimilar argument applied to the other two
lateral faces shows that and
that . By the transitive property,
. Therefore,�ABC � �A�B�C� by SSS�.
12-5 Dilations (pages 499–501)Writing About Mathematics
1. Yes. Since �ABC � �A�B�C�,because corresponding sides are in proportion.
2. (2, 1)
ABArBr 5 BC
BrCr 5 ACArCr
CrArCA 5 ArBr
AB 5 BrCrBC
PArPA 5 CrAr
CA
PBrPB 5 BrCr
BC
ArBrAB 5 PAr
PA 5 PBrPB
PArPA 5 PBr
PB
PArArA 5 PBr
BrB
ArBr y ABCPBP
AP
BOCA 5 OA
DC 5 2
BO ' OAOABO
SQSR 5 2!2
2 5 !2
PQQR 5 2!2
2 5 !2SPQS 5 4
2!2 5 !2
PQ 5 QS 5 2!2
CBDE 5 AB
DC 5 2
326
Developing Skills3. (3, 2) 4. 5. (6, 1)
6. 7. (24, 24) 8. (6, 39)
9. (�12, 21) 10. 11. (�2, 1)12. (�3, �4) 13. (1.5, 1) 14. (�10, �5.5)15. (6, �9) 16. (10, 7.5) 17.18. (�3, 6) 19. (2, 3) 20. (10, 6)21. 22.23. 24.
25.Applying Skills26. Under D
�3, A�(0, �15) and B�(�15, 0). The slopeof AB � the slope of A�B�, and the slope of A�B � the slope of AB�. AA� � BB� � 20, so the diagonals are congruent. Since , thequadrilateral is an isosceles trapezoid.
27. Since is a transversal cutting parallel
lines and , corresponding angles �HBAand �BED are congruent. Similarly, since
is a transversal cutting parallel lines
and , corresponding angles �EBC and �GEFare congruent. Thus, m�HBA � m�BED andm�EBC � m�GEF. By the partition postulate,180 � m�HBA � m�ABC � m�EBC and 180 � m�BED � m�DEF � m�GEF. Thus,m�ABC � 180 � m�HBA � m�EBC andm�DEF � 180 � m�BED � m�GEF. Bysubstituting m�HBA � m�BED and m�EBC� m�GEF, m�ABC � 180 � m�BED �m�GEF. Therefore, m�ABC � m�DEF and �ABC � �DEF.
28. a. A�(10, �15), B�(20, �15), C�(20, 5), D�(10, 5)b. Since and are both horizontal
segments, they are parallel. Since andare both vertical segments, they are
parallel. Therefore, A�B�C�D� is aparallelogram.
c. Since dilations preserve angle measure, allpairs of corresponding angles are congruent.
and
. Therefore, the ratios of the lengths of all pairs of corresponding sides are equal,and ABCD � A�B�C�D�.
d. Since dilations preserve angle measure,�B � �B�. In part c, we showed that
210 5 1
5
ABArBr 5 CD
CrDr 5DADrAr 5 BC
BrCr 5 420 5 1
5
BrCrDrAr
CrDrArBr
EFg
BCg‹
HBEG›
DEg
ABg
‹HBEG
›
AB y ArBr
RO, R1808, D21, or ry-axis + rx-axis
ry-axis + D15
ry-axis + D14
rx-axis + D4rx-axis + D1.5
A10, 203 B
A 1, 158 B
A213, 27
3 BA25
3, 0 B
. Therefore, by SAS�,
�ABC � �A�B�C�.
29. b. A�(6, 3), B�(3, 6), C�(�3, 6), D�(�6, 3),E�(�6, �3), F�(�3, �6), G�(3, �6), H�(6, �3)
c. HA � BC � DE � FG � 2d. H�A� � B�C� � D�E� � F�G� � 6e. A�B� � C�D� � E�F� � G�H�; draw diagonals
and . Under a dilation, angle measureis preserved, so �ABC � �A�B�C� by AA�.Since and AB � ,A�B� � .
f. Since dilations preserve angle measure, allpairs of corresponding angles are congruent.By parts c and d, the ratios of the lengths of allpairs of corresponding sides are equal to 1 : 3.Therefore, the two polygons are similar.
30. a. 10 sq units b. 90 sq unitsc. 160 sq units d. 250 sq unitse. The area of the image of a figure is equal to
the product of the area of the figure and k2,the square of the constant of dilation.
31. a. A(a, b) → A�(ka, kb)
B(a, b � d) → B�(ka, kb � kd)
C(c, b) → C�(kc, kb)
D(c, b � d) → D�(kc, kb � kd)
Since A� and B� have the same x-coordinate,is a vertical segment. Similarly, since C�
and D� have the same x-coordinate, isalso a vertical segment. Vertical segments areparallel. Therefore, the images and are parallel.
b. A(a, b) → A�(a, b)
B(c, d) → B�(kc, kd)
C(a � e, b) → C�(ka � ke, kb)
D(c � e, d) → D�(kc � ke, kd)
The slope of
and the slope of
.
The two slopes are the same. Therefore, theimages and are parallel.
12-6 Proportional Relations AmongSegments Related to Triangles (pages 505–506)
Writing About Mathematics
1. Both are correct. The two ratios are equal: .25 5
CrDrArBr
5 kd 2 kbkc 1 ke 2 ka 2 ke 5 kd 2 kb
kc 2 ka 5 d 2 bc 2 a
DrCr5 d 2 bc 2 a
5kd 2 kbkc 2 ka 5
k(d 2 b)k(c 2 a)BrAr
CrDrArBr
CrDrArBr
3!2!2AB
ArBr 5 BCBrCr 5 1
3
ArCrAC
ABArBr 5 BC
BrCr
327
2. No. Consider two similar triangles �ABC and�A�B�C�. Then, since the ratio of any pair ofcorresponding sides is equal to k, AB � kA�B�,BC � kB�C�, and CA � kC�A�. Thus, theperimeter of the first triangle is AB � BC � CA� k(A�B� � B�C� � C�B�), and the ratio of the
two perimeters is .
Developing Skills3. 6 cm 4. 4 : 3 5. 9 in.6. 12 in. 7. 6 : 7 8. 16 mm
9. a. 7 : 6 b. m, 14 mc. 44 m, m d. Yes
Applying Skills10. Statements Reasons
1. �ABC � �A�B�C� 1. Given.with ratio of similitude k : 1.
2. 2. The ratio of the lengths ofcorresponding sidesof similar polygonsis equal to the ratioof similitude.
3. M is the midpoint of 3. Given.and M� is the mid-
point of .4. AC � 2CM and 4. Definition of
A�C� � 2C�M� midpoint.5. or 5. Substitution (steps
2, 4).
6. �C � �C� 6. In similar triangles,correspondingangles arecongruent.
7. �BCM � �B�C�M� 7. SAS�.
11. Statements Reasons1. �ABC � �A�B�C� 1. Given.2. �C � �C� and 2. In similar triangles,
�B � �B� corresponding an-gles are congruent.
3. is the bisector of 3. Given.
�B and is the bisector of �B�.
4. m�CBE � m�B and 4. Definition of angle
m�C�B�E� � m�B� bisector.12
12
BrErg
BEh
CMCrMr 5 CB
CrBr 5 k1
2CM2CrMr 5 CB
CrBr 5 k1
ArCrAC
ACArCr 5 CB
CrBr 5 k1
5113
1613
k(ArBr 1 BrCr 1 CrBr)ArBr 1 BrCr 1 CrBr 5 k
1
251
(Cont.)
Statements Reasons5. �CBE � �C�E�B� 5. Halves of con-
gruent angles arecongruent.
6. �BCE � �B�C�E� 6. AA�.
12. Let ABCD and A�B�C�D� be two similar parallelograms. Since and �B � �B�, �ABC � �A�B�C� by SAS�.Therefore, because the ratios of the lengths of corresponding sides in similartriangles �ABC and �A�B�C� are equal.
13. Let �ABC � �DEF with ratio of similitude k : 1. Then AC : DF � k : 1, or AC � k(DF). If twotriangles are similar, then the lengths ofcorresponding altitudes have the same ratio asthe lengths of any two corresponding sides. Let
be the altitude to and be the altitudeto . Then BG : EH � k : 1 or BG � k(EH).The area of �ABC � (AC)(BG) and the areaof �DEF � (DF)(EH). By the substitutionpostulate, (AC)(BG) � [k(DF)][k(EH)] � k2(DF)(EH). Therefore, the ratio of the areas is k2 : 1.
14. a. Let ABCD be a trapezoid with parallel bases and . Let E be the intersection point of
the diagonals. Since , alternate interior angles are congruent, so �ABE ��CDE. Since vertical angles are congruent,�DEC � �AEB. Therefore, �ABE � �CDE
b. In similar triangles, the ratio of the lengths ofcorresponding sides is equal to the ratio ofsimilitude, k. Since �ABE � �CDE,
. Therefore, the ratio of similitude is equal to the ratio of the length ofthe parallel sides.
12-7 Concurrence of the Medians of aTriangle (pages 509–510)
Writing About Mathematics
1. Yes. Since the point P divides in the ratio 2 : 1, that is, AP : PM � 2 : 1, is twice thelength of . Therefore, can be divided intotwo congruent segments that are equal to thelength of .
2. Yes. In an isosceles triangle, the altitude to thevertex is also the median. Therefore, in anisosceles triangle, the altitude to the vertex is alsothe perpendicular bisector.
PM
APPMAP
AM
EBED 5 EA
EC 5 ABDC 5 k
AB y CDCDAB
DFEHACBC
ACArCr 5 AB
ArBr 5 BCBrCr
ABArBr 5 BC
BrCr
328
Developing Skills3. (�1, 2) 4. (�1, 1) 5. (1, 3)6. (3, 0) 7. (1, 3) 8. (�2, 4)9. (�4, �1) 10. (7, 1)
Applying Skills11. a. B(�2, 0) and C(2, 0)
b. M(�1, 3) and N (1, 3)c. y � �x � 2d. y � x � 2e. (0, 2)
12. A(3, 3), B(3, �3), C(�6, 0)
12-8 Proportions in a Right Triangle (pages 514–515)
Writing About Mathematics1. �ACD � �BCD when �ABC is an isosceles
right triangle. When �ABC is isosceles, andare the two congruent hypotenuses of right
triangles �ACD and �BCD, respectively. Since � , �ACD � �BCD by HL.
2. �RST is an isosceles right triangle. The altitude iscongruent to itself. Also, it is perpendicular to thehypotenuse, forming congruent right angles. Thealtitude also separates the hypotenuse into twocongruent segments, so the two smaller trianglesare congruent by SAS. Corresponding parts of congruent triangles are congruent, so �and �RST is an isosceles right triangle.
