unit 7: modeling with geometric relationships mathematics 3 ms. c. taylor

108
Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Upload: esther-golden

Post on 27-Dec-2015

222 views

Category:

Documents


0 download

TRANSCRIPT

Page 1: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Unit 7: Modeling with Geometric RelationshipsMathematics 3

Ms. C. Taylor

Page 2: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Warm-UpA reporter wants to know the percentage of

voters in the state who support building a new highway. What is the reporter’s population?

A) the number of people who live in the state

B) the people who were interviewed in the state

C) all voters over 25 years old in the stateD) all eligible voters in the state

Page 3: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Distance around an Arc

An arc of a circle is a segment of the circumference of the circle.

Arc length of a circle in radians:

Arc length of a circle in degrees:

Page 4: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Constructions

https://www.youtube.com/watch?v=fVDr08YbQww

https://www.youtube.com/watch?v=il0EJrY64qE

Page 5: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Warm-Up

Find the arc length of a circle if the diameter is 18 and the angle is .

Page 6: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Theorems Galore!If two lines intersect to form a

linear pair of congruent angles, then the lines are perpendicular.

Proof Hints:• Write an equation based on

the angles forming a linear pair.

• Do some substitution and solve for one of the angles. It should be 90°.

Page 7: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Theorems Galore!If two lines are perpendicular, then they intersect to form four right angles.

Proof Hints:• Use definition of

perpendicular lines to find one right angle.

• Use vertical and linear pairs of angles to find three more right angles.

Page 8: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Theorems Galore!

If two sides of two adjacent acute angles are perpendicular, then the angles are complementary.

Proof Hints:• Use definition of perpendicular to

get the measure of ABC.• Use Angle Addition Postulate,

Substitution, and Definition of complementary angles to finish the proof.

Page 9: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Perpendicular Transversal Theorem

If a transversal is perpendicular to one of two parallel lines, then it is perpendicular to the other.

Theorems Galore!

j Proof Hints:• Use definition of

perpendicular lines to find one right angle.

• Use Corresponding Angles Postulate to find a right angle on the other line.

Page 10: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Theorems Galore!Lines Perpendicular to a

Transversal TheoremIn a plane, if two lines are perpendicular to the same line, then they are parallel to each other.

Proof Hints:• Use definition of

perpendicular lines to find a right angle on each parallel line.

• Use Converse of Corresponding Angles Postulate to prove the lines are parallel.

Page 11: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Proving Vertical Angle TheoremVertical Angles Theorem: Vertical angles are congruent

Page 12: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Proving Vertical Angle Theorem

PROVE 5 7

GIVEN 5 and 6 are a linear pair,6 and 7 are a linear pair

1

2

3

Statements Reasons

5 and 6 are a linear pair, Given6 and 7 are a linear pair

5 and 6 are supplementary, Linear Pair Postulate6 and 7 are supplementary

5 7 Congruent Supplements Theorem

Page 13: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

The Third Angles Theorem below follows from the Triangle Sum Theorem.

Third Angles Theorem

If two angles of one triangle are congruent to two angles of another triangle, then the third angles are also congruent.

If A D and B E, then C F.

Page 14: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

PROPERTIES OF PARALLEL LINESPOSTULATE 15 Corresponding Angles Postulate

1

2

1 2

If two parallel lines are cut by a transversal, then the pairs of corresponding angles are congruent.

Page 15: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

PROPERTIES OF PARALLEL LINESTHEOREM 3.4 Alternate Interior Angles

3

4

3 4

If two parallel lines are cut by a transversal, then the pairs of alternate interior angles are congruent.

Page 16: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

PROPERTIES OF PARALLEL LINESTHEOREM 3.5 Consecutive Interior Angles

m 5 + m 6 = 180°

5

6

If two parallel lines are cut by a transversal, then the pairs of consecutive interior angles are supplementary.

Page 17: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

PROPERTIES OF PARALLEL LINESTHEOREM 3.6 Alternate Exterior Angles

7

8

7 8

If two parallel lines are cut by a transversal, then the pairs of alternate exterior angles are congruent.

Page 18: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

PROPERTIES OF PARALLEL LINESTHEOREM 3.7 Perpendicular Transversal

j k

If a transversal is perpendicular to one of two parallellines, then it is perpendicular to the other.

Page 19: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Proving the Alternate Interior Angles TheoremProve the Alternate Interior Angles Theorem.

