geometry rbc a answers - mr....

Answers

Copyright © Big Ideas Learning, LLC Geometry All rights reserved. Answers

A45

31. yes 32. no 33. yes 34. no

35. a. 1.5 ft

b. $1.60 per ft

c. $2.0

36. a. 12 in. by 18 in.

b. 2216 in.

c. $6.48

d. $13.52

37. 80° 38. 100° 39. 30° 40. 150°

41. 20° 42. 50° 43. 70° 44. 180°

45. 100° 46. 80° 47. 271.5 ft 48. 2363 yd

49. 2717.5 mm 50. 240 cm 51. 254 in.

52. 2735 mi 53. 3101.25 in. 54. 32132230 in.

55. a. $3.75 b. $0.08

56. 3 24y x= − − 57. 4 7y x= −

58. 12 17y x= + 59. 5 11y x= − −

60. 14 11y x= − 61. 2

3 9y x= − −

62. 12 2y x= + 63. 3

8 2y x= −

64. 57 7y x= + 65. 1

4 13y x= − −

66. 2y x= − + 67. 13 3y x= −

68. 3 5y x= − 69. 12 7y x= − +

70. 12 4y x= − + 71. 4 8y x= −

72. 13y x= − 73. 3 20y x= − −

74. 14 3y x= + 75. 2 12y x= − −

76. 4 19y x= − 77. 15 2y x= +

78. 5y = − 79. 12 5y x= − −

80. (5, 0), (2, 3), (4, 7)A B C′ ′ ′ −

81. (3, 6), (0, 9), (2, 1)A B C′ ′ ′ −

82. (1, 4), ( 2, 7), (0, 3)A B C′ ′ ′− −

83. (7, 2), (4, 5), (6, 5)A B C′ ′ ′ −

84. (2, 3), ( 1, 6), (1, 4)A B C′ ′ ′− −

85. (4, 2), (1, 1), (3, 9)A B C′ ′ ′− −

86. (5, 8), (2, 11), (4, 1)A B C′ ′ ′

87. ( 2, 7), ( 5, 10), ( 3, 0)A B C′ ′ ′− − −

Chapter 5 5.1 Start Thinking

If 120 ,m A∠ = ° then 60m B∠ = ° because together

they make a straight angle. If 40 ,m D∠ = ° then

140m E∠ = ° under the same reasoning. If the sum of

, ,m B m C∠ ∠ and m D∠ is 180°, then 80 .m C∠ = °

5.1 Warm Up

1. 1 31m∠ = ° 2. 2 59m∠ = °

3. 3 59m∠ = ° 4. 4 90m∠ = °

5.1 Cumulative Review Warm Up

1. 69x = 2. 6x =

3. 60x = 4. 30x =

5.1 Practice A

1. scalene; right 2. isosceles; acute

3. scalene; not a right triangle

4. isosceles; right triangle

5. 32x = 6. 29x = 7. 30x = 8. 17x =

9. 19.5 , 70.5 , 90° ° °

10. no; An exterior angle will be acute when it is adjacent to the obtuse angle of an obtuse triangle.

11. 145°

5.1 Practice B

1. scalene; obtuse 2. isosceles; acute

3. 14x = 4. 9x =

5. 1169

x = 6. 7; 19.5x y= =

Answers

Geometry Copyright © Big Ideas Learning, LLC Answers All rights reserved. A46

7. The measure of each exterior angle is equal to the sum of the measures of the nonadjacent interior angles. So, 1 , 2 ,m A C m B C∠ = + ∠ = +

and 3 .m A B∠ = +

( )( )

Sum of exterior angles 1 2 3

2

2 180

360

A C B C A B

A B C

= ∠ + ∠ + ∠= + + + + += + += °= °

8. 67; 124x y= = 9. yes; 15 , 60 , 105 .° ° °

10. no; For instance, in a 10 -40 -130° ° ° triangle, at the 130° angle you have a 50° exterior angle which is complementary to the 40° angle.

11. ;C T∠ ≅ ∠ The sum of the angle measures of each

triangle is 180 ,° so if two pairs of corresponding angles are congruent, then the third pair of corresponding angles must also be congruent.

5.1 Enrichment and Extension

1. 63 , 36 , 81 ; acute° ° ° 2. 10; 71x y= =

3. 50; 33x y= = 4. 29; 64x y= =

5. 12.9; 51.4x y= = 6. 1m A m∠ = ∠

5.1 Puzzle Time

THE LETTUCE WAS A “HEAD” AND THE TOMATO WAS TRYING TO “KETCHUP”

5.2 Start Thinking

AB corresponds to ;DE BC corresponds to ;EF AC

corresponds to .DF

, ,A D B E C F∠ ≅ ∠ ∠ ≅ ∠ ∠ ≅ ∠

To get ,DEF multiply the side lengths of ABC by a scale factor of 2 and rotate 180 .ABC °

5.2 Warm Up

1. 9x = 2. 69x =


1.

5.2 Practice A 1. ,A E∠ ≅ ∠ ,B F∠ ≅ ∠ ,C G∠ ≅ ∠ ,D H∠ ≅ ∠

,AB EF≅ ,BC FG≅ ,CD GH≅ and

;DA HE≅ Sample answer: BCDA FGHE≅

2. 8; 19x y= =

3. From the figure, ,J N∠ ≅ ∠ ,K P∠ ≅ ∠

,Q M∠ ≅ ∠ ,JK NP≅ ,KL PL≅ ,LQ LM≅

and .QJ MN≅ Also, KLQ PLM∠ ≅ ∠ by the

Vertical Angles Congruence Theorem (Thm. 2.6). Because all pairs of corresponding angles and all pairs of corresponding sides are congruent,

.JKLQ NPLM≅

4. The figure shows that A R∠ ≅ ∠ and .B S∠ ≅ ∠

So, C T∠ ≅ ∠ by the Third Angles Theorem

(Thm. 5.4). Using the Triangle Sum Theorem (Thm. 5.1), 65 .m T∠ = °

5. ABC and DEF are both equilateral and equiangular.

5.2 Practice B

1. ,A H∠ ≅ ∠ ,B I∠ ≅ ∠ ,C J∠ ≅ ∠ ,D F∠ ≅ ∠,E G∠ ≅ ∠ ,AB HI≅ ,BC IJ≅ ,CD JF≅

,DE FG≅ ;EA GH≅ ABCDE GFJIH≅

2. 65; 6; 146x y z= = =

3. From the figure, ,LP ON≅ ,PM NM≅ and

.LM OM≅ P N∠ ≅ ∠ because all right angles

are congruent to each other. By the Vertical Angles Congruence Theorem (Thm. 2.6),