Developing Skills3. 12 4. 2 5. 20 6. 47. 6 8. 25 9. 9 10. 6
11. 12 12. 6Applying Skills13. 10 in. 14. 32 ft. 15. 2 m, 18 m16. 17. 4, 16 18. 6,19. 18 cm 20. 12 in.21. a. 7 and 28 b. and
12-9 Pythagorean Theorem (pages 519–521)Writing About Mathematics
1. Yes. Let �ABC be obtuse with �C the obtuseangle, AB � c, BC � a, and AC � b. The altitudelies outside the triangle and �ACD is a righttriangle. Let DC � x and AD � y. Then,b2 � x2 � y2. �ABD is also a right triangle, so c2 � (a � x)2 � y2 or c2 � a2 � 2ax � x2 � y2. Bythe substitution postulate, c2 � a2 � 2ax � b2.Since a2 � 2ax � b2 � a2 � b2, c2 � a2 � b2.
2. Yes. The diagonals of a parallelogram bisect eachother and separate the parallelogram into four
7!514!5
12!22!5
STRT
CDCD
CBCA
triangles. If the lengths of the diagonals are 6 and8 and one side of the parallelogram is 5, then a 3-4-5 right triangle is formed. Thus, all of the fourtriangles are right triangles and the diagonals areperpendicular. A parallelogram withperpendicular diagonals is a rhombus.
Developing Skills3. Yes 4. No 5. No6. Yes 7. Yes 8. Yes9. 11.3 cm 10. 52 cm 11. 160 mm
12. 24 ft 13. ft14. a. ft b. 36 ft15. 13 cm 16. 27 ft, 36 ft17. 20 ft, 21 ft, 29 ftApplying Skills18. The diagonals of the rectangle are 26 feet long.
Thus, if he marks the corners of two triangleswith sides that are 10, 24, and 26 feet in length,the two triangles formed are right triangles(because 102 � 242 � 262) and the resultingquadrilateral is a rectangle.
19. 2,448 ft2 20. in.21. If the adjacent sides of a parallelogram are 21
and 28 feet long and the diagonal is 35 feet, then212 � 282 � 352. By the converse of thePythagorean Theorem, a right triangle is formedwith the diagonal and the two adjacent sides, andthe parallelogram has a right angle. Aparallelogram with a right angle is a rectangle.Therefore, the parallelogram is a rectangle.
22. The diagonals of a parallelogram bisect eachother and separate the parallelogram into fourtriangles. If the diagonals of the parallelogramare 48 and 140 centimeters long, then one of thetriangles formed has sides that are 24, 70, and 74 centimeters long. By the converse of thePythagorean Theorem, since 242 � 702 � 742, oneof the triangles formed by the diagonals is a righttriangle, so all of the four triangles formed areright triangles. Therefore, the diagonals areperpendicular, and the parallelogram is arhombus.
23. 7.5 ft 24. ft
12-10 The Distance Formula (pages 524–526)
Writing About Mathematics1. Yes. Since vertical segments have the same x-
coordinate, x1 � x2 � 0 in the distance formula.Similarly, since horizontal segments have thesame y-coordinate, y1 � y2 � 0. In such cases, the
2!3
5!3
!64!61
329
distance formula reduces to �y1 � y2� and �x1 � x2�,respectively, or the length of a vertical and ahorizontal segment.
2. Let x2 � x1 be equal to some number a. If a � 0,then �a� � a, so �a�2 � a2. If a � 0, then �a� � a,so �a�2 � a2 � 0. If a � 0, then �a� � �a, so �a�2 � (�a)2 � a2.
Developing Skills3. 5 4. 13 5.6. 7. 8. 109. 10. 11. 16 or �8
12. 5 or �113. AB � BC � CD � DA � 5. The lengths of the
sides of the quadrilateral are all the same.Therefore, the quadrilateral is a rhombus.
14. a. . Thus, and�PQR is an isosceles triangle.
b. , so and . By the
converse of the Pythagorean Theorem, �PQRis a right triangle.
c. The midpoint, D, of is (4, 0), and. Therefore, the
midpoint of the hypotenuse is equidistantfrom the vertices of �PQR.
15. a. , , and .Therefore, the lengths of the sides of thetriangle are unequal, and the triangle isscalene.
b.and . By the converse of the Pythagorean Theorem, �LMN is a right triangle.
c. The midpoint, D, of is , and
. Therefore, the midpoint of the hypotenuse is equidistantfrom the vertices of �LMN.
16. ; therefore, .
17. a. y � 3x � 13 b. D(�4, 1) c.18. a. ED � DF = FE � 6
b. Yes. The sides all have the same length, so thetriangle is equilateral.
19. a. A�(2, �2), B�(8, 2), C�(4, 8)b. and . Thus, the
length of is not equal to the length of itsimage, , so distance is not preserved under the dilation.
ArBrAB
ArBr 5 2!13AB 5 !13
2!10
DE > FEDE 5 FE 5 !58
DL 5 DN 5 DM 5 52!2
A412, 21
2 BNM
(MN)2 5 (5!2)2 5 50(LM)2 1 (NL)2 5 (2!10)2 1 (!10)2 5 50
NL 5 !10MN 5 5!2LM 5 2!10
DP 5 DQ 5 DR 5 !10PQ
(PQ)2 5 (2!10)2 5 405 40(PR)2 1 (QR)2 5 2(2!5)2PQ 5 2!10
PR > QRPR 5 QR 5 2!5
6!25!22!23!2
2!10
c. , ,, and .
Therefore, , and corresponding sides are in proportion. BySSS�, �ABC � �A�B�C�.
d. In part c, we showed that �ABC � �A�B�C�.Since corresponding angles are congruent insimilar triangles, �A � �A�, �B � �B�, and�C � �C�. Therefore, angle measure ispreserved under this dilation.
20. a. and .Opposite sides are congruent, so ABCD is aparallelogram.
b. A�(6, 0), B�(9, �3), C�(12, 3), D�(9, 6)c. and
. Opposite sides are congruent, soA�B�C�D� is a parallelogram.
d. Yes. Since A�B�C�D� is also a parallelogram,, and the images of the two parallel
lines and are also parallel.
21. and .Opposite sides are congruent, so ABCD is aparallelogram. However, consecutive sides arenot congruent, so ABCD is not a rhombus.
22. , so �ABC is isosceles., so (AB)2 � 80 and � 80. By the converse of the
Pythagorean Theorem, �ABC is a right triangle.
Applying Skills
23. Let , and .
Then, . The lengths of the two sides are the same, so �ABC is an isoscelestriangle.
24. , so the diagonals arecongruent. The slope of , and the slope of
. Since the slopes are negativereciprocals, . Therefore, the diagonalsof a square are congruent and perpendicular.
25. Since and are both horizontal segments,
and BC � DA � a. Thus, ABCD is a
parallelogram. The length of is . If
a2 � b2 � c2, then AB � DA � a, and the two
consecutive sides and are congruent.DAAB
"b2 1 c2AB
BC y DA
DABC
EG ' FHFH 5 21
EG 5 1EG 5 FH 5 a!2
AB 5 BC 5 "a2 1 b2
C 5 (a, 0)A 5 (2a, 0), B 5 (0, b)
5 2(2!10)2(CA)2 1 (CB)2AB 5 4!5
CA 5 CB 5 2!10
BC 5 DA 5 !10AB 5 CD 5 2!5
CDABArBr y CrDr
5 3!5BrCr 5 DrArArBr 5 CrDr 5 3!2
BC 5 DA 5 !5AB 5 CD 5 !2
ABArBr 5 BC
BrCr 5 CACrAr 5 1
2
CrAr 5 2!26ArBr 5 BrCr 5 2!13CA 5 !26AB 5 BC 5 !13
330
A parallelogram with two consecutive congruentsides is a rhombus. Therefore, if a2 � b2 � c2,then ABCD is a rhombus.
26. a.
b.
c.
On the other hand:
Therefore, PM � MQ.
27. a. W�(kw, ky) and X�(kx, kz)
b.
Review Exercises (pages 529–531)1. , , 152. Yes. In �ABC, the measure of the other
acute angle is 90 � 67 � 23°. Therefore,�ABC � �LMN by AA�.
3. 94. and �C � �C. Therefore, since
the ratios of two corresponding sides are inproportion, and the angles included between the two pairs of sides are congruent,�EFC � �ABC by SAS�.
5. 86. a. 16 b. 32 c. d.7. 15.6 cm 8. 12.2 cm9. a. P�(4, �14), Q�(16, 2) b. 1 : 2
c. M(5, 3) d. M�(10, �6)e. N(10, �6) f. Yes, since M� � N
16!58!5
ECAC 5 CF
CB 5 13
2212121
2
5 k ? WX5 k"(x 2 k)2 1 k2(z 2 y)2
5 "k2(x 2 k)2 1 k2(z 2 y)2
WrXr 5 "(kx 2 kw)2 1 (kz 2 ky)2
5 $A x2 2 x12 B 2
1 A y2 2 y12 B 2
5 $A 2x2 2 x1
2 x22 B 2
1 A 2y2 2 y1
2 y22 B 2
5 $A 2x22 2
x1 1 x22 B 2
1 A 2y22 2
y1 1 y22 B 2
MQ 5 $Ax2 2 x1
1 x22 B 2
1 Ay2 2 y1
1 y22 B 2
5 $A x2 2 x12 B 2
1 A y2 2 y12 B 2
5 $A x1 1 x2
2 2x12 B 2
1 A y1 1 y2
2 2y12 B 2
5 $A x1 1 x22 2
2x12 B 2
1 A y1 1 y22
_
2y12 B 2
PM 5 $A x1 1 x22 2 x1 B
2 1 A y1
1 y22 2 y1 B
2
MQ 5 $Ax2 2 x1
1 x22 B 2
1 Ay2 2 y1
1 y22 B 2
PM 5 $A x1 1 x22 2 x1 B
2 1 A y1
1 y22 2 y1 B
2
10. Statements Reasons
1. 1. Given.2. �ABE � �DCE 2. If two parallel lines are
cut by a transversal, thealternate interior anglesare congruent.
3. �CED � �AEB 3.Vertical angles arecongruent.
4. �ABE � �DCE 4. AA�.
11. Statements Reasons
1. �ABE � �DCE 1. Given.2. 2. Definition of
similar polygons.3. 3. If the points at which a
line intersects two sidesof a triangle dividethose sides proportion-ally, then the line isparallel to the thirdside.
12. 12 cm 13. 25 in. 14. 43.8 cm15. a. , ,
b. and.
By the converse of the Pythagorean Theorem,�RST is a right triangle.
16. a. AC � AB � 5, so �ABC is isosceles since ithas two sides with the same length.
b. D(2, 4)c. and d. and
(AB)2 � 52 � 25. By the converse of thePythagorean Theorem, �ADB is a righttriangle. Therefore, is perpendicular to
, so is the altitude to .
17. a. A�(�6, 3), B�(6, �3), C�(0, 9)b. ,
,c. , so corresponding
sides are in proportion. By SSS�,�ABC � �A�B�C�.
d. P(0, 1)e. CP � 2 and PM � 1. Thus, .
f. P�(0, 3) g. Yesh. C�P� � 6 and P�M� � 3. Thus, .
18. 26 cm
CrPrPrMr 5 2
1
CPPM 5 2
1
ABArBr 5 BC
BrCr 5 CACrAr 5 1
3
CrAr 5 6!2ArBr 5 BrCr 5 3!5CA 5 2!2AB 5 BC 5 2!5
BCADBCAD
(DB)2 1 (AD)2 5 (2!5)2 1 (!5)2 5 25DB 5 !5AD 5 2!5
(ST)2 1 (TR)2 5 (!10)2 1 (2!10)2 5 50(RS)2 5 50
TR 5 2!10ST 5 !10RS 5 5!2
y ABg
EFg
CECA 5 CF
CB
y CDg
ABg
331
Exploration (pages 531–532)a. , ,
b.