SOLUTION

GIVEN p || q

p || q Given

Statements Reasons

1

2

3

4

PROVE 1 2

1 3 Corresponding Angles Postulate

3 2 Vertical Angles Theorem

1 2 Transitive property of Congruence

Page 20: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Using Properties of Parallel Lines

SOLUTION

Given that m 5 = 65°, find each measure. Tellwhich postulate or theoremyou use.

Linear Pair Postulatem 7 = 180° – m 5 = 115°

Alternate Exterior Angles Theoremm 9 = m 7 = 115°

Corresponding Angles Postulatem 8 = m 5 = 65°

m 6 = m 5 = 65° Vertical Angles Theorem

Page 21: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Using Properties of Parallel Lines

Use properties ofparallel lines to findthe value of x.

SOLUTION

Corresponding Angles Postulatem 4 = 125°

Linear Pair Postulatem 4 + (x + 15)° = 180°

Substitute.125° + (x + 15)° = 180°

Subtract.x = 40°

Page 22: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Using the Third Angles TheoremFind the value of x.

SOLUTION

In the diagram, N R and L S.

From the Third Angles Theorem, you know that M T.So, m M = m T. From the Triangle Sum Theorem, m M =

180˚– 55˚ – 65˚ = 60˚.

m M = m T

60˚ = (2 x + 30)˚

30 = 2 x

15 = x

Third Angles Theorem

Substitute

Subtract 30 from each side.

Divide each side by 2.

Page 23: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

PROPERTIES OF SPECIAL PAIRS OF ANGLES Congruent Supplements Theorem

If two angles are supplementary to the same angle (or to congruent angles) then they are congruent.

2

3

1

Page 24: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Linear Pair Postulate

If two angles for m a linear pair, then they are supplementary.

m 1 + m 2 = 180°

PROPERTIES OF SPECIAL PAIRS OF ANGLES

Page 25: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

CONGRUENCE OF ANGLESTHEOREM 2.2 Properties of Angle Congruence

Angle congruence is reflexive, symmetric, and transitive.Here are some examples.

TRANSITIVE If A B and B C, then A C

SYMMETRIC If A B, then B A

REFLEX IVE For any angle A, A A

Page 26: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Transitive Property of Angle CongruenceProve the Transitive Property of Congruence for angles.

SOLUTION

To prove the Transitive Property of Congruence for angles, begin by drawing three congruent angles. Label the vertices as A, B, and C.

GIVEN A B, PROVE A C

A

B

C

B C

Page 27: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Transitive Property of Angle CongruenceGIVEN A B,

B C

PROVE A C

Statements Reasons

1

2

3

4

m A = m B Definition of congruent angles

5 A C Definition of congruent angles

A B, Given

B C

m B = m C Definition of congruent angles

m A = m C Transitive property of equality

Page 28: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Using the Transitive PropertyThis two-column proof uses the Transitive Property.

Statements Reasons

2

3

4

m 1 = m 3 Definition of congruent angles

GIVEN m 3 = 40°, 1 2, 2 3

PROVE m 1 = 40°

1

m 1 = 40° Substitution property of equality

1 3 Transitive property of Congruence

Givenm 3 = 40°, 1 2,

2 3

Page 29: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Proving Right Angle Congruence Theorem Right Angle Congruence Theorem

All right angles are congruent.

You can prove Right Angle CongruenceTheorem as shown.

GIVEN 1 and 2 are right angles

PROVE 1 2

Page 30: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Proving Right Angle Congruence Theorem

Statements Reasons

1

2

3

4

m 1 = 90°, m 2 = 90° Definition of right angles

1 2 Definition of congruent angles

GIVEN 1 and 2 are right angles

PROVE 1 2

1 and 2 are right angles Given

Page 31: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

PROPERTIES OF SPECIAL PAIRS OF ANGLES

Congruent Supplements Theorem

If two angles are supplementary to the same angle (or to congruent angles) then they are congruent.

1 23

If m 1 + m 2 = 180°

m 2 + m 3 = 180°

and

1

then

1 3

Page 32: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

PROPERTIES OF SPECIAL PAIRS OF ANGLES

Congruent Complements TheoremIf two angles are complementary to the same angle (or to congruent angles) then the two angles are congruent.

45

6

Page 33: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

PROPERTIES OF SPECIAL PAIRS OF ANGLES

Congruent Complements TheoremIf two angles are complementary to the same angle (or to congruent angles) then the two angles are congruent.

