hscc geom wsk 05 · 2015. 10. 18. · m∠ A + m∠B + m∠C = 180° m∠C = 180° − 32° = 148° m∠A + m∠B + 148 = 180° m∠A + m∠B = 32° The sum of the measures of the

Copyright © Big Ideas Learning, LLC Geometry 141All rights reserved. Worked-Out Solutions

Chapter 5

Chapter 5 Maintaining Mathematical Profi ciency (p. 229)

1. M ( −4 + 0 — 2 , 1 + 7 —

2 ) = M(−2, 4)

AQ = √——

[ 0 − (−4) ] 2 + (7 − 1)2

= √—

(4)2 + (6)2 = √—

16 + 36 = √—

52 ≈ 7.2 units

2. M ( 3 + 9 — 2 , 6 + (−2) —

2 ) = M ( 12 —

2 , 4 —

2 ) = M(6, 2)

GH = √——

(9 − 3)2 + (−2 − 6)2

= √——

(6)2 + (−8)2 = √—

36 + 64 = √—

100 = 10 units

3. M ( −1 + 8 — 2 , −2 + 0 —

2 ) = M ( 7 —

2 , −2 —

2 ) = M ( 7 —

2 , −1 )

UV = √———

[ 8 − (−1) ] 2 + [ 0 − (−2) ] 2

= √——

(8 + 1)2 + (2)2 = √—

92 + 4

= √—

81 + 4 = √—

85 ≈ 9.2 units

4. 7x + 12 = 3x

7x − 7x + 12 = 3x − 7x

12 = − 4x

−3 = x

The solution is x = −3.

5. 14 − 6t = t

14 − 6t + 6t = t + 6t

14 = 7t

2 = t

The solution is t = 2.

6. 5p + 10 = 8p + 1

5p + 10 − 10 = 8p + 1 − 10

5p = 8p − 9

5p − 8p = 8p − 9 − 8p

−3p = −9

p = 3

The solution is p = 3.

7. w + 13 = 11w − 7

w + 13 − w = 11w − 7 − w

13 = 10w − 7

13 + 7 = 10w − 7 + 7

20 = 10w

2 = w

The solution is w = 2.

8. 4x + 1 = 3 − 2x

4x + 1 + 2x = 3 − 2x + 2x

6x + 1 = 3

6x + 1 − 1 = 3 − 1

6x = 2

6 — 6 x = 2 — 6

x = 1 — 3

The solution is x = 1 — 3 .

9. z − 2 = 4 + 9z

z − 2 + 2 = 4 + 9z + 2

z = 6 + 9z

z − 9z = 6 + 9z − 9z

−8z = 6

−8 — −8

z = 6 — −8

z = 6 — −8

z = − 3 —

4

The solution is z = − 3 — 4 .

10. yes; The length can be found using the Pythagorean Theorem.

Chapter 5 Mathematical Practices (p. 230) 1. theorem; It is the Slopes of Perpendicular Lines Theorem

(Thm. 3.14) studied in Section 3.5.

2. theorem; It is the Linear Pair Perpendicular Theorem (Thm. 3.10) studied in Section 3.4.

3. defi nition; This is the defi nition of perpendicular lines.

4. postulate; This is the Two Point Postulate (Post. 2.1) studied in Section 2.3. In Euclidean geometry, it is assumed, not proved, to be true.

5.1 Explorations (p. 231) 1. a. Check students’ work.

b. Check students’ work.

c. The sum of the interior angle measures of all triangles is 180°.

d. Check students’ work; The sum of the measures of the interior angles of a triangle is 180°.

2. a. Check students’ work.


c. Check students’ work.

d. Check students’ work; The sum is equal to the measure of the exterior angle.

e. Check students’ work; The measure of an exterior angle is equal to the sum of the measures of the two nonadjacent interior angles.

142 Geometry Copyright © Big Ideas Learning, LLC Worked-Out Solutions All rights reserved.

Chapter 5

3. The sum of the measures of the interior angles of a triangle is 180°, and the measure of an exterior angle of a triangle is equal to the sum of the measures of the two nonadjacent interior angles.