4 � 4 ✔
Since AFGD and FGCB are both rectangles,corresponding angles are congruent.Therefore, AFGD � FGCB, so AFGD andFGCB are both golden rectangles.
c. Yes d. Results will vary.
Cumulative Review (pages 532–534)Part I
1. 2 2. 4 3. 2 4. 25. 1 6. 4 7. 2 8. 39. 2 10. 2
Part II11. Statements Reasons
1. 1. Given.2. �EBC � �ECB 2. Isosceles Triangle
Theorem3. �ABE is the supple- 3. If two angles form a
ment of �EBC and linear pair, then �DCE is the supple- they are supple-ment of �ECB. mentary.
4. �ABE � �DCE 4. Supplements ofcongruent angles arecongruent.
5. 5. Given.6. �ABE � �DCE 6. SAS.7. 7. Corresponding parts
of congruent trianglesare congruent.
12. Statements Reasons1. at R. 1. Given.2. Points A and B in 2. Given.
plane p3. and 3. If a line is perpen-
dicular to a plane,then it is perpendic-ular to each line inthe plane through theintersection of theline and the plane.
SR ' RBSR ' AR
SR ' plane p
AE > DE
AB > CD
BE > CE
4 0 5 2 1(2)(2) 0 (1 1 !5)(!5 2 1)
1 1 !52 0 2
!5 2 1
AFAD 0 FG
BF
BF 5 !5 2 1AF 5 1 1 !5EC 5 EF 5 !5
(Cont.)
Statements Reasons4. �ARS � �BRS 4. Perpendicular lines
meet to form rightangles and all rightangles are congruent.
5. 5. Reflexive property.6. �ARS � �BRS 6. SAS.7. 7. Corresponding parts
of congruent trianglesare congruent.
Part III13. a. Since the radius is one-half the slant height,
hc � 5.00 ft.
Lateral area �
�b. The radius, slant height, and height of the
cone form a right triangle. Thus,
Volume �
14. a. Since is a horizontal segment, the altitudefrom C to is a vertical line with the same x-coordinate as C. Thus, the equation of thealtitude is x � 2.
b. The slope of . Thus, the altitude fromB to will be a line with a slope of passing through (4, 0). Using the point-slopeformula, the equation of the altitude is:
c. Substitute x � 2 into the equation:
Therefore, the coordinates of D are (2, 1).
d. passes through the points (�1, 0) and
(2, 1). Therefore, the slope of .
The equation of can be found by using the
point-slope formula:
y 5 13x 1 13
y 2 0 5 13(x 1 1)
ADg
ADg
5 1 2 02 1 1 5 1
3
ADg
5 15 21 1 2
y 5 212(2) 1 2
y 5 212x 1 2
y 5 212x 1 2
y 2 0 5 212(x 2 4)
212AC
AC 5 2
ABAB
13pr2hc 5 1
3p(2.50)2(4.33) < 28.3 ft3
hc 5 "h2s 2 r2 5 "5.002 2 2.502 5 4.3301c
49.1 ft2
12pr2hs 5 1
2p(2.50)2(5.00)
SA > SB
SR > SR
332
e. The slope of is �3. Since the slopes of
and are negative reciprocals, they are
perpendicular. Therefore, contains the
altitude from A to .Part IV15. Statements Reasons
1. �ADE is an isosceles 1. Given.triangle.
2. is the median 2. Given.to .
3. 3. In an isosceles triangle, the medianfrom the vertexangle is perpen-dicular to the base.
4. 4. If a transversal is perpendicular toone of two parallellines, it is perpen-dicular to the other.
5. �DFG, �AFG, 5. Definition of per-�CGF, and �BGF pendicular lines.are all right angles.
6. ABCD is a rectangle. 6. Given.7. �FDC, �DCG, 7. The angles of a
�FAB, �GBA are rectangle are all all right angles. right angles.
8. FGCD and FGAB 8. A quadrilateral are both rectangles. with four right an-
gles is a rectangle.9. and 9. In a rectangle,
opposite sides are congruent.
10. 10. Definition of amedian.
11. 11. Transitive property.12. G is the midpoint 12. Definition of
of . midpoint.
16. a.
b.
The two images are the same for bothtransformations. Therefore, thetransformations are equivalent.
R908(x, y) 5 (2y, x)ry5x + rx-axis(x, y) 5 ry5x(x, 2 y) 5 (2y, x)
Cr 5 ry5x + rx-axis(6, 4) 5 ry5x(6, 24) 5 (24, 6)Br 5 ry5x + rx-axis(3, 1) 5 ry5x(3, 21) 5 (21, 3)Ar 5 ry5x + rx-axis(2, 5) 5 ry5x(2, 25) 5 (25, 2)
CB
GC > GB
DF > FA
FA > GBDF > GC
EFGg
' BC
EF ' ADAD
EF
BC
ADg
ADg
BCBC
13-1 Arcs and Angles (pages 541–542)Writing About Mathematics
1. Yes. If two lines intersect at the center of a circle,they form four central angles that are alsovertical angles. Since vertical angles arecongruent, the arcs intercepted by each pair ofvertical central angles are also congruent.
2. Yes. Since the degree measure of a circle is 360,the four arcs each measure 90 degrees, and thecentral angles that intercept the arcs each
measure 90 degrees. Therefore, and are perpendicular.
Developing Skills3. 35 4. 48 5. 90 6. 1407. 180 8. 60 9. 75 10. 100
11. 120 12. 170 13. 91 14. 8915. 91 16. 138 17. 138 18. 13319. 133 20. 227 21. 180 22. 18023. 100 24. 110 25. 35 26. 11527. 115 28. 210 29. 145 30. 14531. 150 32. 215Applying Skills33. In a circle, central angles are congruent if their
intercepted arcs are congruent. Since ,�AOB � �COD. All radii of the same circle are
congruent, so . Therefore,�AOB � �COD by SAS.
34. �AOC and �BOD are vertical angles, so theyare congruent. Radii of the same circle are congruent, so . Then �AOC � �BOD by SAS. Since correspondingparts of congruent triangles are congruent,
.
35. � 90, � 27036. Since and , the diagonals of ABCD, are
perpendicular, ABCD is a rhombus. �AOC and�BOC are right angles since perpendicular linesintersect to form right angles. Radii of the same circle are congruent, so .Therefore, �AOB and �BOC are isosceles righttriangles and m�ABO � m�CBO � 45. Thenm�ABC � m�ABO � m�CBO � 90. Arhombus with a right angle is a square, so ABCDis a square.
Hands-On ActivityIn 1–2, results will vary.
3. By the definition of the degree measure of an arc, since � 60, m�AOB � 60 and sincemABC
OA > OB > OC
BDACmADCCmACC
AC > BD
OA > OB > OC > OD
OA > OB > OC > OD
ABC > CDC
BDg
ACg
333
, m�A�O�B� � 60. Therefore,�AOB � �A�O�B�. Radii of the same circle arecongruent, so and . By definition, �AOB and �A�O�B� are isoscelestriangles with m�OAB � m�OBA � 60 andm�O�A�B� � m�O�B�A� � 60. Then �OAB � �OBA � �O�A�B� � �O�B�A�and �AOB � �A�O�B� by AA�.
13-2 Arcs and Chords (pages 551–552)Writing About Mathematics
1. The apothem is perpendicular to the chord, sothe triangle formed is a right triangle. Also, theapothem is the distance from the center to thechord, so the length of one leg is 3 inches. Thehypotenuse of the triangle is the radius andmeasures 5 inches. By the Pythagorean Theorem,the length of the other leg is 4 inches, so a 3-4-5right triangle is formed. However, the apothemintersects the chord at its midpoint, so the lengthof the chord is 8 inches.
2. No. If the arcs belong to two circles with differentradii, then the arcs are not congruent.
Developing Skills3. 3 in. 4. 4.5 cm 5. 1.5 ft6. 12 mm 7. cm 8. 10 in.9. 24 ft 10. 14 cm 11. 12.4 mm
12. yd 13. radius � 11, diameter � 2214. 5 15. 25 16. 3017. 18. 12 19.20. OB � 30, DE � 6021. OB � 30, OC � 22.Applying Skills
23. We are given and .Since congruent circles are circles with congruentradii, ,and �COD � �AOB � �A�O�B� by SSS.Corresponding parts of congruent triangles arecongruent, so �COD � �AOB � �A�O�B�.
24. We are given and .Since congruent circles are circles with congruent radii, ,and �COD � �AOB � �A�O�B� by SSS. Cor-responding parts of congruent triangles are con-gruent, so �COD � �AOB � �A�O�B�. In acircle or in congruent circles, if central angles arecongruent, then their intercepted arcs arecongruent. Therefore, .CDC > ABC > ArBrC
OA > OB > OC > OD > OrAr > OrBr
CD > AB > ArBr(O > (Or
OA > OB > OC > OD > OrAr > OrBr
CD > AB > ArBr(O > (Or
LM15!3
5!512!5
2!5
!6
OrAr > OrBrOA > OB
mArBrC 5 60
Chapter 13. Geometry of the Circle
25. We are given that diameter intersects chord
at E and bisects at B. Then, .In a circle, congruent chords have congruentcentral angles, so �COE � �DOE. All radii of a circle are congruent, so , and
by the reflexive property of congruence. Therefore, �OEC � �OED bySAS. Corresponding parts of congruent trianglesare congruent, so �DEO � �CEO. If two linesintersect to form congruent adjacent angles, then they are perpendicular, so . Adiameter perpendicular to a chord bisects thechord, so E is the midpoint of and .
26. 5 cm27. In a circle, if the lengths of two chords are
unequal, then the shorter chord is farther from the center. is farther from the center than
, so AB < BC. is farther from the centerthan , so BC < AC. By the transitive property of inequality, AB < BC < AC. If the lengths oftwo sides of a triangle are unequal, then themeasures of the angles opposite these sides areunequal and the larger angle lies opposite thelonger side. Therefore, �B is the largest angle of�ABC.
13-3 Inscribed Angles and Their Measures(pages 555–558)
Writing About Mathematics
1. Draw diameter . Choose any point C on thecircle and draw and . Inscribed �ACBintercepts a semicircle, so it is a right angle.Triangle ACB is a right triangle with hypotenuse
and legs and .2. Yes. The measure of an inscribed angle of a circle
is equal to one-half the measure of itsintercepted arc. Therefore, m�ABC � 25 andm�PQR � 25, so �ABC � �PQR.