4

If m 4 + m 5 = 90°

m 5 + m 6 = 90°

and

then

4 6

56

4

Page 34: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Proving Congruent Supplements Theorem

Statements Reasons

1

2

GIVEN 1 and 2 are supplements

PROVE 2 3

3 and 4 are supplements

1 4

1 and 2 are supplements Given

3 and 4 are supplements

1 4

m 1 + m 2 = 180° Definition of supplementary angles

m 3 + m 4 = 180°

Page 35: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Proving Congruent Supplements Theorem

Statements Reasons

3

GIVEN 1 and 2 are supplements

PROVE 2 3

3 and 4 are supplements

1 4

4

5 m 1 + m 2 = Substitution property of equalitym 3 + m 1

m 1 + m 2 = Transitive property of equalitym 3 + m 4

m 1 = m 4 Definition of congruent angles

Page 36: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Proving Congruent Supplements Theorem

Statements Reasons

GIVEN 1 and 2 are supplements

PROVE 2 3

3 and 4 are supplements

1 4

6

7

m 2 = m 3 Subtraction property of equality

2 3 Definition of congruent angles

Page 37: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor
Page 38: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

2

3

4

5

6 SAS Congruence Postulate DRA DRG

1

Proving Triangles Congruent

GivenDR AG

If 2 lines are , then they form 4 right angles.

DRA and DRGare right angles.

Right Angle Congruence Theorem DRA DRG

GivenRA RG

Reflexive Property of CongruenceDR DR

Statements Reasons

D

GA R

GIVEN

PROVE DRA DRG

DR AG

RA RG

Page 39: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Congruent Triangles in a Coordinate Plane

sOLUTION

Use the SSS Congruence Postulate to show that ABC FGH.

AC FH

AB FGAB = 5 and FG = 5

AC = 3 and FH = 3

Page 40: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Congruent Triangles in a Coordinate Plane

d = (x 2 – x1 ) 2 + ( y2 – y1 )

2

= 3 2 + 5

2

= 34

BC = (– 4 – (– 7)) 2 + (5 – 0 )

2

d = (x 2 – x1 ) 2 + ( y2 – y1 )

2

= 5 2 + 3

2

= 34

GH = (6 – 1) 2 + (5 – 2 )

2

Use the distance formula to find lengths BC and GH.

Page 41: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Congruent Triangles in a Coordinate Plane

All three pairs of corresponding sides are congruent, ABC FGH by the SSS Congruence Postulate.

BC GHBC = 34 and GH = 34

Page 42: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Goal 1

Identifying Congruent FiguresTwo geometric figures are congruent if they have exactly the same size and shape.

Each of the red figures is congruent to the other red figures.

None of the blue figures is congruent to another blue figure.

Page 43: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Identifying Congruent FiguresWhen two figures are congruent, there is a correspondence between their angles and sides such that corresponding angles are congruent and corresponding sides are congruent.

For the triangles below, you can write , which reads “triangle ABC is congruent to triangle PQR.” The notation shows the congruence and the correspondence.

ABC PQR

There is more than one way to write a congruence statement, but it is important to list the corresponding angles in the same order. For example, you can also write .BCA QRP

Corresponding Angles Corresponding Sides

A P B Q C R

BC QR

RPCA

AB PQ

Page 44: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Naming Congruent PartsThe two triangles shown below are congruent. Write a congruence statement. Identify all pairs of congruent corresponding parts.

SOLUTION

Angles:

Sides:

D R, E S, F T

, , RS DE TRFD ST EF

The diagram indicates that .

The congruent angles and sides are as follows.

DEF RST

Page 45: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Example

Using Properties of Congruent FiguresIn the diagram, NPLM EFGH.

Find the value of x.

SOLUTION

You know that .GHLM

So, LM = GH.

8 = 2 x – 3

11 = 2 x

5.5 = x

Page 46: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Using Properties of Congruent FiguresIn the diagram, NPLM EFGH.

SOLUTION

Find the value of x.

You know that .GHLM

So, LM = GH.

8 = 2 x – 3

11 = 2 x

5.5 = x

Find the value of y.

You know that N E.

So, m N = m E.

72˚ = (7y + 9)˚

63 = 7y

9 = y

SOLUTION

Page 47: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

SOLUTION

Paragraph Proof

From the diagram, you are given that all three corresponding sides are congruent.

, NQPQ ,MNRP QMQR and

Because P and N have the same measures, P N.

By the Vertical Angles Theorem, you know that PQR NQM.

