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GeometryChapter 5 Resource Book 327

LESSON

5.3 Practice CFor use with pages 324–330

Find the value of x.

1.

12x 1 27

15x 1 9

2. (8x 2 63)8

(4x 2 27)8

3.

2x 1 338x 2 12

Can you conclude that ###$ EH bisects ∠ FEG? Explain.

4.

F

H

G

E

5.

F

H

G

E 6.

F

H

GE

Can you fi nd the value of x?

7.

9

x 2 2

8.

11

3x 1 2

9.

428

x8

Find the indicated measure given that point G is the incenter of n ACE.

10. Find DG. 11. Find BG.

A F E

G

B D

C

13

12 CB

D G

AFE

15

9

LE

SS

ON

5.3

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GeometryChapter 5 Resource Book328

12. Error Analysis Explain why the conclusion is not correct given the information in the diagram.

S

R

U

T

V

P

RV 5 PV

Find the value of x that makes N the incenter of the triangle.

13.

FD M

E

L

K

N5x

15

25

14.

LC

B

M

N

A

K

26102x

15. Window You are hanging a sun catcher in a triangular window. Use the diagram to explain how to fi nd the correct length of string ( } BD ) needed to display the ornament

at the incenter of the window.

A

B

C

D

16. Proof Write a two-column or paragraph proof.

GIVEN: D is on the bisector of ∠ BAC.

AD

B

C

} DB ⊥ } AB , } DC ⊥ } AC

PROVE: } DB > }

DC

LESSON

5.3 Practice C continuedFor use with pages 324–330

LES

SO

N 5

.3

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A61Geometry

Chapter 5 Resource Book

Review for Mastery

1. 23 2. 10 3. ###$ PQ bisects }

RS , so PR 5 PS. Because Q is on the perpendicular bisector of

} RS ,

QR 5 QS by Theorem 5.2. 4. No; If T were on

@##$ PQ , then T would be equidistant from R and S. T is 14 units from R and 15 units from S. 5. 9

Problem Solving Workshop:Worked Out Example

1. (4.5, 3.5) 2. (4.5, 6.1)

Challenge Practice

1. x 5 2, y 5 3 2. x 5 3, y 5 2 3. x 5 12, y 5 8 4. x 5 7, y 5 13

5.

Statements Reasons

1. } GJ is ⊥ bisector of } HK . 1. Given2. } HJ > } JK 2. Defi nition of segment bisector3. } MH > } MK , 3. Perpendicular

} GH > } GK Bisector Theorem

4. } GM > }

GM 4. Refl exive Property of Congruence5. nGHM > nGKM 5. SSS Congruence Postulate6. ∠GHM > ∠GKM 6. Corresp. parts of

> n are >.

6. Begin by drawing a line segment from point B to point F as shown. You are given

} FC is the

perpendicular bisector of }

AB and } FE is the perpendicular bisector of } BD . By the PerpendicularBisector Theorem, you know that AF 5 FB andFD 5 FB. Using the Transitive Property of Equality, you can conclude that AF 5 FD. By the defi nition of congruent segments you know that }

AF > } FD .

A

C

B E D

F

7.

Statements Reasons

1. } UW > } UY , } UV > } UZ 1. Given2. } UX is ⊥ bisector of } WY . 2. Given3. } WX > } XY 3. Defi nition of segment bisector

Statements Reasons

4. ∠UWX > ∠UYX 4. Base Angles Theorem5. nUWX > nUYX 5. SAS Congruence Postulate6. ∠UXW and ∠UXY 6. Defi nition of are right angles. ⊥ lines

7. } UX > } UX 7. Refl exive Property of Congruence

8. nUVX > nUZX 8. HL Congruence Theorem9. X is the midpoint of 9. Defi nition of } VZ . midpoint

Lesson 5.3

Practice Level A

1. 7 2. 208 3. 5 4. 4 5. Yes; Angle Bisector Theorem 6. No; You do not know if

} DC is

perpendicular to ###$ BC or if }

DA is perpendicular to ###$ BA . 7. Yes; Angle Bisector Theorem

8. No; You do not know if }

DC is perpendicular to ###$ BC or if } DA is perpendicular to ###$ BA . 9. 15 10. 5

11. The incenter of the triangular back yard because the incenter of a triangle is equidistant from the sides of the triangle. 12. 12 ft

Practice Level B

1. 19 2. 288 3. 868 4. No; you don’t know that

} DA and

} DC are ⊥ to the rays. 5. No; you

don’t know that D is ⊥ to rays. 6. Yes; Converse of Angle Bisector Theorem 7. 7 8. 3 9. 8

10. Yes; x 5 9 by Angle Bisector Theorem.

11. No; you need to know that the congruent segments are ⊥ to the rays. 12. No; you need to know that the two segments are congruent.