Developing Skills3. 44 4. 36 5. 85 6. 1007. 140 8. 24 9. 90 10. 120
11. 190 12. 25013. a. 160 14. a. 50 15. a. 34
b. 44 b. 100 b. 68c. 56 c. 50 c. 84d. 112 d. 80 d. 34e. 200 e. 160 e. 42
CBCAAOB
CBCAAOB
ACBCBC
AB
DE > CECD
AOB ' CD
OE > OEOC > OD
CBC > BDCCDCCD
AOB
334
16. a. 40 17. a. 54b. 80 b. 48c. 80 c. 50d. 40 d. 55e. 100 e. 50
18. a. 60 19. a. 120b. 90 b. 120c. 210 c. 120d. 45 d. 60e. 105 e. 60f. 30 f. 60
Applying Skills20. a. Since vertical angles are congruent,
�LPR � �SPM. If two inscribed angles of acircle intercept the same arc, then they arecongruent. Therefore, �RLP � �MSP and�LPR � �SPM by AA�.
b. 8 cm
21. If and , .
Therefore, . In a circle, congruent arcs have congruent chords, so and �ABCis isosceles.
22. a. In a parallelogram, opposite angle arecongruent. Therefore, �ABC � �ADC orm�ABC � m�ADC. The measure of aninscribed angle of a circle is equal to one-halfthe measure of its intercepted arc, so
m�ABC � and m�ADC � .
By the substitution postulate,
or .
b.c. By the reasoning in part a, m�BAD �
m�BCD � 90 and ABCD is equiangular. Anequiangular parallelogram is a rectangle, soABCD is a rectangle.
23. It is given that ,
so . Since
� 360, � 180. The measure of an inscribed angle of a circle is equal to one-half themeasure of its intercepted arc, so m�DEF � 90and �DEF is a right triangle.
24. It is given that . Since and are diameters, and �ABC and �BCD areinscribed in semicircles. Therefore, �ABC and�BCD are right angles and congruent. If twoinscribed angles of a circle intercept the samearc, then they are congruent, so �BAC � �CDB.Thus, �ABC � �DCB by ASA.
BODAOCAB > CD
mDEFCmDEFC 1 mFGDCmDEFC 5 mFGDC
mDEC 1 mEFC 5 mFGDC
mABCC 5 mADCC 5 180
mABCC 5 mADCC5 12mADCC12mABCC
12mADCC1
2mABCC
AB > CAABC > CAC
mCAC 5 130mBCC 5 130mABC 5 100
25. In a circle, congruent chords have congruent arcs.
Since , . Vertical angles are congruent, so �AEB � �DEC. If two inscribedangles of a circle intercept the same arc, thenthey are congruent, so �BAE � �CDE.Therefore, �ABC � �DCB by AAS.
26. Let ABCD be inscribed in circle O with .
Draw . If parallel lines are cut by a transversal, then alternate interior angles arecongruent, so �ACD � �BAC or m�ACD �m�BAC. The measure of an inscribed angle of acircle is equal to one-half the measure of its
intercepted arc, so m�ACD � and
m�BDC � . By the substitution postulate,
or . Therefore,
. In a circle, congruent arcs have congruent chords, so and ABCD is anisosceles trapezoid.
27. We are given that . Draw . If parallel lines are cut by a transversal, then alternateinterior angles are congruent, so �ACD � �BAC or m�ACD � m�BAC. Themeasure of an inscribed angle of a circle is equalto one-half the measure of its intercepted arc, so
m�ACD � and m�BDC � . By
the substitution postulate, or
. Therefore, .
28. Since , . The measure of an inscribed angle of a circle is equal to one-halfthe measure of its intercepted arc, so
m�ACD � and m�BDC � .Halves of equal quantities are equal, so m�ACD� m�BDC and �ACD � �BDC. If two linesare cut by a transversal so that the oppositeinterior angles formed are congruent, then they are parallel. Therefore, .
29. In a circle, all radii are congruent, so . Vertical angles are
congruent, so �AOB � �COD. Then �AOB � �COD by SAS. Corresponding partsof congruent triangles are congruent, so �BAC � �DCA. If two lines are cut by atransversal so that the opposite interior anglesformed are congruent, then they are parallel.Therefore, .AB y CD
OA > OB > OC > OD
AB y CD
12mBCC1
2mADC
mADC 5 mBCCADC > BCCADC > BCCmADC 5 mBCC12mADC 5 1
2mBCC12mBCC1
2mADC
ACAB y CD
AD > BCADC > BCC
mADC 5 mBCC12mADC 5 1
2mBCC12mBCC
12mADC
AC
AB y CD
AB > CDABC > CDC
335
30. In Exercise 26, we proved that a trapezoidinscribed in a circle must be isosceles. Since
, ABCD is a trapezoid and thus isosceles. It follows that . Two chords are congruent if and only if their arcs are
congruent. Thus, , so
. Similarly,and DBCE is an isosceles trapezoid. Again,two chords are congruent if and only if their arcs are congruent. Thus, .Since , DCFE is also an isosceles
trapezoid with . Thus,
, and
.
31. In a circle, two chords are congruent if and only
if their arcs are congruent. Since is not
congruent to , is not congruent to .Since opposite sides of ABCD are not congruent,ABCD is not a parallelogram.
13-4 Tangents and Secants (pages 564–567)Writing About Mathematics
1. No. Since l is tangent to circle O at A, .Since m is tangent to circle O at B, .Both and are segments of . If two lines are each perpendicular to the same line,then they are parallel. Therefore, l || m and thelines do not intersect.
2. A polygon inscribed in a circle intersects thecircle at each of its vertices, whereas a circleinscribed in a polygon intersects each of its sidesat exactly one point.
Developing Skills
3. AB � 10, BC � 15, CA � 15; since BC � CA,so �ABC is isosceles.
4. AB � 40, BC � 50, CA � 30; since 302 � 402 �502, �ABC is a right triangle.
5. a. 25 6. a. 7 7. a. 24b. 20 b. 14 b. 16c. 40 c. 10 c. 36
8. a. 7 9. a. 41 10. a. 12b. b. 40 b. 12c. c. 40 c. 5
d. 1311. a. 3 12. a. 4 d. 8
b. 6 b. 3 e. 11c. 8 c. 3 f. 12d. 10
!13!13
BC > CA
AOBOBOAm ' OB
l ' OA
CDABCDCABC
CAC > DBC > DFC > ECCDFC > DEC 1 EFC > FCC 1 EFC > ECC
ED > FC
EF y DCECC > DBC > CAC
CB y EDBAC > DAC 1 BAC > BDCCAC > CBC 1 DAC > BCC
DA > BCAB y DC
13. a. cm b. 1.73 cm14. a. 80 b. 70 c. 50 d. 160
e. 80 f. 10 g. 20 h. 100i. 110 j. 130 k. 360
Applying Skills
15. and are tangent to circle O at Q and R,respectively. Tangent segments drawn to a circlefrom an external point are congruent, so
. by the reflexive property ofcongruence. Radii of a circle are congruent, so
. Therefore, �OPQ � �OPR by SSS.Corresponding parts of congruent triangles arecongruent, so �OPQ � �OPR. By definition,
bisects �RPQ.
16. and are tangent to circle O at Q and R,respectively. Tangent segments drawn to a circlefrom an external point are congruent, so
. by the reflexive property ofcongruence. Radii of a circle are congruent, so
. Therefore, �OPQ � �OPR by SSS.Corresponding parts of congruent triangles arecongruent, so �QOP � �ROP. By definition,
bisects �QOR.
17. a. Tangent segments drawn to a circle from anexternal point are congruent, so .By the isosceles triangle theorem,�PQR � �PRQ.
b. From part a, and �PQR � �PRQ.by the reflexive property of
congruence, so �QPS � �RPS by SAS.Corresponding parts of congruent triangles are congruent, so and QS � RS.Also, �QSP � �RSP. If two lines intersect toform congruent adjacent angles, then they are perpendicular. Therefore, .
c. The length of the altitude to the hypotenuse ofa right triangle is the mean proportionalbetween the lengths of the segments of the hypotenuse. Thus, if x � OS, then or
. Solving for x gives x � 2 or 8.Since OS < SP, OS � 2 and SP � 8.
18. a. We are given perpendicular tangents. Since , m�ACB � 90.
If two tangents are drawn to a circle from anexternal point, then the line segment from thecenter of the circle to the external pointbisects the angle formed by the tangents.Therefore, m�ACO � 45. A tangent line is
AC ' BCAC and BC
2x2110x 5 16
x4 5 4
10 2 x
OP ' QR
QS > RS
SP > SPPQ > PR
PQ > PR
OPh
OQ > OR
PO > POPQ > PR
PRg
PQg
POh
OQ > OR
PO > POPQ > PR
PRg
PQg
!3
336
perpendicular to a radius at a point of tangency, so and m�OAC � 90.The sum of the measures of the angles in atriangle is 180, so m�AOC � 45. Since �OAC � �AOC, .
b. Since �AOC is a 45-45-degree right triangle,
OC � .
c. From part a, m�ACB � 90 and .Radii of a circle are congruent, so .By the same reasoning in part a, . Bythe transitive property of congruence,
. Since AOBC isequilateral and has a right angle, it is a square.
19. Isosceles �ABC with vertex A is circumscribedabout circle O, and D, E, and F are the points of tangency of , , and , respectively. By the isosceles triangle theorem, �DBE � �FCE. Iftwo tangents are drawn to a circle from anexternal point, then the line segment from thecenter of the circle to the external point bisectsthe angle formed by the tangents. Therefore,m�OBE � and m�OCE � ,so �OBE � �OCE. Since is tangent to circle O at E, , forming congruent rightangles �BEO and �CEO. by thereflexive property of congruence, so �BEO ��CEO by AAS. Corresponding parts ofcongruent triangles are congruent, so and E is the midpoint of .
20. a. If a line is tangent to a circle, then it isperpendicular to a radius at a point on the
circle. Therefore, and . If two lines are each perpendicular to the sameline, then they are parallel, so . We are given that OA < O�B. Since one pair ofopposite sides of OABO� is not bothcongruent and parallel, OABO� is not a
parallelogram. Therefore, cannot be
parallel to .
b. From part a, and .Therefore, �OAC and �O�BC are rightangles and congruent. By the reflexiveproperty of congruence, �C � �C. Therefore,�OAC � �O�BC by AA�.
c. AC � 8, AB � 4, OC � , O�C � ,
OO� � "17
3!172!17
ABg
' OrBABg
' OA
ABg
OOrg
OA y OrB
ABg
' OrBABg
' OA
BCBE > CE
OE > OEBC ' OE
BC
12m/FCE1
2m/DBE
ACBCAB
AO > OB > BC > AC
BC > BOOA > OB
AC > AO
"2OA
AC > AO
AC ' OA
21. a. We are given that OA � O�B, so .If a line is tangent to a circle, then it isperpendicular to a radius at that point on the
circle. Therefore, and ,forming congruent right angles �OAC and�O�BC. Vertical angles are congruent, so�OCA � �O�CB. �OCA � �O�CB by AAS.Corresponding parts of congruent triangles are congruent, so and OC � O�C.
b. From part a, �OCA � �O�CB.Corresponding parts of congruent trianglesare congruent, so and AC � BC.
Hands-On Activitya. Results will vary.b. (1) By definition, the center, P, is the
intersection of the bisectors of the angles of aregular polygon. If a point lies on thebisector of an angle, then it is equidistantfrom the sides of the angle. Thus, the center Pis equidistant from the sides of the regularpolygon. The distance from a point to a lineis equal to the length of the perpendicularfrom the foot. Thus, the apothems are allcongruent.