By the Third Angles Theorem, R M.

Decide whether the triangles are congruent. Justify your reasoning.

So, all three pairs of corresponding sides and all three pairs of corresponding angles are congruent. By the definition of congruent triangles, .PQR NQM

Proving Triangles are Congruent

Page 48: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Proving Two Triangles are Congruent

A B

C D

E

|| , DCAB ,DCAB E is the midpoint of BC and AD.

Plan for Proof Use the fact that AEB and DEC are vertical angles to show that those angles are congruent. Use the fact that BC intersects parallel segments AB and DC to identify other pairs of angles that are congruent.

GIVEN

PROVE .AEB DEC

Prove that .AEB DEC

Page 49: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Proving Two Triangles are Congruent

Statements Reasons

EAB EDC, ABE DCE

AEB DEC

E is the midpoint of AD,E is the midpoint of BC

,DEAE CEBE

Given

Alternate Interior Angles Theorem

Vertical Angles Theorem

Given

Definition of congruent triangles

Definition of midpoint

|| ,DCAB DCAB

SOLUTION

AEB DEC

A B

C D

E

Prove that .AEB DEC

Page 50: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

You have learned to prove that two triangles are congruent by the definition of congruence – that is, by showing that all pairs of corresponding angles and corresponding sides are congruent.

Theorem 4.4 Properties of Congruent Triangles

Reflexive Property of Congruent Triangles

D

E

F

A

B

C

J K

L

Every triangle is congruent to itself.

Symmetric Property of Congruent Triangles

Transitive Property of Congruent Triangles

If , then .ABC DEF DEF ABC

If and , then .JKLABC DEF DEF ABC JKL

Proving Triangles are Congruent

Page 51: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

SSS AND SAS CONGRUENCE POSTULATES

If all six pairs of corresponding parts (sides and angles) are congruent, then the triangles are congruent.

and thenIf Sides are

congruent

1. AB DE

2. BC EF

3. AC DF

Angles are congruent

4. A D

5. B E

6. C F

Triangles are congruent

ABC DEF

Page 52: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Using the SSS Congruence Postulate

Prove that PQW TSW.

Paragraph Proof

SOLUTION

So by the SSS Congruence Postulate, you

know that

PQW TSW.

The marks on the diagram show that PQ TS,

PW TW, and QW SW.

Page 53: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

SSS AND SAS CONGRUENCE POSTULATESPOSTULATE: Side-Angle-Side (SAS) Congruence Postulate

then PQS WXY

If two sides and the included angle of one triangle arecongruent to two sides and the included angle of a second triangle, then the two triangles are congruent.

If Side PQ WX

Side QS XY

Angle Q XA

S

S

Page 54: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

1

Using the SAS Congruence Postulate

Prove that AEB DEC.

2

3 AEB DEC SAS Congruence Postulate

21

AE DE, BE CE Given

1 2 Vertical Angles Theorem

Statements Reasons

Page 55: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

2

3

4

5

6 SAS Congruence Postulate DRA DRG

1

Proving Triangles Congruent

GivenDR AG

If 2 lines are , then they form 4 right angles.

DRA and DRG are right angles.

Right Angle Congruence Theorem DRA DRG

GivenRA RG

Reflexive Property of CongruenceDR DR

Statements Reasons

D

GA R

GIVEN

PROVE DRA DRG

DR AG

RA RG

Page 56: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Congruent Triangles in a Coordinate Plane

AC FH

AB FGAB = 5 and FG = 5

SOLUTION

Use the SSS Congruence Postulate to show that ABC

FGH.

AC = 3 and FH = 3

Page 57: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Congruent Triangles in a Coordinate Plane

d = (x 2 – x1 ) 2 + ( y2 – y1 )

2

= 3 2 + 5

2

= 34

BC = (– 4 – (– 7)) 2 + (5 – 0 )

2

d = (x 2 – x1 ) 2 + ( y2 – y1 )

2

= 5 2 + 3

2

= 34

GH = (6 – 1) 2 + (5 – 2 )

2

Use the distance formula to find lengths BC and GH.

Page 58: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Congruent Triangles in a Coordinate Plane

All three pairs of corresponding sides are congruent,

ABC FGH by the SSS Congruence Postulate.

BC GHBC = √34 and GH = √34

Page 59: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Theorems

Theorem 6.6: If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram.

A

D

B

C

ABCD is a parallelogram.