.LMP OMN∠ ≅ ∠ L O∠ ≅ ∠ by the Third

Angles Theorem (Thm. 5.4). Because corresponding sides and angles are congruent,

.LPM ONM≅

A

D

EF

B C2 cm51°

51°

39°

39°

2.5 cm 3.2 cm5 cm6.4 cm

4 cm

x

y8

4

−8

−4

8 12−4

X

Y

X′

Y′

Answers


A47

4. Because RSTU UVQR≅ and VUR∠ is a right

angle, SUR∠ is a right angle. Using vertical

angles, 47 ,m UPV∠ = ° and by the Triangle Sum

Theorem (Thm. 5.1), 2 5 43x y+ = and

43 .m RSV∠ = ° Because it forms a linear pair

with a 47° angle, 133 .m UPS∠ = ° Using angle

addition and corresponding parts of congruent figures, VST∠ is a right angle. Then

( )360 90 90 133

47

4 3 .

m PUT

x y

∠ = − + +

== −

Solving the system 2 5 43

4 3 47

x y

x y

+ =− =

gives 14x = and 3.y = Finally,

90 43 133 .m RST∠ = + = °

5.

yes; To prove the triangles congruent, use the facts that AC CA= and opposite sides of a rectangle are congruent to show that 3 pairs of corresponding sides are congruent. Then use the Alternate Interior Angles Theorem (Thm. 3.2) to show two pairs of angles congruent. The third pair of corresponding angles are congruent right angles.


1.

2.

3. a. yes; You are given .ADB CDA CDB≅ ≅

So, .AB BC CA≅ ≅ Because all three sides of ABC are congruent, it is an equilateral triangle.

b. 120°

c. 30 , 30° °

d. The angle measures are equal because CDB is isosceles.

e. The measure of each of the congruent angles of each small triangle is 30 .° By the Angle Addition Postulate (Post. 1.4), the measure of each angle of ABC is 60 .°

5.2 Puzzle Time

CRAB CAKES

5.3 Start Thinking

Sample answer:

ABC ADE≅ because all angles and sides are

congruent. There is no further information needed because the given directions make it impossible to

construct ADE ≅ .ABC

5.3 Warm Up

1. AC 2. FH

A

B

D

C

A B

C

D

E

x

y

4

6

2

−2

4−2−4 (0, 0)(3, 0)(−3, 0)

(−3, 4) (3, 4)

STATEMENTS REASONS

1. ABD CDB∠ ≅ ∠ 1. Given

2. ADB CBD∠ ≅ ∠ 2. Given

3. AD BC≅ 3. Given

4. AB DC≅ 4. Given

5. BD BD≅ 5. Reflexive Property of Equality

6. BAD BCD∠ ≅ ∠ 6. Triangle Sum Theorem

7. ABD CDB≅ 7. All corresponding parts are congruent.

STATEMENTS REASONS

1. ||AB DC 1. Given

2. AB DC≅ 2. Given

3. is the midpoint

of and .

E

AC BD 3. Given

4. AE EC≅ 4. Definition of midpoint

5. BE ED≅ 5. Definition of midpoint

6. EAB ECD∠ ≅ ∠ 6. Alternate interior angles

7. ABD BDC∠ ≅ ∠ 7. Alternate interior angles

8. AEB CED∠ ≅ ∠ 8. Vertical Angles

Theorem

9. AEB CED≅ 9. All corresponding parts are congruent.

Answers



1. ED 2. EC

5.3 Practice A

1. yes; You know that two sides and the included angle are congruent, so you can use the SAS Congruence Theorem (Thm. 5.5).

2. no; The congruent vertical angles are not the included angles, so SAS cannot be used.

3. ;ABO CBO≅ ,OB OB≅ and because they are

radii, .AO CO≅ It is given that .AOB COB∠ ≅

So, SAS is satisfied.

4. ;ABE CBE≅ Because ,AB BC≅

,ABE CBE∠ ≅ ∠ and ,BE BE≅ SAS is satisfied.

5. no; The SAS Congruence Theorem (Thm. 5.5) applies after a translation, reflection, or rotation, because those are congruence transformations, but not after a dilation, which changes the size of the figure.

6.

5.3 Practice B

1. yes; You know that two sides are congruent, and using the Third Angles Theorem (Thm. 5.4) you can find that the included angles are congruent, so SAS applies.

2. yes; One pair of sides is marked congruent, you can use segment addition to show a second pair of sides congruent, and you can use the Third Angles Theorem (Thm. 5.4) to show their included angles congruent. So, SAS applies.

3. , ,PLM PMN and ;PNL The sides ,PL ,PM

and PN are congruent because they are radii of the same circle. The three angles at P are congruent, so you can use SAS with these angles and the sides including them.

4. ,RVU ,SXW and ;TZY Show that each

obtuse angle measures ( )360 60 90 90− + +

and that the sides including these angles are all congruent because they are sides of squares that border an equilateral triangle. So, SAS applies.

5. 7; 24x y= =

6. AED is equilateral and equiangular, so

,AE DE≅ and .CAD BDA∠ ≅ ∠ ,EB EC≅ so

EB DE EC DE+ = + by the Addition Property of Equality. Substituting AE for DE gives

.EB DE EC AE+ = + Using the Segment Addition Postulate (Post. 1.2), BD EB DE= +and .CA EC AE= + So, BD CA= by

substitution. Also, AD DA≅ by the Reflexive Property of Segment Congruence (Thm. 2.1).

ACD DBA≅ by the SAS Congruence

Theorem (Thm. 5.5).


1. NP OP≅ because they have an equal length of 6;

MP MP≅ by the Reflexive Property of Segment

Congruence (Thm. 2.1); MO MN≅ because they

have an equal length of 3 2. So, PMO PMN≅ by the SSS Congruence

Theorem (Thm. 5.8). It is possible to prove PMO PMN≅ by the SAS Congruence

Theorem (Thm. 5.5) because you already know that

MP MP≅ and .MO MN≅ You also know that

NO MP⊥ because the slopes are opposite reciprocals. So, ,OMP NMP∠ ≅ ∠ leading to

PMO PMN≅ by the SAS Congruence

Theorem (Thm. 5.5).

2. yes; HL Congruence Theorem (Thm. 5.9)

STATEMENTS REASONS

1. and ABD CBD∠ ∠ are right angles.