(2) Each circle constructed in part a used one ofthe apothems as the radius. Since theapothems are all congruent, every apothemis also a radius of the circle, so the foot ofeach apothem is on the circle.
(3) By construction, each side of the regularpolygon is perpendicular to one of theapothems at the foot of the apothem. If a lineis perpendicular to a radius at a point on acircle, then it is tangent to the circle. Sincethe foot of each apothem is on the circle andeach apothem is a radius of the circle, thesides of the regular polygon are tangent tothe circle.
c.
13-5 Angles Formed By Tangents, Chords,and Secants (pages 572–574)
Writing About Mathematics
1. Yes. Let be tangent to the circle O at A. Then
and m�AOB � 90. Let be a secant that intersects circle O at A and C. Thenm�OAB � m�OAC � m�CAB. Therefore,m�OAC � 90 and the radius is not
perpendicular to the secant .ACg
OA
ACg
ABg
' OA
ABg
r2 5 a2 1 A s2 B
2
AC > BC
OC > OrC
ABg
' OrBABg
' OA
OA > OrB
337
2. Yes. Vertical angles are congruent, and eachcentral angle is equal to the measure of itsintercepted arc. One-half of the sum of two equalintercepted arcs is the intercepted arc.
Developing Skills3. 35 4. 30 5. 306. 166 7. 170 8. 609. 50 10. 45 11. 50
12. 130 13. 130 14. 18015. 20 16. 100 17. 8018. 30 19. 60
20. ,21. 55 22. 140 23. 3524. 140 25. 70 26. 22027. a. 125 b. 55 c. 55
d. 90 e. 90 f. 125g. m�AED � . Therefore,
. A line through the center of acircle that is perpendicular to a chord bisectsthe chord, so bisects .
28. a. 150 b. 150 c. 30d. 30 e. 75 f. 105
Applying Skills
29. a. We are given that . If two parallel lines are cut by a transversal, then thealternate interior angles formed arecongruent. Therefore, �OCP � �OEA and�OPC � �OAE, and �OPC � �OAE byAA�.
b. , , ,
, , m�P � 3030. Vertical central angles are congruent and
intercept congruent arcs. Therefore, �FOE ��GOD and . We are given that
. By the partition postulate,
and
. Thus, by substitution,
. The measure
of an angle formed by a tangent and a secant is
equal to one-half the difference of the inter-
cepted arcs, so
and . By substitu-
tion,
, so �A � �C.Therefore, and �AOC is isosceles.OA > OC
12(mEFBC 2 mGBC) 5 m/C
m/A 5 12(mDGBC 2 mFBC) 5
m/C 5 12(mEFBC 2 mGBC)
m/A 5 12(mDGBC 2 mFBC)
5 mDGBC 2 mDGC 5 mDGCmEFCmFBC 5 mEFBC 25 mGBC
mDGBC 2 mDGCmEFBC 2 mEFC 5 mFBCmEFBC 5 mDGBC
mEFC 5 mDGC
mACC 5 105mBDC 5 75
mFBC 5 60mCFC 5 45mADC 5 75
AB y CPg
ACBD
BD ' AC
12(55 1 125) 5 90
mRSQC 5 225mRQC 5 135
31. Let D be a point on major . Since
m�AOB � 120, � 120 and � 240.The measure of an angle formed by two tangentsintersecting outside the circle is equal to one-halfthe difference of the measures of the intercepted
arcs, so m�P � . Tangent segments drawn to a circle from an externalpoint are congruent, so �PAB � �PBA. Sincethe sum of the measures of the angles of atriangle is 180, m�PAB � m�PBA � 60.Therefore, �ABP is equiangular. If a triangle isequiangular, then it is equilateral, so �ABP isequilateral.
32. We are given secant intersecting circle O at
A and B and chord . Assume that m�CBD �
. �ABD is an inscribed angle, so m�ABD
� . Since �ABD and �CBD form a linear
pair, m�CBD � m�ABD � 180. By the
substitution postulate,
or . However,
since .
Our assumption is false and m�CBD � .
13-6 Measures of Tangent Segments,Chords, and Secant Segments (pages 579–581)
Writing About Mathematics1. Yes. If two chords intersect, the product of the
measures of the segments of one chord is equalto the product of the segments of the other. Since AB � 24, and M is the midpoint of , AM �BM � 24. Because 12 12 � 144, any chord ofcircle O that intersects at M is separated by M into two segments such that the product of thelengths is also 144.
2. No. If two secant segments are drawn to a circlefrom an external point, then the product of thelengths of one secant segment and its externalsegment is equal to the product of the lengths ofthe other secant segment and its externalsegment. Since (AP)(BP) � (CP)(DP) and AP > CP, BP must be less than CP.
Developing Skills3. 8 4. 6 5. 106. 7 7. 8 8. 289. 4 10. 8 11. 4
AB
AB
12mBDC
mBDC 1 mADC 1 mABC 5 360mADC 2 360
mBDC 1 mBDC 1 mADC 5 360
12mBDC 1 12mADC 5 180
12mADC
12mBDC
BD
ABCg
12(240 2 120) 5 60
mADBCmABCADBC
338
12. 27 13. 20 and 6 14. 14 and 215. 16 16. 6 17. 418. 819. AC � 24, AB � 6, BC � 1820. AC � 20, AB � 5, BC � 1521. AC � 16, AB � 4, BC � 1222. AC � 25, AB � 9, BC � 1623. 20 24. 12 25. 30
26. ,
27. 8 28.29. AD � 3, AE � 6 30. AD � 7, AE � 1231.
13-7 Circles in the Coordinate Plane (pages 584–587)
Writing About Mathematics1. Yes. Consider a circle with center (h, k) and
radius r. Two points on the circle are (h � r, k)and (h � r, k). The chord connecting these pointsgoes through the center, so it is a diameter. Thesepoints have the same y-coordinate, so thisdiameter is horizontal. Two other points on thecircle are (h, k � r) and (h, k � r). The chordconnecting these points goes through the center,so it is a diameter. These points have the same y-coordinate, so this diameter is vertical.
2. Yes. Dividing each side by 3 gives .This is the equation of a circle centered at theorigin with a radius of 2.
Developing Skills
3.4.5.6.7.8.9.
10.11.12.13.14.15.16.17.18.19.20.21.22. x2 1 (y 1 2)2 5 36
(x 1 1)2 1 y2 5 1(x 1 4)2 1 (y 1 4)2 5 25(x 2 5)2 1 (y 1 5)2 5 9(x 2 4)2 1 (y 2 3)2 5 16(x 1 3)2 1 (y 2 2)2 5 4(x 2 10)2 1 (y 1 1)2 5 73(x 2 1)2 1 (y 2 6.5)2 5 76.25(x 1 6)2 1 (y 2 1)2 5 9x2 1 (y 2 12)2 5 25(x 1 1)2 1 (y 2 3)2 5 16(x 2 2)2 1 (y 2 9)2 5 16x2 1 y2 5 16x2 1 y2 5 4(x 1 3)2 1 (y 1 3)2 5 4(x 2 6)2 1 y2 5 81(x 2 4)2 1 (y 1 2)2 5 100(x 1 2)2 1 y2 5 36(x 2 1)2 1 (y 2 3)2 5 9x2 1 y2 5 9
x2 1 y2 5 4
152
8!2
AC 5 15$215AB 5 3$21
5
23.
24.
25.
26.
27.
xO1
y
1(0, 0)
xO 1
y
�1 ( , )�34
52
1
y
�1
( , 1)�32
O x
xO1
y
1
(�4, 4)
xO1
1y
(2, �5)
339
28.
29.30. Yes. The equation of the circle is x2 � y2 � 32.
Substituting (4, 4) into the equation gives 42 � 42
� 32. Therefore, (4, 4) is a point on the circle.31. Yes. It is equivalent to ,
which is the equation for a circle centered at (�2, 1) with a radius of 5.
Applying Skills32. a. y � �x � 2 b. x � 0
c. d. (0, 2)
e. The perpendicular bisector of a side isequidistant from the endpoints of each side, sothe circumcenter, which is the intersection ofthe three perpendicular bisectors, isequidistant from the vertices of �ABC.
f. Yes. Let (0, 2) be the center of the circle andA(2, 6) be a point on the circle. Then the radiusis and the equation of the circle is x2 � (y � 2)2 � 20. Substituting B(�4, 0) and into the equation gives
, so B is on the circle.Substituting C(4, 0) into the equation gives 42
� (0 � 2)2 � 16 � 4 � 20, so C is on the circle.33. a. Yes. The x-coordinate of the incenter is 3.
6(3) � 8y � 10
18 � 8y � 10
y � �1
The incenter is at (3, �1).b. The incenter is on each of the angle bisectors,
and a point on the bisector of an angle isequidistant from the sides of the angle.
c. Yes. The equation of is y � 2, so (3, 2) is on. Since the incenter of the triangle isPR
PR
5 16 1 4 5 20242 1 (0 2 2)2
!20
y 5 13x 1 2
(x 1 2)2 1 (y 2 1)2 5 25
(x 2 2)2 1 (y 2 3)2 5 169
xO 1
y
1(1, 1)
equidistant from sides of the circle, andalso intersect the circle at one point.
d.34. The equation of the perpendicular bisector of
is y � �2x � 4 and the equation of theperpendicular bisector of is y � �1.
�1 � �2x � 4
2x � �3
x �
The circumcenter of �ABC is (�1.5, �1), whichis also the center of the circle. The radius equalsthe distance between (�1.5, �1) and
A(�1, 3) � .The equation of the circle is
.
35. a. ft b. c.d. 6 ft e. 12 ft
36. a. (x � 13)2 � (y � 13)2 � 169b. (8, 1) and (8, 25); (18, 1) and (18, 25)c. (8, 13) or (18, 13)d. 1,800 ft and 800 ft
13-8 Tangents and Secants in theCoordinate Plane (pages 592–593)
Writing About Mathematics1. Yes. The equation of the circle is
(x � r)2 � (y � k)2 � r2. The equation of the y-axis is x � 0. Solving for y gives:
Thus, the y-axis intersects the circle in the point(0, k), so it is tangent to the circle.
2. No. Since the slope of is not the negative reciprocal of the slope of , it is not perpen-dicular to at the point of intersection.
Therefore, is not tangent at A.Developing Skills
3. a. (0, 6) b. Tangent4. a. (6, 8) and (8, 6) b. Secant5. a. (3, 4) and (4, 3) b. Secant6. a. (1, 3) and (�1, �3) b. Secant7. a. (0, �3) and (3, 0) b. Secant8. a. (2, 2) and (�2, �2) b. Secant9. a. (4, 3) and (�3, �4) b. Secant
10. a. (2, 4) and (4, 2) b. Secant11. a. (�3, 3) b. Tangent
ABg
CACA
ABg
y 5 k
(y 2 k)2 5 0
r2 1 (y 2 k)2 5 r2
(0 2 r)2 1 (y 2 k)2 5 r2
A s, s!33 B(s, s!3)12!3
(x 1 1.5)2 1 (y 1 1)2 5 16.25
"(21 1 1.5)2 1 (3 1 1)2 5 !16.25
232
BCAB
(x 2 3)2 1 (y 1 1)2 5 9QR
PQ
340
12. a. (5, 5) b. Tangent13. a. (2, 2) b. Tangent14. a. (0, �4) and (2, �2) b. Secant15. x � 3 16. y � �417. y � x � 4 18. y � 2x � 10Applying Skills19. a. y � 3x � 5 b. x � 0 c. P(0, �5)
d. PB � 0 and PE � 0, so(PA)(PB) � (PD)(PE) � 0.