Page 60: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Theorems

Theorem 6.7: If both pairs of opposite angles of a quadrilateral are congruent, then the quadrilateral is a parallelogram.

ABCD is a parallelogram.

A

D

B

C

Page 61: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Theorems

Theorem 6.8: If an angle of a quadrilateral is supplementary to both of its consecutive angles, then the quadrilateral is a parallelogram.

ABCD is a parallelogram.

A

D

B

C

(180 – x)° x°

Page 62: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Theorems

Theorem 6.9: If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

ABCD is a parallelogram.

A

D

B

C

Page 63: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Ex. 1: Proof of Theorem 6.6

Statements:

1. AB ≅ CD, AD ≅ CB.

2. AC ≅ AC

3. ∆ABC ≅ ∆CDA

4. BAC ≅ DCA, DAC ≅ BCA

5. AB║CD, AD ║CB.

6. ABCD is a

Reasons:

1. Given

C

D

B

A

Page 64: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Ex. 1: Proof of Theorem 6.6Statements:

1. AB ≅ CD, AD ≅ CB.

2. AC ≅ AC

3. ∆ABC ≅ ∆CDA

4. BAC ≅ DCA, DAC ≅ BCA

5. AB║CD, AD ║CB.

6. ABCD is a

Reasons:

1. Given

2. Reflexive Prop. of Congruence

C

D

B

A

Page 65: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Ex. 1: Proof of Theorem 6.6Statements:

1. AB ≅ CD, AD ≅ CB.

2. AC ≅ AC

3. ∆ABC ≅ ∆CDA

4. BAC ≅ DCA, DAC ≅ BCA

5. AB║CD, AD ║CB.

6. ABCD is a

Reasons:

1. Given

2. Reflexive Prop. of Congruence

3. SSS Congruence Postulate

C

D

B

A

Page 66: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Ex. 1: Proof of Theorem 6.6Statements:

1. AB ≅ CD, AD ≅ CB.

2. AC ≅ AC

3. ∆ABC ≅ ∆CDA

4. BAC ≅ DCA, DAC ≅ BCA

5. AB║CD, AD ║CB.

6. ABCD is a

Reasons:

1. Given

2. Reflexive Prop. of Congruence

3. SSS Congruence Postulate

4. CPCTC

C

D

B

A

Page 67: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Ex. 1: Proof of Theorem 6.6Statements:

1. AB ≅ CD, AD ≅ CB.

2. AC ≅ AC

3. ∆ABC ≅ ∆CDA

4. BAC ≅ DCA, DAC ≅ BCA

5. AB║CD, AD ║CB.

6. ABCD is a

Reasons:

1. Given

2. Reflexive Prop. of Congruence

3. SSS Congruence Postulate

4. CPCTC

5. Alternate Interior s Converse

C

D

B

A

Page 68: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Ex. 1: Proof of Theorem 6.6Statements:

1. AB ≅ CD, AD ≅ CB.

2. AC ≅ AC

3. ∆ABC ≅ ∆CDA

4. BAC ≅ DCA, DAC ≅ BCA

5. AB║CD, AD ║CB.

6. ABCD is a

Reasons:

1. Given

2. Reflexive Prop. of Congruence

3. SSS Congruence Postulate

4. CPCTC

5. Alternate Interior s Converse

6. Def. of a parallelogram.

C

D

B

A

Page 69: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Using properties of parallelograms.

Method 1

Use the slope formula to show that opposite sides have the same slope, so they are parallel.

Method 2

Use the distance formula to show that the opposite sides have the same length.

Method 3

Use both slope and distance formula to show one pair of opposite side is congruent and parallel.

Page 70: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Ex. 2: Proving Quadrilaterals are Parallelograms

As the sewing box below is opened, the trays are always parallel to each other. Why?

2.75 in. 2.75 in.

2 in.

2 in.

Page 71: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Ex. 2: Proving Quadrilaterals are Parallelograms

Each pair of hinges are opposite sides of a quadrilateral. The 2.75 inch sides of the quadrilateral are opposite and congruent. The 2 inch sides are also opposite and congruent. Because opposite sides of the quadrilateral are congruent, it is a parallelogram. By the definition of a parallelogram, opposite sides are parallel, so the trays of the sewing box are always parallel.

2.75 in. 2.75 in.

2 in.

2 in.

Page 72: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Another Theorem ~

Theorem 6.10—If one pair of opposite sides of a quadrilateral are congruent and parallel, then the quadrilateral is a parallelogram.