13. 16 14. 7 15. 5 16. 8 17. Directly between points L and R so that

} SG bisects ∠ LSR;

the distance between you and each goalpost is equal which minimizes the amount you have to move in either direction. 18. 35 ft

Practice Level C

1. 6 2. 9 3. 7.5 4. No; you don’t know that H is equidistant to rays. 5. Yes; Converse of the Angle Bisector Theorem 6. Yes; Angle Bisector Theorem 7. yes 8. no 9. no 10. 5 11. 9

12. } RV is not perpendicular to }

SQ and } PV is not perpendicular to

} QU . 13. 4 14. 12

Lesson 5.2, continuedA

NS

WE

RS

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A62GeometryChapter 5 Resource Book

Practice Level C

1. 6 2. 9 3. 7.5 4. No; you don’t know that H is equidistant to rays. 5. Yes; Converse of the Angle Bisector Theorem 6. Yes; Angle Bisector Theorem 7. yes 8. no 9. no 10. 5 11. 9

12. } RV is not perpendicular to }

SQ and } PV is not perpendicular to

} QU . 13. 4 14. 12

15. Use the Concurrency of Angle Bisectors of a Triangle Theorem to fi nd the incenter of n ABC. Then measure the length of angle bisector.

16. 1. D is on the bisector of ∠ BAC.; (Given) 2. ∠ BAD > ∠ CAD; (Defi nition of angle bisector)3. } DB ⊥ } AB , } DC ⊥ } AC ; (Given) 4. ∠ ABD and ∠ ACD are right angles.; (Perpendicular lines form right angles.) 5. ∠ ABD > ∠ ACD; (All right angles are congruent.) 6. } AD >

} AD ; (Refl exive

Property of Congruence) 7. n ADB > n ADC; (AAS Congruence Theorem) 8. } DB ù

} DC ;

(Corresponding parts of congruent triangles are congruent.)

Review for Mastery

1. 7 2. 5 3. 9 4. 7 5. 3

Problem Solving Workshop:Mixed Problem Solving

1. a.

x

y

1

1

L

C

P

T b. (4.5, 2.5)

c. about 2.55 mi 2. a. 15 in. b. 7.5 in. c. 60 in.

3. 60 in. 4. ∠ NLP, ∠ NMP, ∠ MPJ; } MN and }

JL are parallel, so ∠ KNM > ∠ NLP. } MP and } KL are parallel, so ∠ KNM > ∠ NMP. } MN and } JL are parallel, so ∠ MPJ > ∠ NMP > ∠ KNM.

5. The triangle is obtuse. Sample answer:

6. a. l is the perpendicular bisector of }

AB . b. ###$ PG should bisect ∠ APB. c. m∠ APB increases; more diffi cult; The goalie has a greater area to defend because the distances from the goalie to the sides of ∠ APB (the shooting angle) increase.

Challenge Practice

1. x 5 6 2. x 5 7

3. No, the angle bisectors pass through one vertex and the incenter. In order for the incenter to lie at a vertex, the angle bisector would have to coincide with one of the sides of the angle, makingan angle of 08 which does not make a triangle.

4. No, the incenter is the center of the inscribed circle. This circle touches each side of the triangle.

5. First draw isosceles nABC with an angle bisector ###$ BD at the vertex angle B.

1 2

B

A D C

3 4

Statements Reasons

1. nABC is isosceles. 1. Given 2. } AB >

} BC 2. Defi nition of

isosceles triangle 3. ###$ BD bisects ∠B. 3. Given 4. ∠1 > ∠2 4. Defi nition of angle bisector 5. } BD > } BD 5. Refl exive Property of Congruence 6. nABD > nCBD 6. SAS Congruence Postulate

7. ∠3 > ∠4 7. Corresp. parts of > n are >.

8. ∠3 > ∠4 are a linear 8. Defi nition of linear pair. linear pair 9. ∠3 > ∠4 are 9. Linear Pair supplementary. Postulate10. m∠3 5 m∠4 5 908 10. If two angles are

equal and supplementary, then they measure 908.

11. } BD ⊥ }

AC 11. If two lines intersect to form a linear pair of > ?, then the lines are ⊥.

12. } AD > }

DC 12. Corresp. parts of> n are >.

13. ###$ BD is the ⊥ 13. Perpendicular bisector of

} AC . Bisector Theorem

Lesson 5.3, continuedA

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