20. a. y � x � 2 b. y � 10 c. P(8, 10)d. , , PD � 8
✔
21. a. y � �x � 6 b. y � x � 6c. (6, 0) d. PA � PB � 18
22. The line and the circle intersect at a single point:(18, �4).
23. a. Both points are solutions to the equation.
b. 424. a. The equation of is . The
equation of is x � 2. The equation of is
y � 10. intersects the circle at .
intersects the circle at (2, 7). intersectsthe circle at (�1, 10).
b. The sides of the triangle each intersect thecircle at exactly one point.
c. Yes. Each side of the triangle is tangent to thecircle.
Review Exercises (pages 598–599)1. a. 100 2. a. 100 3. a. 40
b. 50 b. 110 b. 130c. 80 c. 75 c. 65d. 90 d. 30 d. 50e. 90 e. 100 e. 95
4. PB � 6, BC � 18 5. 106. AE � 3, EC � 8 7. (1)8. (3) 9. 60
10. 15 cm 11. AB � 4, AC � 12
12. ,13. Let ABCD be inscribed in circle O with .
Draw . If parallel lines are cut by a transversal, then alternate interior angles are congruent, so �ACD � �BAC or m�ACD � m�BAC. The measure of aninscribed angle of a circle is equal to one-half the measure of its intercepted arc, so
m�ACD � and m�BDC � . By 12mBCC1
2mADC
ACAB y CD
mBCC 5 160mABC 5 mACC 5 100
ACBC
(2325, 51
5)AB
ACBCy 5 24
3x 1 23AB
(x 2 2)2 1 (y 2 3)2 5 25
64 5 64(2!2)(16!2) 5 82
(PA)(PB) 5? (PD)2
PB 5 16!2PA 5 2!2
the substitution postulate, or
. Therefore, . In a circle, congruent arcs have congruent chords, so
and ABCD is an isosceles trapezoid.
14. a. Since , alternate interior angles �B and �C are congruent so m�B � m�C.
m�B � and m�C � . By the
substitution postulate, . If two inscribed angles of a circle intercept the samearc, then they are congruent. Therefore,�A � �C and �D � �B. By the transitiveproperty of congruence, �A � �B and �C � �D. If two angles of a triangle arecongruent, then the sides opposite these angles are congruent, so and
. Therefore, �ABE and �CDE areisosceles.
b. From part a, . Therefore,
.
c. From part a, �A � �C and �D � �B.Therefore, �ABE � �CDE by AA�.
15. We are given that .
Therefore, � 90.The measure of an inscribed angle of a circle isequal to one-half the measure of its intercepted
arc, so m�A � .Therefore, m�A is a right angle. In a circle, if twoarcs are congruent, then their chords are congruent. Therefore,and ABCD is a rhombus. A rhombus with a rightangle is a square, so ABCD is a square.
16. Let ABC be a triangle with right angle �B and M the midpoint of . Then, . Let �ABC be inscribed in a circle. Since a right angle
is inscribed in a semicircle, is a semicircle and is a diameter. Since , M is thecenter of the circle. All radii of the same circleare congruent, so and MA �
MC � MB, and the midpoint of the hypotenuseis equidistant from the vertices of the triangle.
17. Since , PAB � PCD. If two secant segments are drawn to a circle from an externalpoint, then the product of the lengths of onesecant segment and its external segment is equalto the product of the lengths of the other secantsegment and its external segment. Therefore,
PAB > PCD
MA > MC > MB
MA > MCACABCC
MA > MCAC
AB > BC > CD > DA
12(mBCC 1 mCDC) 5 1
2(180) 5 90
mABC 5 mBCC 5 mCDC 5 mDACABC > BCC > CDC > DAC
ACC > BDCmACC 5 mBDC
CE > DEAE > BE
mACC 5 mBDC12mBDC1
2mACC
AB y CD
AD > BC
ADC > BCCmADC 5 mBCC12mADC 5 1
2mBCC
341
(PAB)(PA) � (PCD)(PD). By the divisionpostulate, PA � PD. Then, by the subtractionpostulate, PAB � PA � PCD � PD, so AB � CD and . Two chords areequidistant from the center of a circle if and onlyif the chords are congruent. Therefore, and
are equidistant from the center of the circle.
18. 12 cm
Exploration (pages 599–600)a. Place the point of your compass on a point P on a
line . With any convenient radius, draw arcs
that intersect at C and at D. With C and D as centers and the compass open to a radiusequal to CD, draw arcs that intersect at a point E.Connect C, D, and E to form �CDE. Congruent radii were used to draw , , and , so
and �CDE is equilateral.
b. The interior angles of a regular hexagon eachmeasure 120°. The angle bisectors of the anglesof the polygon intersect in the center of thepolygon and form congruent equilateraltriangles. Use the construction from part a toconstruct equilateral �ABP. Then, constructequilateral triangle �BCP. Next, constructequilateral triangle �CDP. Continue this processuntil you have constructed equilateral �FAP.Since the equilateral triangles all share two sides,they are all congruent. Thus, the sides of theresulting hexagon, ABCDEF, are all congruent.Each interior angle of the hexagon is the unionof two angles of two adjacent equilateraltriangles. Since the angles of an equilateral allmeasure 60°, each interior angle of the hexagonmeasures 60 � 60 � 120°. Therefore, the hexagonformed is a regular hexagon.
c. The diagonals of a square bisect each other, areperpendicular, and are congruent. Let O be thecenter of the circle. Draw a diameter of the circle. Construct the perpendicular bisector ofthe diameter. Let C and D be the points wherethe perpendicular bisector intersects the circle.Since and are radii, , and theperpendicular bisector passes through O. But
and are also radii. Therefore, and bisect each other, are congruent, and areperpendicular, so ACBD is a square.
CDABOCODDC
AO > OBOBAO
AOB
CD > DE > CECEDECD
PBh
PAh
ABg
CDAB
AB > CD
d. From the construction in part c, , , ,
and are all congruent since their central angles are all right angles. Thus, the arcs formedby bisecting these arcs are all congruent (sincehalves of congruent arcs are congruent). Twochords are congruent if and only if their arcs arecongruent. Therefore, the chords formed by thearcs are all congruent, so the sides of the polygonformed are all congruent. Draw segments fromthe vertices of the polygon to the center of thecircle. These segments are all congruent becausethey are radii of the same circle. Thus, congruentisosceles triangles are formed by SSS. The baseangles of these triangles are congruent, so itfollows that the interior angles of the octagon are all congruent. Therefore, the octagon formed is a regular octagon.
f. Construct six congruent and adjacent equilateraltriangles with the center of the circle as a vertexand using two radii as sides.
Cumulative Review (pages 600–603)Part I
1. 3 2. 1 3. 44. 3 5. 1 6. 37. 3 8. 2 9. 1
10. 3Part II11. a. (2, 2) b. (�2, 0)
c. Yes. The slope of is and the slope of
is .d. Yes. By the reflexive property of congruence,
�B � �B. If two parallel lines are cut by atransversal, then the corresponding angles arecongruent. Since , �CAD � �EDBand �ABC � �DBE by AA�.
12. Statements Reasons1. ABCD with 1. Given.
, and and bi-
sect each other.2. ABCD is a 2. If the diagonals of a
parallelogram. quadrilateral bisect eachother, then the quadri-lateral is a parallelogram.
3. ABCD is a 3. If the diagonals of arectangle. parallelogram are con-
gruent, then the parallel-ogram is a rectangle.
BDACAC > BD
DE y AC
12
AC12DE
DACBDCCBCACC
342
Part III13. Statements Reasons
1. and 1. Given.
2. �AEF and �CEF 2. Definition of are right angles. perpendicular lines.
3. �AEF � �CEF 3. Right angles arecongruent.
4. 4. Reflexive propertyof congruence.
5. 5. Given.6. �AEF � �CEF 6. SAS.7. 7. Corresponding
parts of congruenttriangles arecongruent.
14.
x � 15 � x � 3
x cannot be 3 since 3 � 7 � �4.The lengths of the sides of the triangle are 15, 8,and 17.
Part IV15. a. (�5, 2); ry=x (5, 2) � (2, 5), R90 (2, 5) � (�5, 2)
b. ry-axisc. No. Rotation is a direct isometry and line
reflection is an opposite isometry.
16. a. We are given that AD : DC � 1 : 3 and BE :
EC � 1 : 3. Let AD � x, so DC � 3x. Let BE
� y, so EC � 3y. AC � AD � DC � 4x and
BC � BE � EC � 4y. and
. Therefore, AC : DC � BC : EC.
b. From part a, AC : DC � BC : EC. By thereflexive property of congruence, �C � �C.Therefore, �ABC � �DEC by SAS�.
BCEC 5
4y3y 5 4
3
ACDC 5 4x
3x 5 43
(x 2 15)(x 2 3) 5 0
x2 2 18x 1 45 5 0
x2 1 x2 2 14x 1 49 5 x2 1 4x 1 4
x2 1 (x 2 7)2 5 (x 1 2)2
FA > FC
EA > EC
EF > EF
EFg
' CDg
EFg
' ABg
14-1 Constructing Parallel Lines (pages 607–609)
Writing About Mathematics1. The distance between two parallel lines is
defined as the length of the perpendicular fromany point on one line to the other line. Therefore,since CE is the perpendicular with length 2PQ,
every point on is at a distance 2PQ from .2. If two coplanar lines are each perpendicular to
the same line, then they are parallel. Thus,
and . The distance between two parallel lines is defined as the length of theperpendicular from any point on one line to theother line. Thus, the distance from any point on
to and is .Developing Skills
3. a. Use Construction 7.b. Mark any point Q on l. Use Construction 6 to
construct the perpendicular to m through Q.Let R be the intersection of this perpendicularand line m. Then use Construction 3 to construct the perpendicular bisector of .
c. Construct a line parallel to l (and above l) that
is the same distance away as m. Extend to a point S such that QR � RS. Then useConstruction 7 to construct a line parallel to lthrough S.
d. Yes. If two of three lines in the same plane areeach parallel to the third line, then they areparallel to each other.
4. a. Use Construction 1 to construct a segment congruent to the segment with length b. Use
Construction 5 to construct and perpendicular to with C and D on thesame side of . Set the compass radius to a.
With A as the center, mark off point X on .With B as the center and using the same
radius, mark off point Y on . Draw .ABYX is a rectangle.
b. Use Construction 1 to construct a segmentcongruent to the segment with length a.
Use Construction 5 to construct and perpendicular to with C and D on thesame side of . Set the compass radius to a.