ABCD is a

parallelogram.

A

B C

D

Page 73: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Ex. 3: Proof of Theorem 6.10Given: BC║DA, BC ≅ DAProve: ABCD is a

Statements:

1. BC ║DA

2. DAC ≅ BCA

3. AC ≅ AC

4. BC ≅ DA

5. ∆BAC ≅ ∆DCA

6. AB ≅ CD

7. ABCD is a

Reasons:

1. Given

C

D

B

A

Page 74: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Ex. 3: Proof of Theorem 6.10Given: BC║DA, BC ≅ DAProve: ABCD is a

Statements:

1. BC ║DA

2. DAC ≅ BCA

3. AC ≅ AC

4. BC ≅ DA

5. ∆BAC ≅ ∆DCA

6. AB ≅ CD

7. ABCD is a

Reasons:

1. Given

2. Alt. Int. s Thm.

C

D

B

A

Page 75: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Ex. 3: Proof of Theorem 6.10Given: BC║DA, BC ≅ DAProve: ABCD is a

Statements:

1. BC ║DA

2. DAC ≅ BCA

3. AC ≅ AC

4. BC ≅ DA

5. ∆BAC ≅ ∆DCA

6. AB ≅ CD

7. ABCD is a

Reasons:

1. Given

2. Alt. Int. s Thm.

3. Reflexive Property

C

D

B

A

Page 76: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Ex. 3: Proof of Theorem 6.10Given: BC║DA, BC ≅ DAProve: ABCD is a

Statements:

1. BC ║DA

2. DAC ≅ BCA

3. AC ≅ AC

4. BC ≅ DA

5. ∆BAC ≅ ∆DCA

6. AB ≅ CD

7. ABCD is a

Reasons:

1. Given

2. Alt. Int. s Thm.

3. Reflexive Property

4. Given

C

D

B

A

Page 77: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Ex. 3: Proof of Theorem 6.10Given: BC║DA, BC ≅ DAProve: ABCD is a

Statements:

1. BC ║DA

2. DAC ≅ BCA

3. AC ≅ AC

4. BC ≅ DA

5. ∆BAC ≅ ∆DCA

6. AB ≅ CD

7. ABCD is a

Reasons:

1. Given

2. Alt. Int. s Thm.

3. Reflexive Property

4. Given

5. SAS Congruence Post.

C

D

B

A

Page 78: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Ex. 3: Proof of Theorem 6.10Given: BC║DA, BC ≅ DAProve: ABCD is a

Statements:

1. BC ║DA

2. DAC ≅ BCA

3. AC ≅ AC

4. BC ≅ DA

5. ∆BAC ≅ ∆DCA

6. AB ≅ CD

7. ABCD is a

Reasons:

1. Given

2. Alt. Int. s Thm.

3. Reflexive Property

4. Given

5. SAS Congruence Post.

6. CPCTC

C

D

B

A

Page 79: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Ex. 3: Proof of Theorem 6.10Given: BC║DA, BC ≅ DAProve: ABCD is a

Statements:

1. BC ║DA

2. DAC ≅ BCA

3. AC ≅ AC

4. BC ≅ DA

5. ∆BAC ≅ ∆DCA

6. AB ≅ CD

7. ABCD is a

Reasons:

1. Given

2. Alt. Int. s Thm.

3. Reflexive Property

4. Given

5. SAS Congruence Post.

6. CPCTC

7. If opp. sides of a quad. are ≅, then it is a .

C

D

B

A

Page 80: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Ex. 4: Using properties of parallelograms

Show that

A(2, -1), B(1, 3), C(6, 5) and D(7,1) are the vertices of a parallelogram.

6

4

2

-2

-4

5 10 15

D(7, 1)

C(6, 5)

B(1, 3)

A(2, -1)

Page 81: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Ex. 4: Using properties of parallelograms Method 1—Show that opposite

sides have the same slope, so they are parallel.

Slope of AB. 3-(-1) = - 4

1 - 2 Slope of CD.

1 – 5 = - 4

7 – 6 Slope of BC.

5 – 3 = 2

6 - 1 5 Slope of DA.

- 1 – 1 = 2

2 - 7 5 AB and CD have the same

slope, so they are parallel. Similarly, BC ║ DA.

6

4

2

-2

-4

5 10 15

D(7, 1)

C(6, 5)

B(1, 3)

A(2, -1)

Because opposite sides are parallel, ABCD is a parallelogram.