1. Given

2. bisects .BD AC 2. Given

3. AB BC≅ 3. Definition of a segment bisector

4. BD BD≅ 4. Reflexive Property of Segment Congruence (Thm. 2.1)

5. ABD CBD≅ 5. SAS Congruence Theorem (Thm. 5.5)

Answers


A49

3.

4.

5. ( ) ( )4, 10 , 15, 3X Y

5.3 Puzzle Time

OBTUSE

5.4 Start Thinking

Sample answer:

.ABD CBD≅ The length of 8.BD = EGH EGF≅ because all side lengths and angles

are congruent.

5.4 Warm Up

1. 89x = 2. 83x =

3. 21x = 4. 80x =


1. A triangle is an isosceles triangle if and only if the legs are of equal length.

2. A Chinese puzzle is a tangram if and only if it is made up of seven pieces.

3. A parallelogram is a rectangle if and only if it has four right angles.

5.4 Practice A

1. 25x = 2. 6x =

3. 20; 30x y= = 4. 30; 4x y= =

5. Because they are vertical angles, .DCE BCA∠ ≅ ∠ Because they both measure

70 ,° .B D∠ ≅ ∠ So, A E∠ ≅ ∠ by the Third

Angles Theorem (Thm 5.4). But E D∠ ≅ ∠

because they are base angles of an isosceles triangle, so 70 .m A∠ = ° Because they have equal

measures, A∠ and B∠ are base angles of an

isosceles triangle.

6. yes; An isosceles triangle is obtuse when the vertex angle is obtuse and the base angles are acute.

x

y

4

−2

−2 2−4 DC

A

GH

E F

B

STATEMENTS REASONS

1. AC BD⊥ 1. Given

2. is the midpoint

of .

D

AC 2. Given

3. is a rightangle.

ADB∠ 3. Definition of perpendicular lines

4. 90m ADB∠ = ° 4. Definition of right angle

5. is a right angle.

CDB∠ 5. Definition of perpendicular lines

6. 90m CDB∠ = ° 6. Definition of right angle

7. m ADB m CDB∠ = ∠ 7. Substitution

8. AD DC≅ 8. Definition of midpoint


10. ADB CDB≅ 10. SAS Congruence Theorem (Thm. 5.5)

STATEMENTS REASONS

1. DE BF= 1. Given

2. AE CF= 2. Given

3. AE DB⊥ 3. Given

4. CF BD⊥ 4. Given


AEB∠ 5. Definition of perpendicular lines


CFD∠ 6. Definition of perpendicular lines

7. AEB CFD∠ ≅ ∠ 7. Right Angles Congruence Theorem (Thm. 2.3)

8. BE BF FE= + 8. Segment Addition Postulate (Post. 1.2)

9. FD FE ED= + 9. Segment Addition Postulate (Post. 1.2)

10. FD FE BF= + 10. Substitution

11. FD BE= 11. Substitution

12. AE CF≅ 12. Definition of congruent segments

13. FD BE≅ 13. Definition of congruent segments

14. AEB CFDΔ ≅ Δ 14. SAS Congruence Theorem (Thm. 5.5)

Answers


5.4 Practice B

1. 36x = 2. 90x =

3. 14; 7x y= = 4. 29; 3x y= =

5. It is given that CBD CDB∠ ≅ ∠ and

.CAE CEA∠ ≅ ∠ By the Converse of the Base

Angles Theorem (Thm. 5.7), BC DC≅ and

.AC EC≅ By the Reflexive Property of Angle Congruence (Thm. 2.2), .C C∠ ≅ ∠

ACD ECB≅ by the SAS Congruence

Theorem (Thm. 5.5). Because congruent parts of

congruent triangles are congruent, .AD EB≅

6.


1. ( )12

r p q= +

2.

3. 12.9; 51.4x y= = 4. 29; 64x y= =

5.

It is possible to partition a right triangle into two isosceles triangles. Suppose ABC is a right triangle with right angle C. Because m CAD∠ is

less than ,m ACB∠ it is possible to construct D

such that ,m ACD m CAD∠ = ∠ as shown above.

Because 90m ACD m DCB m CAD∠ + ∠ = ° = ∠

m DBC+ ∠ and ,m CAD m ACD∠ = ∠ it follows

that .m DCB m DBC∠ = ∠ So, DBC is an

isosceles triangle.

5.4 Puzzle Time

NEITHER IT’S BEST TO WRITE WITH A PEN

5.5 Start Thinking

In ,JKL 20 , 80 ,m J m K∠ = ° ∠ = ° 80 .m L∠ = °

JKL is an isosceles triangle. The angles in PQR

are congruent. It is not possible to create two triangles with the same side lengths but different angles. As discussed in Section 5.4, the side lengths and angles are related.

5.5 Warm Up

1. RUT∠ 2. STR∠ 3. TRS∠

4. UTR∠ 5. SRT∠ 6. RST∠


1. 1 54 ;m∠ = ° straight angle

2 54 ;m∠ = ° Corresponding Angles Theorem

(Thm. 3.1)

2. 2 131 ;m∠ = ° Corresponding Angles Theorem

(Thm. 3.1) 1 49 ;m∠ = ° straight angle

5.5 Practice A

1. congruent; Two pairs of sides are marked congruent and the third pair of sides is shared, so the SSS Congruence Theorem (Thm. 5.8) applies.

a° a°

3a° 3a°

4a°

2a°2a°

4a°6a°

6a°

5a° 5a°

7a°7a°

(180 − 2a)°

(180 − 6a)°

(180 − 4a)° (180 − 8a)°

(180 − 10a)°

(180 − 14a)°

(180 − 12a)°

STATEMENTS REASONS

1. , EBC ECB

AE DE

∠ ≅ ∠

≅

1. Given

2. EB EC≅ 2. Converse of the Base Angles Theorem (Thm. 5.7)

3. AEB DEC∠ ≅ ∠ 3. Vertical Angles Congruence Theorem (Thm. 2.6)

4. AEB DEC≅ 4. SAS Congruence Theorem (Thm. 5.5)

5. AB DC≅ 5. Corresponding parts of congruent triangles are congruent.

1 in.

0.5 in.K L

J

1 in.

75.5° 75.5°

29°1 in.

0.5 in.Q R

P

1 in.

75.5° 75.5°

29°

A

C B

D

Answers


A51

2. not congruent; The hypotenuse and a leg of one triangle are congruent to the legs of the other triangle, so the triangles cannot be congruent.

3. congruent; The hypotenuses are shared and a pair of legs is congruent, so the HL Congruence Theorem (Thm. 5.9) applies.