With A as the center, mark off point X on .With B as the center and using the same
ACh
ABAB
BDg
ACg
AB
XYBDh
ACh
ABAB
BDg
ACg
AB
QRh
QR
12GHCD
gABg
GHg
GHg
y CDg
GHg
y ABg
ABg
CDg
343
radius, mark off point Y on . Draw .ABYX is a square.
c. Use Construction 1 to construct a segment congruent to the segment with length a. UseConstruction 2 to construct an angle congruent
to �A on . On the side of this angle not containing point B, use Construction 1 toconstruct a segment congruent to thesegment with length b. Use Construction 7 toconstruct a line parallel to through pointC. On this line, use Construction 1 to construct
congruent to the segment with length asuch that is on the same side of . Draw
. ABDC is a parallelogram.
d. Repeat part c, but construct to be congruent to the segment with length a.ABDC is a rhombus.
5. a. Construct the perpendicular bisector of atC. Construct the perpendicular bisector of at D and of at E. Then D, C, and E divide
into four congruent parts.b. Set the compass radius to AB. Draw a circle.c. Set the compass radius to AC. Draw a circle.
6. a. Use Construction 7.b. Use Construction 3 to find the midpoint M of
. Draw .Then is the median to .c. Use Construction 3 to find the midpoint N of
. Draw . Then is the median to .d. Yes. Any two medians of a triangle intersect in
the same point, and two points determine aline. Thus, the line drawn from C to the median of is the same as that drawn fromC to P.
7. a. Use Construction 3.b. Use Construction 3.c. Use Construction 3 to find the midpoint M of
.d. P and M. Any two perpendicular bisectors of
the sides of a triangle intersect in the samepoint. Thus, the line containing P and M is the perpendicular bisector of .
8. a. Use Construction 6 to construct a line perpendicular to through B intersecting
at D. is the altitude to .b. Use Construction 6 to construct a line
perpendicular to through A intersectingat E. is the altitude to .BCAEBC
BC
ACBDACAC
AB
AB
AB
BCANANBC
ACBMBMAC
ABCB
ACAB
AC
BDACCD
CD
AB
AC
ABh
AB
XYBDh
Chapter 14. Locus and Construction
c. P and C. Any two altitudes of the sides of atriangle intersect in the same point, and two
points determine a line. Thus, contains the altitude to .
9. a. Use Construction 4 to construct the angle
bisector, , of �CBA. Let D be the
intersection of and . Draw . Then is the angle bisector from B in �ABC.
b. Use Construction 4 to construct the angle
bisector, , of �CAB. Let E be the
intersection of and . Draw . Then
is the angle bisector from A in �ABC.c. P and C. Any two angle bisectors of a triangle
intersect in the same point, and two pointsdetermine a line. Thus, the line determined byP and C contains the angle bisector from C in�ABC.
10. b. Draw any point D on . Set the compass radius to AD. With D as the center, mark off a
point X on . With X as the center and using the same compass radius, mark off point
E on . Then DE � 2AD.c. Use Construction 7 to construct a line parallel
to through D.d. If a line is parallel to one side of a triangle and
intersects the other two sides, then the pointsof intersection divide the sides proportionally.
Therefore, since is parallel to and intersects the sides of �ABE,
.
11. a. Answers will vary.
b. Draw any point N on . Set the compass radius to PN. With L as the center, mark off
point L2 on . With L2 as the center and using the same compass radius, mark off point
L3 on . Repeat this procedure until you have drawn L8. The point Q is L8, and thepoint S is L5.
c. Use Construction 7 to draw a line parallel tothrough S.
d. If a line is parallel to one side of a triangle andintersects the other two sides, then the pointsof intersection divide the sides proportionally.
Thus, since is parallel to and intersects the sides of �PQR, PS : SQ � PT : TR � 3 : 5.If two line segments are divided propor-tionally, then the ratio of the length of a partof one segment to the length of the whole is
QRSTg
QR
PLh
PLh
PLh
AF : FB 5 AD : DE 5 1 : 2
EBDFg
EB
ADh
ADh
ACh
AE
AEBCAYh
AYh
BDBDACBX
hBXh
ABPCg
344
equal to the ratio of the corresponding lengthsof the other segment. Thus, PS : PQ � PT : PR� 3 : 8. Since �P � �P, �PST ~ �PQR bySAS~ with a constant of proportionality of .
14-2 The Meaning of Locus (pages 612–613)Writing About Mathematics
1. No. The locus of all points equidistant from theendpoints of a segment is the line that is theperpendicular bisector of the segment. Differentpoints on the line will be different distances fromthe endpoints.
2. The perpendicular to the line or ray through thepoint
Developing Skills3. The circle with the point as the center and a
radius of 10 centimeters4. The perpendicular bisector of 5. The line that is parallel to both lines and midway
between them
6. A pair of lines parallel to each 4 inches away
from such that is midway between them7. The line that is parallel to both lines and midway
between them8. The line that is parallel to both sides and midway
between them9. The line containing the diagonal between the two
other vertices10. The point that is the intersection of both
diagonals11. The circle with the same center and a radius of 1
inch12. The circle with the same center and a radius of 5
inches13. A pair of circles, both with the same center as the
given circle and one with a radius of 1 inch andthe other with a radius of 5 inches
14. a. The circle with the same center and a radius of(r � m)
b. The circle with the same center and a radiusof (r � m)
c. A pair of circles, both with the same center asthe given circle and one with a radius of (r � m) and the other with a radius of (r � m)
15. The circle with the same center as the givencircles and a radius of 14 centimeters
16. The perpendicular bisector of . (Note: Thetriangles are drawn on both sides of .)
17. A pair of lines parallel to each 3 feet away
from such that is midway between themABg
ABg
ABg
ABAB
ABg
ABg
ABg
AB
38
Applying Skills18. The circle centered at the base of the hour hand
with a radius equal to the length of the hour hand19. The line parallel to the track at a distance from
the track equal to the radius of the train wheel20. The line parallel to the curbs that is equidistant
from the curbs21. The circle with the stake as the center and a
radius of 6 meters22. The line that bisects the angle formed by the two
roads23. The line that is the perpendicular bisector of the
two floats on the lake24. The line parallel to the horizontal line at a
distance from the horizontal line equal to theradius of the dime
25. The circle with the same center as the circulartrack and a radius of 32 feet
14-3 Five Fundamental Loci (pages 615–616)Writing About Mathematics
1. Yes. The locus of points equidistant to and
is the line that is parallel to both and midway between them. This line is the perpendicularbisector of , and so every point on this line isequidistant to P and S.
2. Two lines that are equidistant from twointersecting lines are the angle bisectors of theangles formed by the two intersecting lines. Letone of the angles formed by the intersecting lineshave measure 2x. Then both of the adjacentangles have measure (180 � 2x). Thus, themeasures of the angles formed by the anglebisectors are x and (90 � x), and so the measureof the angle formed by the two angle bisectors isx � (90 � x), or 90°.
Developing Skills3. The perpendicular bisector of the segment
formed by the two points4. The circle with a radius of 6 inches and the center
equal to the midpoint of the segment5. The perpendicular bisector of the base of the
isosceles triangle6. The line containing the angle bisector of the
vertex angle of the isosceles triangle7. The two distinct perpendicular bisectors of the
sides of the square8. The line parallel to both bases and midway
between them9. The circle with a radius of 4 inches and the center
equal to the midpoint of the base
SP
RSg
PQg
345
10. The locus is the union of the two segments thatare both congruent to and parallel to the altitudeand 6 centimeters away from the altitude, and thetwo semicircles of radius 6 centimeters centeredat the endpoints of the altitude.
11. a. The locus is the line parallel to both lines andmidway between them.
b. Draw point P on any of the given lines. Thelocus is the circle centered at P with a radiusof 3 centimeters.
c. 212. a. The locus is the perpendicular bisector of .
b. The locus is a circle centered at M with aradius of .
c. 2d. A square. The diagonals bisect each other
since they are both radii of the same circle.Therefore, the quadrilateral formed is a paral-lelogram. The diagonals are congruent sincethey are both diameters of the same circle.Therefore, the parallelogram is a rhombus. Thediagonals are perpendicular since one is con-tained in the perpendicular bisector of theother. Therefore, the rhombus is a square.
14-4 Points at a Fixed Distance inCoordinate Geometry (pages 618–619)
Writing About Mathematics1. Yes. It intersects the circle in two points.2. Yes. The locus of points of two intersecting lines
is a pair of lines that bisect the angles formed bythe intersecting lines. A point is on the anglebisector of an angle if it is equidistant from the
sides of the angle. Since and are each tangent to the circle, they each intersect the circlein exactly one point. At these two points, the radiidrawn are perpendicular to the lines and meet in
the point O. Thus, O is equidistant from and
, and so O is on the locus.PBg
PAg
PBg
PAg
12AB
AB
Developing Skills3. 4.5.6.7.8.9. x � 12 and x � 2 10. x � 0 and x � 2
11. y � 6 and y � �2 12. y � 13 and y � 113. (�4, �3) and (3, 4) 14. (12, 5) and (�5, �12)15. (�8, �5) and (6, 9) 16. (�2, 0) and (0, 2)17. a. y � 1 and y � 7
b. x � 1 and x � 3c. (1, 7), (3, 7), (1, 1), (3, 1)
18. a.b. y � �4 and y � 2c. (5, 2) and (�1, 2)
19. a.b. y � 3 and y � �3c. (�4, 3), (4, 3), (�4, �3), (4, �3)
20. a.b. x � 8 and x � �8c. (8, 6), (8, �6), (�8, 6), (�8, �6)
21. a.
b. x � 2 and x � �2c. (2, 0), , , (�2, 0)
22. a.b. x � �1 and x � 4c. (�1, 10), (�1, 0), (4, 5)
14-5 Equidistant Lines in CoordinateGeometry (pages 622–624)
Writing About Mathematics1. Yes. The locus of points equidistant from two
intersecting lines is a pair of lines that are theangle bisectors of the angles formed by theintersecting lines. The angle bisectors of theangles formed by y � x and y � �x are the x- and y-axes.
2. Yes. The locus of points equidistant from twoparallel lines is the line parallel to the lines andmidway between them. The slope of this line is x,and the y-intercept is the average of the y-intercepts of the given lines, that is,b � � 6. Therefore, the locus is y � x � 6.
Developing Skills3. x � 5 4. y � �15. y � x 6.
7. 8.9. x � 3 10. y � �3
11. y � x � 6 12. y � �x � 2
y 5 213x 2 3y 5 21
2x 2 12
y 5 12x 1 1
2 1 102
(x 1 1)2 1 (y 2 5)2 5 25(22, 24!2)(22, 4!2)
(x 1 4)2 1 y2 5 36(x 1 4)2 1 y2 5 4
x2 1 y2 5 100
x2 1 y2 5 25
(x 2 2)2 1 (y 2 2)2 5 9
(x 1 3)2 1 (y 2 5)2 5 18(x 2 3)2 1 (y 1 1)2 5 10(x 2 1)2 1 (y 2 1)2 5 49x2 1 (y 1 2)2 5 9
(x 1 1)2 1 y2 5 1x2 1 y2 5 16
346
13. y � 2x � 3 14. y � �2x � 815. (1, 6) and (1, 0) 16. (4, �1) and (�4, �1)Applying Skills
17. a.
b.