Page 82: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Ex. 4: Using properties of parallelograms Method 2—Show that

opposite sides have the same length.

AB=√(1 – 2)2 + [3 – (- 1)2] = √17

CD=√(7 – 6)2 + (1 - 5)2 = √17

BC=√(6 – 1)2 + (5 - 3)2 = √29

DA= √(2 – 7)2 + (-1 - 1)2 = √29

AB ≅ CD and BC ≅ DA. Because both pairs of opposites sides are congruent, ABCD is a parallelogram.

6

4

2

-2

-4

5 10 15

D(7, 1)

C(6, 5)

B(1, 3)

A(2, -1)

Page 83: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Ex. 4: Using properties of parallelograms

Method 3—Show that one pair of opposite sides is congruent and parallel.

Slope of AB = Slope of CD = -4

AB=CD = √17

AB and CD are congruent and parallel, so ABCD is a parallelogram.

6

4

2

-2

-4

5 10 15

D(7, 1)

C(6, 5)

B(1, 3)

A(2, -1)

Page 84: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Proving quadrilaterals are parallelograms:

Show that both pairs of opposite sides are parallel.

Show that both pairs of opposite sides are congruent.

Show that both pairs of opposite angles are congruent.

Show that one angle is supplementary to both consecutive angles.

Page 85: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

.. continued..

Show that the diagonals bisect each other

Show that one pair of opposite sides are congruent and parallel.

Page 86: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Engineering

Deshon uses an expandable gate to keep his new puppy in the kitchen. As the gate expands or collapses, the shapes that form the gate always remain parallelograms. Explain why this is true.

Page 87: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Drafting Before computer drawing program become

available, blueprints for buildings or mechanical parts were drawn by hand. One of the tools drafters used, is a parallel ruler. Holding one of the bars in the place and moving the other allowed the drafter to draw a line ll to the first in many position on the page. Why does the parallel ruler guarantee that the second line will be ll to the first?

Page 88: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Arts The Navoja people are well known for their skills in

weaving. Eye- Dazzler rugs became popular with Navoja weavers in the 1880s.What types of shapes do you see most?

Page 89: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Side-Side-Side (SSS) Similarity Theorem

If the corresponding side lengths of two triangles are proportional, then the triangles are similar.

Page 90: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

SOLUTION

Compare ABC and DEF by finding ratios of corresponding side lengths.

Shortest sides

ABDE

43

86 ==

Is either DEF or GHJ similar to ABC?

Longest sidesCAFD

43

1612 ==

Remaining sides

BCEF

43

12 9 ==

All of the ratios are equal, so ABC ~ DEF.ANSWER

Use the SSS Similarity Theorem

Page 91: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Use the SSS Similarity Theorem (continued)

Longest sides

CAJG

1616 == 1

Remaining sides

BCHJ

65

1210 ==

The ratios are not all equal, so ABC and GHJ are not similar.

ANSWER

Compare ABC and GHJ by finding ratios of corresponding side lengths.

Shortest sides

ABGH

88 == 1

Page 92: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Use the SSS Similarity Theorem

Which of the three triangles are similar? Write a similarity statement.

Page 93: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

SOLUTION

Use the SSS Similarity Theorem

Find the value of x that makes ABC ~ DEF.

STEP 1 Find the value of x that makes corresponding side lengths proportional. 4

12 = x –1 18 Write proportion.

4 18 = 12(x – 1)

72 = 12x – 12

7 = x

Cross Products Property

Simplify.

Solve for x.

Page 94: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Use the SSS Similarity Theorem (continued)Check that the side lengths are

proportional when x = 7.STEP 2

BC = x – 1 = 6

618

412 =AB

DEBCEF=

?

DF = 3(x + 1) = 24

824

412 =

ABDE

ACDF=

?

When x = 7, the triangles are similar by the SSS Similarity Theorem.

ANSWER

Page 95: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Use the SSS Similarity Theorem

Find the value of x that makes PQRXYZ ~

X Z

Y

P R

Q

20

12

x + 6 30

3(x – 2)

21

Page 96: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Side-Angle-Side (SAS) Similarity Theorem

If an angle of one triangle is congruent to an angle of a second triangle and the lengths of the sides including these angles are proportional, then the triangles are similar.

Page 97: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Use the SAS Similarity TheoremLean-to Shelter

You are building a lean-to shelter starting from a tree branch, as shown. Can you construct the right end so it is similar to the left end using the angle measure and lengths shown?