4.

5.

6. not congruent

7. a. You know the shared sides are congruent, so you need to measure each of the other sides of the two triangular faces to determine whether the SSS Congruence Theorem (Thm. 5.8) applies.

b. a regular hexagon

5.5 Practice B

1. The triangles share one pair of sides and two other pairs of sides are marked congruent, so the SSS Congruence Theorem (Thm. 5.8) can be used.

2. The triangles are right triangles and the hypotenuses and a pair of legs are shown to be congruent, so the HL Congruence Theorem (Thm. 5.9) can be used.

3. The triangles have three pairs of congruent sides and each triangle has a right angle, so they can be proven congruent using the SSS Congruence Theorem (Thm. 5.8), the HL Congruence Theorem (Thm. 5.9), or the SAS Congruence Theorem (Thm. 5.5).

4.

5.

6. Sample answer: Sketch the lines in a coordinate plane. The slopes of lines a and b show that they form vertical right angles of the triangles. The Distance Formula shows that the hypotenuses have different lengths, so the triangles are not congruent.

STATEMENTS REASONS

1. ;

bisects .

AB AD

AC BD

≅ 1. Given

2. AC AC≅ 2. Reflexive Property of Segment Congruence (Thm. 2.1)

3. BC DC≅ 3. Definition of segment bisector

4. ABC ADC≅ 4. SSS Congruence Theorem (Thm. 5.8)

STATEMENTS REASONS

1. and JL GF

KL HF

≅≅

1. Given

2. and

are right angles.J G∠ ∠

2. Given

3. and JKL GHF are right triangles.

3. Definition of a right triangle

4. JKL GHF≅ 4. HL Congruence Theorem (Thm. 5.9)

STATEMENTS REASONS

1. , PS RS

SQ PR

≅

⊥

1. Given

2. and are right angles.

PQS RQS∠ ∠ 2. Definition of perpendicular lines

3. and are right triangles.

PQS RQS 3. Definition of right triangles

4. QS QS≅ 4. Reflexive Property of Segment Congruence (Thm. 2.1)

5. PSQ SQR≅ 5. HL Congruence Theorem (Thm. 5.9)

STATEMENTS REASONS

1. ,

,

BC ED

AB FE

AD FC

≅

≅

≅

1. Given

2. BC ED= 2. Definition of congruent segments

3. ,BD BC CD

EC ED CD

= += +

3. Segment Addition Postulate (Post. 1.2)

4. BD ED CD= + 4. Substitution Property of Equality

5. BD EC= 5. Substitution Property of Equality

6. BD EC≅ 6. Definition of congruent segments

7. ABD FEC≅ 7. SSS Congruence Theorem (Thm. 5.8)

Answers



1. a. 7x = b. 4x = c. 4 or 4x = −

2.

3.

4.

If one side of RSTΔ is congruent to one side of

,DEF such that ,RS MN≅ then you know that the triangles are congruent because equilateral triangles have three congruent sides. So, all sides in

RST would be equal to all sides in ,MNO making them congruent by the SSS Congruence Postulate.

5. ( ) ( )2 , 2 , 2 , 2J e c f d K e a f b− − − −

5.5 Puzzle Time

A TEENAGER

5.6 Start Thinking

Sample answer:

;XYZ ABC≅/ Any size triangle can have the angle

measures indicated. One example would be to draw the triangle on a coordinate plane and then dilate it with the origin as the center. This example shows that knowing two triangles have the same angle measures is not enough to prove congruence.

5.6 Warm Up

1. SAS Congruence Theorem (Thm. 5.5)

2. none

3. SAS Congruence Theorem (Thm. 5.5)

4. none


1.

5.6 Practice A 1. yes; AAS Congruence Theorem (Thm. 5.11)

X Z

Y

70° 60°

50°

A C

B

70° 60°

50°

x

y

8

−4

84−2

X

X

Z

Y

X′

X″

Y″

Z″

Y′

Z′

R T

S

M O

N

STATEMENTS REASONS

1. WA WT≅ 1. Given

2. is the midpoint of .S AT 2. Given

3. AS ST≅ 3. Definition of midpoint

4. WS WS≅ 4. Reflexive Property of Segment Congruence (Thm. 2.1)

5. WSA WSTΔ ≅ Δ 5. SSS Congruence Theorem (Thm. 5.8)

6. 1180

m WAS m∠ + ∠= °

6. Definition of perpendicular lines

7. AEB CFD∠ ≅ ∠ 7. Definition of linear pair

8. 2180

m WTS m∠ + ∠= °

8. Definition of linear pair

9. WAS WTS∠ ≅ ∠ 9. Corresponding parts of congruent triangles are congruent.

10. m WAS m WTS∠ = ∠ 10. Definition of congruence

11. 12

m WAS mm WTS m

∠ + ∠= ∠ + ∠

11. Substitution

12. 12

m WAS mm WAS m

∠ + ∠= ∠ + ∠

12. Substitution

13. 1 2m m∠ = ∠ 13. Subtraction

14. 1 2∠ ≅ ∠ 14. Definition of congruent angles

STATEMENTS REASONS

1. GR GT≅ 1. Given

2. RS ST≅ 2. Given

3. GS GS≅ 3. Reflexive Property of Segment Congruence (Thm. 2.1)

4. GRS GST≅ 4. SSS Congruence Theorem (Thm. 5.8)

Answers


A53

2. no; Although two pairs of sides and one pair of angles are congruent, they are not the corresponding parts needed for the SAS Congruence Theorem (Thm. 5.5).

3. yes; ASA Congruence Theorem (Thm. 5.10)

4.

5.

6. Given: , ;AB CD ABD CDB BD DB≅ ∠ ≅ ∠ ≅

by the Reflexive Property of Segment Congruence (Thm. 2.1); So, ABD CDB≅ by the SAS

Congruence Theorem (Thm. 5.5).

Given: , ;AB EF ABD F ABD EDF≅ ∠ ≅ ∠ ∠ ≅ ∠by the Vertical Angles Congruence Theorem (Thm. 2.6); So, ABD EFD≅ by the AAS


Given: , ;CD EF BDC F BD DF≅ ∠ ≅ ∠ ≅

because congruent parts of congruent triangles are congruent; So, CDB EFD≅ by the SAS


There is not enough information to show that AGD is congruent to any other triangle.

5.6 Practice B

1. yes; ASA Congruence Theorem (Thm. 5.10)

2. no; The sides and angles marked congruent do not correspond to be used for either ASA or AAS.