✔
c. Distance from (3, 1) to (�2, 6) �Distance from (5, 5) to (�2, 6) �
18. a. y � x � 1b.
✔
c. y � �x � 5 d. A(1, 4)e. B(5, 0) f.
19. The locus of points equidistant from twointersecting lines is a pair of lines that bisect theangles formed by the intersecting lines. Thus, thelocus of points equidistant from y � x and y � �x are the x- and y-axes. Translationspreserve angle measure. Thus, if the x- and y-axesare the angle bisectors of y � x and y � �x, thentheir images will be the angle bisectors of theimages of y � x and y � �x. Under , the images of y � x and y � �x are y � x � 1 and y � �x � 1, respectively. The images of the anglebisectors are the y-axis and the line y � 1 underthe same translation. Therefore, the y-axis andthe line y � 1 are the angle bisectors of y � x � 1and y � �x � 1, and so they are the locus ofpoints equidistant from these two lines.
20. The locus of points equidistant from twointersecting lines is a pair of lines that bisect theangles formed by the intersecting lines. Thus, thelocus of points equidistant from y � 3x and y � �3x are the x- and y-axes. Translationspreserve angle measure. Thus, if the x- and y-axisare the angle bisectors of y � 3x and y � �3x,then their images will be the angle bisectors ofthe images of y � 3x and y � �3x. Under ,the images of y � 3x and y � �3x are y � 3x � 2and y � �3x – 2, respectively. The images of the angle bisectors are the y-axis and the liney � �2 under the same translation. Therefore,the y-axis and the line y � �2 are the anglebisectors of y � 3x � 2 and y � �3x – 2, and sothey are the locus of points equidistant fromthese two lines.
T0,22
T0,1
PA 5 PB 5 2!2
2 5 2
2 5? 3 2 1
y 5 x 2 1
!50!50
6 5 66 5? 1 1 5
6 5? 212(22) 1 5
y 5 212x 1 6
y 5 212x 1 5
21. a. (0, b) b. (0, c)c.d. Since the first line is parallel to the second
line, alternate interior angles are congruent.Thus, . Since verticalangles are congruent, �BMA � �B�MA�.Since M is the midpoint of , .Therefore, by ASA, �ABM � �A�B�M.
e. Since corresponding parts of congruent triangles are congruent, . The distance between two lines is the length of aperpendicular segment joining both lines.Since is perpendicular to the given lines,BB� is the distance between the given lines.Thus, since M is midway between B� and B, itlies on the line equidistant to the given lines.
14-6 Points Equidistant from a Point and aLine (pages 629–630)
Writing About Mathematics1. No. The solutions to the equation
�x2 � 2x � 8 � 0 are x � �2 and x � 4. Thegraph of y � �x2 � 2x � 8 intersects the x-axis at (�2, 0) and (4, 0). When the graph intersectsthe y-axis, the x-coordinate is 0.
2. Yes. The tangent of a parabola of the formis a horizontal line passing
through its turning point. Since (1, 0) is theturning point and is on the x-axis, the x-axis is thetangent of the parabola.
Developing Skills3. (3, �8); x � 3 4. (1, 2); x � 15. (�2, �5); x � �2 6. (1, 6); x � 1
7. (�4, 20); x � �4 8.
9. (3, 5), (0, 2) 10. (�2, 3), (1, 0)
x
O�2
2
y
xO�1 1
y
A 52, 217
4 B ; x 5 52
y 5 ax2 1 bx 1 c
BBr
BM > BrM
AM > ArMAAr
/BrArM 5 /BAM
M 5 A 0 1 02 , b 1 c
2 B 5 A0, b 1 c2 B
347
11. (4, 3), (1, 0) 12. (�4, 5), (1, 0)
13. (5, 7), (1, �1) 14. (2, 2), (�2, �6)
15. (2, 3), (6, �5) 16. (2, 0), (�2, �8)
17. 18.
19. 20.
Hands-On Activity
1.2.
Review Exercises (pages 631–632)1. a. Answers will vary. Example: Draw any
segment and then construct its perpendicularbisector. The angles formed are right angles.
b. Use Construction 4 to bisect the angle formedin part a.
c. Use Construction 1 to construct congruent to the segment with length a. Use Construc-tion 2 to construct �ABX congruent to the
AB
y 5 ax2 2 2hax 1 ah2 1 ky 5 ax2 2 6ax 1 9a 1 5
y 5 2 112x2 1 12x 2 34y 5 x2
8
y 5 2x2y 5 x2
xO1�1
y
xO1
�1
y
xO2
�2
y
xO1
�1
y
xO�1
�1
y
xO�1
1
y
45° angle from part b. Set the compass radiusto b. With B as the center, mark off a point C
along . Use Construction 7 to construct a line parallel to through A and a lineparallel to through C. Let D be the intersection of these two lines. Then ABCD isa parallelogram with m�B � 45.
2. Draw a ray and a segment . Use Construction 1 to construct a segment congruent
to on . Call the other endpoint Q1.
Repeat this construction on , , ,
and . Let the final endpoint be called Q.Rename Q2 as S. Then PS : SQ � 2 : 3.
3. The perpendicular bisector of the segmentformed by the two points
4. The two points that are on the perpendicularbisector of and centimeters from thesegment
5. The two points that are on the perpendicularbisector of the segment and 2 centimeters fromthe segment
6. The two points that are the intersection of theline parallel to the two lines and 2.5 inches fromeach of them and the circle centered at the givenpoint with radius 4 inches
7.8. (4, 4) and (0, 4)9. x � 2
10. y � 511. x-axis and y-axis12.13. (4, 3) and (3, 4)14. a–b.
c. (3, �4) and (�1, 4)
xO�1
y
�1
y 5 212x
(x 1 1)2 1 (y 2 2)2 5 9
!7AB
Q4X›
Q3X›
Q2X›
Q1X›
PXh
YZ
YZPXh
ABBC
BXh
348
15. (4, 3) and (�3, �4)
16. (2, �3) and (2, 1)
17. (�2, �1) and (1, 2)
18. (4, 2) and (1, 5)
19. 1 point 20. (3, 1)
Exploration (page 632)a. An ellipse is the locus of points such that the sum
of the distances from two fixed points is aconstant, k. Let k represent the length of thestring. Let P be any point on the curve that isdrawn. Then, , the length of thestring. Since P is an arbitrary point on the curve,all of the points on the curve are such that thesum of the distances from and is k.Therefore, the curve drawn is an ellipse.
F2F1
F1P 1 PF2 5 k
xO1
1
y
xO 1
1
y
xO1
y
�1
xO1
y
�1
b. The ellipse becomes more circular.c. The distance between each point and the curve
can increase infinitely while keeping thedifference between the two distances constant.
Cumulative Review (pages 633–635)Part I
1. 2 2. 1 3. 3 4. 15. 1 6. 3 7. 4 8. 39. 2 10. 3
Part II11. In �ABC, �A � �A by the reflexive property of
congruence. If two parallel lines are cut by a trans-versal, then the corresponding angles are congru-ent. Since , �ADE � �ABC. Therefore,�ABC � �ADE and the sides are in proportion.
EC � AC � AE � 20 � 8 � 1212. The circumference of the base � 75. Then the
radius is:
Part III13. a–b.
c. (3, 8) and (�1, 4)
y � 3 � 5 � 8y � �1 � 5 � 4
x 5 3, 21(x 2 3)(x 1 1) 5 0
x2 2 2x 2 3 5 0x2 2 x 1 2 5 x 1 5
xO1
1
y
< 1,800 cubic feet
55,625
p
5 4p A 5,6254p2 B
5 13p A 75
2p B 212
Volume 5 13pr2h
r 5 752p
2pr 5 75
AE 5 8
615 5 AE
20
ADAB 5 AE
AC
DE y BC
349
14. a.
y � 4 � 2 � 2
The line intersects the circle only at (2, 2) so istangent to the circle.
b. The slope of the tangent line is �1. The centerof the circle is (0, 0), so the slope of the radius to the point of tangency is . Since these slopes are negative reciprocals, the tangentand the radius to the point of tangency areperpendicular.
Part IV15. a. Radii of congruent circles are congruent.
If two points are each equidistant from theendpoints of a line segments, then the pointsdetermine the perpendicular bisector of theline segment.
b. Radii of congruent circles are congruent.�ABC � �DEF by SSS.Corresponding parts of congruent trianglesare congruent.
/ABC > /DEF/ACB > /DFE/FDE > /CAB
BC > EFAB > DEAC > DF
/AME > /EMB > /AMD > /BMD/EAD > /EBD/AEB > /ADB
/AEM > /BEM > /ADM > /BDM/EAM > /DAM > /EBM > /DBM
AE > AD > BE > BDDM > MEAM > MB
AE > AD > BE > BD
22 5 1
x 5 2
(x 2 2)2 5 0
x2 2 4x 1 4 5 0
2x2 2 8x 1 8 5 0
x2 1 16 2 8x 1 x2 5 8
x2 1 (4 2 x)2 5 8
y 5 4 2 xx 1 y 5 4
16. a. Statements Reasons1. ABCD inscribed in 1. Given.
circle O with and a
diameter2. m�D � 90 and 2. An angle inscribed
m�B � 90 in a semicircle is aright angle.
3. �A and �B are 3. If two parallel linessupplements. are cut by a trans-�C are �D are versal, then the in-supplements. terior angles on the
same side of thetransversal aresupplementary.
4. m�A � 90 and 4. Definition of sup-m�C � 90 plementary angles.
5. ABCD is a 5. If a quadrilateral is rectangle. equiangular, then it
is a rectangle.
b. Statements Reasons1. Diagonals and 1. Given.
of ABCDintersect at E.�ABE � �CDE
2. �EAB � �ECD 2. Corresponding an-gles of similar trian-gles are congruent.
3. 3. If two lines are cut by a transversal sothat the alternate in-terior angles formedare congruent, thenthe lines areparallel.
4. �ABE is not con- 4. Given.gruent to �CDE.
5. is not con- 5. The ratios of cor-gruent to . responding sides of
similar triangles arein proportion.
CEAE
AB y CD
BDAC
ACAB y CD
350
6. E is not the mid- 6. Definition of point of . midpoint.
7. does not 7. Definition of bisect . bisector.
8. ABCD is not a 8. The diagonals of a parallelogram. parallelogram
bisect each other.9. is not parallel 9. Definition of
to . parallelogram.10. ABCD is a 10. Definition of
trapezoid. trapezoid.
c. Statements Reasons1. �ABC is equi- 1. Given.
lateral. D, E, and Fare the midpoints of , , and
, respectively.2. and 2. A line segment
joining the mid-points of two sidesof a triangle is parallel to the thirdside.
3. DECF is a 3. Definition of parallelogram. parallelogram.
4. 4. Definition of equilateral triangle.
5. BD � and 5. Definition of
BE � midpoint.
6. 6. Halves of congru-ent segments arecongruent.
7. ABCD is a 7. Definition of rhombus. rhombus.
BD > BE
12BC
12AB
AB > BC
DF y ECDE y FCAC
BCAB
ADBC
ACBD
AC