Page 98: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Use the SAS Similarity Theorem (continued)

Both m A and m F equal = 53°, so A F. Next, compare the ratios of the lengths of the sides that include A and F.

~

SOLUTION

Shorter sides Longer sidesABFG

32

96 == AC

FH32

1510 ==

The lengths of the sides that include A and F are proportional.

ANSWER

So, by the SAS Similarity Theorem, ABC ~ FGH. Yes, you can make the right end similar to the left end of the shelter.

Page 99: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Choose a methodTell what method you would use to show that the triangles are similar.

Find the ratios of the lengths of the corresponding sides.

Shorter sides Longer sides

SOLUTION

CACD

35

1830 ==

BCEC

35

915 ==

The corresponding side lengths are proportional. The included angles ACB and DCE are congruent because they are vertical angles. So, ACB ~ DCE by the SAS Similarity Theorem.

Page 100: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Choose a method

A. SRT ~ PNQ

Explain how to show that the indicated triangles are similar.

B. XZW ~ YZX

Explain how to show that the indicated triangles are similar.

Page 101: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

AA Similarity (Angle-Angle)

A D

If 2 angles of one triangle are congruent to 2 angles of another triangle, then the triangles are similar.

E

DA

B

CF

B E

ABC ~ DEFConclusion:

andGiven:

Page 102: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

SSS Similarity (Side-Side-Side)

If the measures of the corresponding sides of 2 triangles are proportional, then the triangles are similar.

E

DA

B

CF

Given:

Conclusion:

5

11 22

8 1610

BC

EF

AB

DE

AC

DF

8

16

5

10

11

22

ABC ~ DEF

Page 103: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

SAS Similarity (Side-Angle-Side)

ABC ~ DEF

If the measures of 2 sides of a triangle are proportional to the measures of 2 corresponding sides of another triangle and the angles between them are congruent, then the triangles are similar.

Given:

Conclusion:

E

DA

B

CF

5

11 22

10

AB ACA D and

DE DF

Page 104: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Similarity is reflexive, symmetric, and transitive.

1. Mark the Given.2. Mark …

Shared Angles or Vertical Angles3. Choose a Method. (AA, SSS , SAS)Think about what you need for the chosen method and be sure to include those parts in the proof.

Steps for proving triangles similar:

Proving Triangles Similar

Page 105: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Problem #1:

Pr :

Given DE FG

ove DEC FGC

CD

E

G

F

Step 1: Mark the given … and what it implies

Step 2: Mark the vertical angles

Step 3: Choose a method: (AA,SSS,SAS)Step 4: List the Parts in the order of the method with reasons

Step 5: Is there more? Statements Reasons

Given

Alternate Interior <s

AA Similarity

Alternate Interior <s

1. DE FG2. D F 3. E G

4. DEC FGC

AA

Page 106: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Problem #2

Step 1: Mark the given … and what it implies

Step 2: Choose a method: (AA,SSS,SAS)Step 4: List the Parts in the order of the method with reasons

Step 5: Is there more? Statements Reasons

Given

Division Property

SSS Similarity

Substitution

SSS

: 3 3 3

Pr :

Given IJ LN JK NP IK LP

ove IJK LNP

N

L

P

I

J K

1. IJ = 3LN ; JK = 3NP ; IK = 3LP

2. IJ

LN=3,

JK

NP=3,

IK

LP=3

3. IJ

LN=

JK

NP=

IK

LP

4. IJK~ LNP

Page 107: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Problem #3

Step 1: Mark the given … and what it implies

Step 3: Choose a method: (AA,SSS,SAS)

Step 4: List the Parts in the order of the method with reasons

Next Slide………….

Step 5: Is there more?

SAS

: int

int

Pr :

Given G is the midpo of ED

H is the midpo of EF

ove EGH EDF

E

DF

G H

Step 2: Mark the reflexive angles

Page 108: Unit 7: Modeling with Geometric Relationships Mathematics 3 Ms. C. Taylor

Statements Reasons

1. G is the Midpoint of H is the Midpoint of

Given

2. EG = DG and EH = HF Def. of Midpoint

3. ED = EG + GD and EF = EH + HF Segment Addition Post.

4. ED = 2 EG and EF = 2 EH Substitution

Division Property

Substitution

Reflexive Property

SAS Postulate7. GEHDEF

8. EGH~ EDF

6. ED

EG=

EF

EH

5. ED

EG=2 and

EF

EH =2

ED

EF