3. yes; AAS Congruence Theorem (Thm. 5.11)

4.

5.

STATEMENTS REASONS

1. , PS RT PQ TQ≅ 1. Given

2. and S R

P T

∠ ≅ ∠∠ ≅ ∠

2. Alternate Interior Angles Theorem (Thm. 3.2)

3. PSQ TRQ≅ 3. AAS Congruence Theorem (Thm. 5.11)

STATEMENTS REASONS

1. bisects , BD ADC

BD AC

∠

⊥

1. Given


3. ADB CDB∠ ≅ ∠ 3. Definition of angle bisector


ABD CBD∠ ∠ 4. Definition of perpendicular lines

5. ABD CBD∠ ≅ ∠ 5. Right Angles Congruence Theorem (Thm. 2.3)

6. ABD BCD≅ 6. ASA Congruence Theorem (Thm. 5.10)

STATEMENTS REASONS

1. bisects ,BD AE

A E∠ ≅ ∠ 1. Given

2. AC CE≅ 2. Definition of segment bisector

3. ACB ECD∠ ≅ ∠ 3. Vertical Angles Congruence Theorem (Thm. 2.6)

4. ABC EDC≅ 4. ASA Congruence Theorem (Thm. 5.10)

STATEMENTS REASONS

1. ,

,

I J

IM JN

KL MN

∠ ≅ ∠

≅

1. Given

2. M N∠ ≅ ∠ 2. Corresponding Angles Theorem (Thm. 3.1)

3. KL MN= 3. Definition of congruent segments

4. KL LMLM MN

+= +

4. Addition Property of Equality

5. and KM KL LM

LN LM MN

= +

= +

5. Segment Addition Postulate (Post. 1.2)

6. KM LN= 6. Substitution Property of Equality

7. KM LN≅ 7. Definition of segment congruence

8. IKM LJN≅ 8. AAS Congruence Theorem (Thm. 5.11)

Answers


6. Draw RT . By the definition of a parallelogram,

|| and || .RS QT QR TS QTR SRT∠ ≅ ∠ and

QRT STR∠ ≅ ∠ by the Alternate Interior Angles

Theorem (Thm. 3.2). RT TR≅ by the Reflexive Property of Segment Congruence (Thm. 2.1).

QRT STR≅ by the ASA Congruence Theorem

(Thm. 5.10). So, QR TS≅ and RS QT≅

because corresponding parts of congruent triangles are congruent.


1. 12

any real value except 2; or 0;m m m= = − =

The triangles are congruent by AAS or ASA.

2. Sample answer I: Label point ( )1, 6 as A, point

(5, 4) as B, point (4, 3) as C, point (1, 0) as D, and point (5, 2) as E. You can calculate

4.472, 1.414,AB DE CE CB= ≈ = ≈ and

4.242.AC CD= ≈ So, ABC DEC≅ by

the SSS Congruence Theorem (Thm. 5.8).

Sample answer II:

Label point (1,6) as A, point (5,4) as B, point (4,3) as C, point (1,0) as D, and point (5,2) as E. 1.414CE CB= ≈ and 4.242.AC CD= ≈ Additionally, the slope of ,BD AE⊥ so, 90 .m ACB m DCE∠ = ∠ = ° By the SAS Congruence Theorem (Thm. 5.5), .ABC DEC≅

3. BC and AD are parallel, because their slopes are

equal, with AC being a transversal. The Alternate Interior Angles Theorem (Thm. 3.2) applies to

make .BCA CAD∠ ≅ ∠ AB and CD are parallel

because their slopes are equal, with AC being a transversal. The Alternate Interior Angles Theorem (Thm. 3.2) applies to make .BAC ACD∠ ≅ ∠

AC AC≅ by the Reflexive Property of Segment Congruence (Thm. 2.1). So, ABC CDA≅ by

the ASA Congruence Theorem (Thm. 5.10).

4.

5.

x

y

2

−2

2−2

x

y

4

6

2

4 6

(1, 6)

(5, 4)

(4, 3)

(1, 0)

(5, 2)

STATEMENTS REASONS

1. HB EB

BHG BEA

HGJ EAD

JGB DAB

≅∠ ≅ ∠∠ ≅ ∠∠ ≅ ∠

1. Given

2. m BHG m BEA

m HGJ m EAD

m JGB m DAB

∠ = ∠∠ = ∠∠ = ∠

2. Definition of congruency

3. m HGJ m JGB

m HGB

∠ + ∠= ∠

3. Angle Addition Postulate (Post. 1.4)

4. m EAD m DAB

m EAB

∠ + ∠= ∠

4. Angle Addition Postulate (Post. 1.4)

5. m EAD m DAB

m HGB

∠ + ∠= ∠

5. Substitution

6. m HGB m EAB∠ = ∠ 6. Substitution

7. HGB EAB∠ ≅ ∠ 7. Definition of congruency

8. BHG BEA≅ 8. AAS Congruence Theorem (Thm. 5.11)

STATEMENTS REASONS

1. ABC ABDFCA EDA

≅∠ ≅ ∠ 1. Given

2. CA DA≅ 2. Corresponding parts of congruent triangles are congruent.

3. CAF DAE∠ ≅ ∠ 3. Vertical Angles Congruence Theorem (Thm. 2.6)

4. CAF DAE≅ 4. ASA Congruence Theorem (Thm. 5.10)

Answers


A55

6.

5.6 Puzzle Time

A WATCH

5.7 Start Thinking

Answer should include, but is not limited to: Any construction of a small triangle (preferably small enough to fit on a notebook page) with labels for side lengths and angle measurements.

5.7 Warm Up

1. Transitive Property of Segment Congruence

2. Reflexive Property of Angle Congruence

3. Symmetric Property of Angle Congruence

4. Reflexive Property of Segment Congruence

5. Symmetric Property of Segment Congruence

6. Transitive Property of Angle Congruence


1. , ,ABC CBA B∠ ∠ ∠ 2. , , 2JKL LKJ∠ ∠ ∠

3. , ,HMN NMH M∠ ∠ ∠ 4. , ,MPR RPM P∠ ∠ ∠

5.7 Practice A

1. The figure shows that , ,DAC E B C∠ ≅ ∠ ∠ ≅ ∠

and .AB DC≅ So, ACD EBA≅ by the AAS

Congruence Theorem (Thm. 5.11) and EB AC≅ because corresponding parts of congruent triangles are congruent.