4. m∠ A + m∠B + m∠C = 180° m∠C = 180° − 32° = 148° m∠A + m∠B + 148 = 180° m∠A + m∠B = 32° The sum of the measures of the two nonadjacent interior

angles is 32° and the third angle which is adjacent tot he exterior angle has a measure of 180° − 32° = 148°. These are known because of the conjectures made in Explorations 1 and 2.

5.1 Monitoring Progress (pp. 232–235) 1. Sample answer: Obtuse isosceles triangle:

Acute scalene triangle:

2. AB = √——

(0 − 3)2 + (0 − 3)2 = √—

9 + 9 = √—

18 ≈ 4.2

BC = √——

[ 3 − (−3) ] 2 + (3 − 3)2 = √—

(3 + 3)2 = √—

62 = 6

AC = √——

[ 0 − (−3) ] 2 + (0 − 3)2 = √—

32 + 32 = √—

9 + 9

= √—

18 ≈ 4.2

Because AC = AB, that indicates that △ABC is isosceles.

Slope of — AB = 3 − 0 — 3 − 0

= 3 — 3 = 1

Slope of — BC = 3 − 3 — −3 − 3

= 0 — −6

= 0

Slope of — AC = − 3 − 0 —

3 − 0 = −3 —

3 = −1

Because the product of the slopes of — AB and — AC equals −1, that indicates that — AB ⊥ — AC , therefore △ABC is a right triangle. So, △ABC is a right isosceles triangle.

3. 40° + 3x° = (5x − 10)° 50 = 2x

x = 25

m∠1 + 3(25°) + 40° = 180° m∠1 + 75° + 40° = 180° m∠1 + 115° = 180° m∠1 = 65°

4. 2x° + (x − 6)° = 90° 3x − 6 = 90

3x = 96

x = 32

2x° = 2(32°) = 64° (x − 6)° = 32° − 6° = 26° The angle measurements are 26° and 64°.

5.1 Exercises (pp. 236–238)

Vocabulary and Core Concept Check 1. no; By the Corollary to the Triangle Sum Theorem (Cor.

5.1), the acute angles of a right triangle are complementary. Because their measures have to add up to 90°, neither angle could have a measure greater than 90°.

2. The measure of an exterior angle of a triangle is equal to the sum of the measures of the two nonadjacent interior angles.

Monitoring Progress and Modeling with Mathematics 3. Two sides are congruent and one angle is a right angle.

So, △XYZ is a right isosceles triangle.

4. All sides are congruent and therefore, all angles are congruent. So, △LMN is an equiangular equivalent triangle.

5. None of the sides are congruent and one angle is obtuse. So, △JKH is an obtuse scalene triangle.

6. None of the sides are congruent and all angles are acute. So, △ABC is an acute scalene triangle.

7. AB = √——

(6 − 2)2 + (3 − 3)2 = √—

42 = 4

BC = √——

(2 − 6)2 + (7 − 3)2 = √—

(−4)2 + 42

= √—

16 + 16 = √—

32 ≈ 5.7

AC = √——

(2 − 2)2 + (7 − 3)2 = √—

42 = 4

△ABC is isosceles because AB = AC.

Slope of AB = 3 − 3 — 6 − 2

= 0 — 4 = 0

Slope of BC = 7 − 3 — 2 − 6

= 4 — −4

= −1

Slope of AC = 7 − 3 — 2 − 2

= 4 — 0 = undefi ned

Because — AB has a slope of 0 and — AC has an undefi ned slope, — AB ⊥ — AC . There is a right angle at ∠A, which makes △ABC a right triangle. So, △ABC is a right isosceles triangle.


Chapter 5

8. AB = √——

(6 − 3)2 + (9 − 3)2 = √—

32 + 62 = √—

9 + 36

= √—

45 ≈ 6.7

BC = √——

(6 − 6)2 + [ 9 − (−3) ] 2 = √——

(0)2 + (9 + 3)2

= √—

122 = 12

AC = √——

(6 − 3)2 + [ 3 − (−3) ] 2 = √——

32 + (3 + 3)2

= √—

9 + 36 = √—

45 ≈ 6.7

△ABC is isosceles because AB = AC.