2. The figure shows that ACB DCB∠ ≅ ∠ and

.AC DC≅ Use the Reflexive Property of Segment Congruence (Thm. 2.1) to show

.BC BC≅ Then ABC DBC≅ by the SAS

Congruence Theorem (Thm. 5.5). So, A D∠ ≅ ∠because corresponding parts of congruent triangles are congruent.

3. Show that PQRS is a parallelogram by definition, then show QPS RSP∠ ≅ ∠ by the Alternate

Interior Angles Theorem (Thm. 3.2). Show PS SP≅

by the Reflexive Property of Segment Congruence

(Thm. 2.1). The figure shows .PQ SR≅ So,

PQS SRP≅ by the SAS Congruence Theorem

(Thm. 5.5), and PR SQ≅ because corresponding

parts of congruent triangles are congruent.

4. The figure shows that HI JI≅ and .HK JK≅ Use the Reflexive Property of Segment Congruence

(Thm. 2.1) to show that .IK IK≅ Then HIK JIK≡ by the SSS Congruence Theorem

(Thm. 5.8), and H J∠ ≅ ∠ because corresponding parts of congruent triangles are congruent.

5. Place a stake at L so that .LM MN⊥ Find K, the

midpoint of .ML Locate the point J so that

,ML JL⊥ and J, K, and N are collinear. Then find JL; Given: M∠ and L∠ are right angles, K is the

midpoint of .ML Paragraph proof: Because they are right angles, .M L∠ ≅ ∠ MK LK= by the definition of the midpoint of a segment.

JKL NKM∠ ≅ ∠ by the Vertical Angles

Congruence Theorem (Thm. 2.6). JKL NKM≅

by the ASA Congruence Theorem (Thm. 5.10). LJ MN= because corresponding parts of congruent triangles are congruent.

6. 5DE = ; Sample answer: 44m C∠ = ° by the Triangle Sum Theorem (Thm. 5.1), so you have

, , andA D C F AC DF∠ ≅ ∠ ∠ ≅ ∠ ≅ .

ABC DEF≅ by the ASA Congruence

Theorem (Thm. 5.10). Therefore, DE AB= because corresponding parts of congruent triangles are congruent.

STATEMENTS REASONS

1. AE BF

CE DF

AB CD≅

1. Given

2. EAC FBD∠ ≅ ∠ 2. Corresponding Angles Theorem (Thm. 3.1)

3. ECA FDB∠ ≅ ∠ 3. Corresponding Angles Theorem (Thm. 3.1)

4. BC BC≅ 4. Reflexive Property of Segment Congruence (Thm. 2.1)

5. AB BC AC+ = 5. Segment Addition Postulate (Post. 1.2)

6. CD BC DB+ = 6. Segment Addition Postulate (Post. 1.2)

7. AB BC

CD BC

+

= +

7. Addition Property of Equality

8. AC DB= 8. Substitution

9. AC DB≅ 9. Definition of congruent segments

Answers


5.7 Practice B

1. The figure shows that ,G J∠ ≅ ∠ and .GI JH=Because they are vertical angles, .GKI JKH∠ ≅ ∠

So, GKI JKH≅ by the AAS Congruence

Theorem (Thm. 5.11) and GK JK≅ because corresponding parts of congruent triangles are congruent.

2. The figure shows that BE CE= and .AD CB⊥Because they are right angles, .AEB AEC∠ ≅ ∠

AE AE≅ by the Reflexive Property of Segment Congruence (Thm. 2.1). AEB AEC≅ by the

SAS Congruence Theorem (Thm. 5.5). Because corresponding parts of congruent triangles are congruent, .BA CA=

3. The figure shows that ,BA FG≅ ,A G∠ ≅ ∠ and

B∠ and F∠ are right angles. So, show that

ABC GFE≅ by the ASA Congruence

Theorem (Thm. 5.10). Then BCA FEG∠ ≅ ∠because corresponding parts of congruent triangles are congruent, which leads to congruence of their respective vertical angles, .DCH DEH∠ ≅ ∠

Then DH DH≅ by the Reflexive Property of Segment Congruence (Thm. 2.1) and right angles

are formed by ,DH CE⊥ so CDH EDH≅ by the AAS Congruence Theorem (Thm. 5.11).

So, DC DE≅ because corresponding parts of congruent triangles are congruent.

4. Use the Converse of the Base Angles Theorem

(Thm. 5.7) to show that .SV SU≅ Show that RVS TUS∠ ≅ ∠ because they form linear pairs

with congruent angles. Then RVS TUS≅ by

the SAS Congruence Theorem (Thm. 5.5), which leads to 1 2∠ ≅ ∠ because corresponding parts of

congruent triangles are congruent.

5. a. Sample answer: Mark point C along the edge of

your roof. Then find the midpoint D of .AC

Locate point E so that AC CE⊥ and E, D, and

B are collinear. Measure .CE This is the same as AB.

b. Sample answer: The method works because you have right angles, congruent segments, and vertical angles which lead to ABD CED≅ by the ASA Congruence Theorem (Thm. 5.10),

and CE corresponds to AB in these congruent triangles.


1. 2; 5x x= = 2. 0; 4x x= =

3. 20, 120, 6x y z= = = ±

4. In ABC and ,DEF you can use the Distance

Formula to find that 5 units,AB DE= =5.385,BC EF= ≈ and 10 units.AC FD= =

So, you can conclude that ABC DEF≅ by

the SSS Congruence Theorem (Thm. 5.8). So, A D∠ ≅ ∠ because corresponding parts of

congruent triangles are congruent.

5.

STATEMENTS REASONS

1. is the midpoint

of .

L

JN

PJ QN

PL QL

≅

≅

1. Given

2. JL LN≅ 2. Definition of midpoint

3. PLJ PLN≅ 3. SSS Congruence Theorem (Thm. 5.8)


PKJ QMN∠ ∠ 4. Given

5. PKJ QMN∠ ≅ ∠ 5. Right Angles Congruence Theorem (Thm. 2.3)

6. KJP MNQ∠ ≅ ∠ 6. Corresponding parts of congruent triangles are congruent.

7. PKJ QMN≅ 7. AAS Congruence Theorem (Thm. 5.11)

8. MQN KPJ∠ ≅ ∠ 8. Corresponding parts of congruent triangles are congruent.

A B

EC

D

Answers


A57

6.

5.7 Puzzle Time

TO KEEP THEIR INSIDES IN

5.8 Start Thinking

Answer should include, but is not limited to: Any construction using dynamic geometry software consisting of ABC with whole-number degree angle measures, centered at the origin. A second triangle, larger or smaller, should also be centered at the origin and contain the same angle measures. The two triangles both have the same angle measures but are not the same size, so that they are not congruent.

5.8 Warm Up

1. 11.7 2. 3.2 3. 18

4. 7.3 5. 12.6 6. 21.8


1. 30 EF+ 2. 4 GH• 3. AB JK−

5.8 Practice A

1.

It is easy to find the width and length of the rectangle by placing a vertex at the origin.

2.

It is easy to see the length of each leg by placing the vertex with the right angle at the origin.

3.

It is easy to see the width and length of the rectangle by placing a vertex at the origin.

4.

It is easy to see the length of the base and the height, and to determine that the triangle is isosceles by placing one vertex at the origin.

STATEMENTS REASONS

1.

2 3

R S∠ ≅ ∠∠ ≅ ∠

1. Given

2. TV TV≅ 2. Reflexive Property of Segment Congruence (Thm. 2.1)

3. RTV SVT≅ 3. AAS Congruence Theorem (Thm. 5.11)

4. RT SV≅ 4. Corresponding parts of congruent triangles are congruent.

5. 5 6∠ ≅ ∠ 5. Vertical Angles Congruence Theorem (Thm. 2.6)

6. RTU SVU≅ 6. AAS Congruence Theorem (Thm. 5.11)

7. RU SU≅ 7. Corresponding parts of congruent triangles are congruent.

x

y4

−2

42 6

(0, 2) (6, 2)

(6, 0)(0, 0)

x

y

(0, w)

( , 0)(0, 0)

( , w)

x

y

4

2

42 6

(0, 4)

(0, 0)(4, 0)

x

y

(b, 0)

, h)(

(0, 0)

b2

Answers


5.

( )

2 24, , midpoint of :

, 12

2, undefined, midpoint of :

, 1

, 0, midpoint of : , 02

AB a mAB ABa

a

BC mAB AB

a

aAC a mAB AB

= + =

= =

= =

ABC is a right triangle because AC is horizontal

and BC is vertical.

When 2,a ABC= is also isosceles because

then 2.AC BC= =

6.

2 2

, undefined, midpoint of :

0, 2

, , midpoint of :

, 2 2

, 0, midpoint of : , 02

JK a mJK JK

a

aKL a b mKL KL

b

b a

bJL b mJL JL

= =

= + = −

= =

JKL is a right triangle because JL is horizontal

and JK is vertical. JKL is not isosceles because .a b≠

7. ( ) ( ) 2 20, 0 , , 0 ;O C h OB h k= +

8. ( ) ( ) ( ) 2

2 2

2 , 0 , 0, 0 , 2, 4 ; 4 4 1,

2 4 2 1

G h O D k FD k

DE k h h

− = +

= + − +

9.

( )

( ) ( )

( ) ( )

2 2 2 2

2 2

Coordinate proof:

Segments and have the same length.

0 ; 0


0 ; 0


0 0 ;

0 0

CO DO

CO h h DO h h

AC BD

AC k k BD k k

AO BO

AO h k h k

DO h k h

= − − = = − =

= − = = − =

= − − + − = +

= − + − = 2 2

by the SSS Congruence

Theorem (Thm. 5.8).

k

ACO BDO

+

≅

10. Sample answer: yes; In this position, it is relatively easy to find the lengths of the horizontal base and the side adjacent to the base with its vertex at ( )0, 0 .

11. Sample answer: ( ) ( ) ( )0, 0 , 0, 3 , 4 , 0 ; 5R S a T a a

5.8 Practice B

1.

It is easy to find the width and length of the rectangle by placing a vertex at the origin with one side on each axis.

2.

It is easy to see the length of one leg by placing a vertex at the origin with a side on the horizontal axis.

3.

It is easy to see the length of one side by placing a vertex at the origin with a side on the horizontal axis.

x

y

C(a, 0)

B(a, 2)

A(0, 0)

2

4

x

y

L(b, 0)J(0, 0)

K(0, a)

x

y

(0, k) (2k, k)

(0, 0) (2k, 0)

x

y

(3, 0)

(3, k)

(0, 0)

2 4 6

x

y

(0, 0) (a, 0)

(b, c)

Answers


A59

4.

( )2 2

2 2

2 , 0, midpoint of : , 0

, , midpoint of :

, 2 2

, , midpoint of :

3,

2 2

JL a mJL JL a

bJK a b mJK JK

a

a b

bKL a b mKL KL

a

a b

= =

= + =

= + = −

JKL is isosceles, because .JK KL≅

5.

( )

( )

( )

5 , 0, midpoint of : 2.5 , 0

14 5, , midpoint of : 4 , 2

24

5 , , midpoint of : 6.5 , 23

PQ a mPQ PQ a

PR a mPR PR a a

QR a mQR QR a a

= =

= =

= =

PQR is isosceles, because .PQ QR≅

6. ( )0, 0 ,F ( )0, ,G k ( )2 , ,H k k− ( ), 0 ;J k

2 2,GH k= 5FH k=

7. ( )0, 0 , , 2 ,2

kA B k

3, 2 ,

2

kC k

( ), 0 ;E k ,BC k=

117

2CD k=

8. parallelogram; 5

,2

mWX = − 1,

2mXY =

5,

2mYZ = − and

1.

2mZW = Because both pairs

of opposite sides are parallel, WZYZ is a parallelogram by definition. It is not a trapezoid because it has two pairs of parallel sides instead of one. It is not a rectangle because the slopes are not negative reciprocals of each other, which indicates that the sides are not perpendicular.

9. Given: Vertex C of ABC is on the line the perpendicular bisector of .AB

Prove: ABC is isosceles.

AC and BC have the lengths

( )2 2 2 2 ,AC a b a b= − + = +

2 2 .BC a b= +

So, ,AC BC= and by the definition of congruent

segments, .AC BC≅ Therefore, ABC is isosceles by the definition of isosceles triangle.


1. Call the point S the midpoint of ,RP with the

coordinates , .2 2

b a a bS

− + + =

The slope of

( ),

b aRP

b a

− −=

− − and the slope of ,

a bSQ

b a

− −=−

making them opposite reciprocals and forming the

90° angles RSQ and PSQ. SQ SQ≅ by the

Reflexive Property of Segment Congruence

(Thm. 2.1). The length of both RS and SP is equal

to 2 2

,2 2

a b b a+ − +

so RSQ PSQ≅

by the SAS Congruence Theorem (Thm. 5.5). The

slope of a

RQb

= − and the slope of .b

QPa

=

They are opposite reciprocals and form the right angle PQR.

0,x =

x

y

J(0, 0) L(2a, 0)

K(a, b)

x

y

P(0, 0) Q(5a, 0)

R(8a, 4a)

x

y

A(−a, 0) B(a, 0)

C(0, b)

Answers


2. The slope of CD is equal to the slope of ,AB but .AD BC≠ So, it is a trapezoid, but not an

isosceles trapezoid.

3. a. coordinates of : , ;2 2

s tP

coordinates of

: ,2 2

s r tQ

+

b. equation of ( );2

tPR y x r

s r= = −

−

equation of t

QO y xr s

= =+

4. The trapezoid is isosceles because .TR PA≅ The base is on the x-axis, and the y-axis bisects the bases. Label point ( )0, 2 ,E c= point ( )0, 0 ,G =

point ( ), ,D b a c= − − and point ( ), .F b a c= +You can then show that

( )22 .DE EF FG DG c b a= = = = + +

All sides are equal, making DEFG a rhombus.

5. ( ),b b− 6. ( )0, a

7. ( ) ( ), or ,a b b a b b− − −

8. ( ),b a b−

5.8 Puzzle Time

AIR BAG

Cumulative Review

1. 3

, 32

M −

2. 5 3,

2 2M −

3. 5, 2

2M

4. ( )2, 0M 5. ( ) 5, 4M 6. 11, 1

2M

7. 7

, 12

M −

8. 31,

2M −

9. 9, 1

2M −

10. 1

0,2

M −

11. ( )1, 3M − 12. 5, 2

2M

13. 5 , 0

2M −

14. 15, 4

2M −

15. 3, 2

2M −

16. 15 9,

2 2M

17. 1

7,2

M − −

18. ( )5, 6M

19. ( )5, 3M − 20. 3 3

,2 2

M

21. 8.54

22. 12.37 23. 10.44 24. 13.34

25. 7.21 26. 6.32 27. 15.03

28. 6.32 29. 12.53 30. 5

31. 7 32. 10 33. 8.25

34. 6.40 35. 13.60 36. 7.62

37. 6.71 38. 7.62 39. 12.53

40. 15.81 41. 324 in. 42. 3234 in.

43. 21 44. 11 45. 18

46. 9 47. 10 48. 20

49. 8 50. 34 51. 2750 ft

52. a. 2216 ft

b. 2 containers

c. $31.98

53. 87° 54. 69° 55. 74°

56. 61° 57. 115° 58. 78°

59. 20x = 60. 6x = − 61. 5x = −

62. 9x = − 63. 8x = 64. 4x =

65. 2x = − 66. 7x = 67. 9x =

68. 11x = 69. 6x = 70. 5x =

71. 5x = 72. 4x = 73. 3x = −

74. 14x = 75. 1x = 76. 10x =

77. 6x = 78. 4x = 79. 4x = −

80. 6x = − 81. 32x = − 82. 8x = −

83. 3 9y x= − + 84. 7 11y x= −

85. 8 13y x= − 86. 3 6y x= −

87. 2 6y x= − + 88. 5 7y x= −

89. 0.125 3y x= − + 90. 12 8y x= +

91. 0.125 8y x= − 92. 2 24y x= − −

Answers


A61

93. 94

24y x= − 94. 8 2y x= −

95. a. 2224 ft b. 12 ft by 14 ft c. 2168 ft

96. a. 10 ft b. 24 ft c. 224 ft

Chapter 6 6.1 Start Thinking

The roof lines become steeper; The two top chords will get longer as the king post gets longer, but the two top chords will always be congruent to each other. The angles formed by the top chords and the king post are congruent. The angles formed by the top chords and the bottom chord are congruent.

6.1 Warm Up

1. 5 2. 8 3. 2 4. 1−


1.

2.

6.1 Practice A

1. P lies on the perpendicular bisector of ;RS The

markings show that TP

satisfies the definition of a perpendicular bisector.

2. P lies on the perpendicular bisector of ;RS Because

T is equidistant from the endpoints R and S, T lies

on the perpendicular bisector of RS by the Converse of the Perpendicular Bisector Theorem (Thm. 6.2). Because only one line can be

perpendicular to RS at U, TU

must be the

perpendicular bisector of ,RS and P is on .TU

3. P lies on the angle bisector of ;DEF∠ P is

equidistant from sides ED

and EF

of angle ,DEF∠ so P is on the angle bisector of DEF∠ by

the Converse of the Angle Bisector Theorem (Thm. 6.4).

4. 20; Because D is on the perpendicular bisector of

,AC D is equidistant from A and C.

5. 17; By the Perpendicular Bisector Theorem (Thm. 6.1), .GJ GH= Solving 4 5 2 11x x+ = + gives 3,x = so 2 11 17.x + =

6. 14; Q is on the angle bisector of ,PSR∠ so Q is

equidistant from and .SP SR

7. 76 ;° By the Converse of the Angle Bisector

Theorem (Thm. 6.4), GE

bisects .DGF∠

Solving 3 8 5 12x x+ = − gives 10.x = So,

( ) ( )3 10 8 5 10 12 76 .m DGF∠ = + + − ° = °

8. 4 7y x= − +

9. Because any point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment, you can draw an isosceles triangle by drawing segments from each endpoint of the segment to the same point on the perpendicular bisector.

10. yes; In a right triangle, the bisector of the right angle is also the perpendicular bisector of the hypotenuse when the right angle is isosceles. The figure shows that when right of C ABC∠

is bisected by , CD ACD BCD≅ by either

the ASA Congruence Theorem (Thm. 5.10) or the SAS Congruence Theorem (Thm. 5.5). Then the corresponding sides

andAC BC must be

congruent, so ABC is isosceles.

STATEMENTS REASONS

1. P is the midpoint

of MN and .TQ

1. Given

2. MP NP≅ 2. Definition of segment midpoint

3. PT PQ≅ 3. Definition of segment midpoint

4. MPQ NPT∠ ≅ ∠ 4. Vertical Angles Congruence Theorem (Thm. 2.6)

5. MQP NTP≅ 5. SAS Congruence Theorem (Thm. 5.5)

STATEMENTS REASONS

1. ,AB DC≅

AC DB≅

1. Given

2. BC BC≅ 2. Reflexive Property of Segment Congruence (Thm. 2.1)

3. ABC DCB≅ 3. SSS Congruence Theorem (Thm. 5.8)

A

C B

D

geometry rbc a answers - mr....

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