Slope of AB = 9 − 3 — 6 − 3

= 6 — 3 = 2

Slope of BC = −3 − 9 — 6 − 6

= −12 — 0 = undefi ned

Slope of AC = −3 − 3 — 6 − 3

= −6 — 3 = −2

None of the slopes of the sides of the triangle are negative reciprocals of each other. So, the triangle is not a right triangle. △ABC is an isosceles triangle.

9. AB = √——

(4 − 1)2 + (8 − 9)2 = √—

32 + (−1)2 = √—

9 + 1

= √—

10 ≈ 3.2

BC = √——

(2 − 4)2 + (5 − 8)2 = √——

(−2)2 + (−3)2 = √—

4 + 9

= √—

13 ≈ 3.6

AC = √——

(2 − 1)2 + (5 − 9)2 = √—

12 + (−4)2

= √—

1 + 16 = √—

17 ≈ 4.1

△ABC is a scalene triangle.

Slope of AB = 8 − 9 — 4 − 1

= −1 — 3

Slope of BC = 5 − 8 — 2 − 4

= −3 — −2

= 3 — 2

Slope of AC = 5 − 9 — 2 − 1

= −4 — 1 = −4

None of the slopes of the sides of the triangle are negative reciprocals of each other. So, the triangle is not a right triangle. △ABC is a scalene triangle.

10. AB = √———

[ 0 − (−2) ] 2 + (−3 − 3)2 = √—

22 + (−6)2

= √—

4 + 36 = √—

40 ≈ 6.3

BC = √———

(3 − 0)2 + [ −2 − (−3) ] 2 = √——

(3)2 + (−2 + 3)2

= √—

9 + 1 = √—

10 ≈ 3.2

AC = √———

[ 3 − (−2) ] 2 + (−2 − 3)2 = √——

(3 + 2)2 + (−5)2

= √—

25 + 25 = √—

50 ≈ 7.1

△ABC is a scalene triangle.

Slope of AB = −3 − 3 — 0 − (−2)

= −6 — 2 = −3

Slope of BC = −2 − (−3) — 3 − 0

= −2 + 3 — 3 = 1 —

3

Slope of AC = −2 − 3 — 3 − (−2)

= −5 — 3 + 2

= −5 — 5 = −1

Because AB has a slope of −3 and BC has a slope of 1 — 3 , — AB ⊥ — BC . There is a right angle at ∠B, which makes △ABC a right triangle. So, △ABC is a right scalene triangle.

11. m∠1 + 78° + 31° = 180° m∠1 + 109° = 180° m∠1 = 71° The triangle is an acute triangle.

12. m∠1 + 30° + 40° = 180° m∠1 + 70° = 180° m∠1 = 110° The triangle is an obtuse triangle.

13. m∠1 + 38° + 90° = 180° m∠1 + 128° = 180° m∠1 = 52° The triangle is a right triangle.

14. m∠1 + 60° + 60° = 180° m∠1 + 120° = 180° m∠1 = 60° The triangle is an equiangular triangle.

15. m∠2 = 75° + 64° = 139°

16. x° + 45° = (2x − 2)° x = 2x − 47

−x = −47

x = 47

(2x − 2)° = 2 ⋅ 47 − 2 = 94 − 2 = 92

The exterior angle has a measure of 92°.

17. 24° + (2x + 18)° = (3x + 6)° 42 + 2x = 3x + 6

2x = 3x −6

−x = −36

x = 36

(3x + 6)° = 3 ⋅ 36 + 6 = 114


18. (x + 8)° + 4x° = (7x − 16)° 5x + 8 = 7x − 16

−2x + 8 = −16

−2x = −24

x = 12

(7x − 16)° = 7 ⋅ 12 − 16 = 84 − 16 = 68


19. 3x° + 2x° = 90° 5x = 90

x = 18

3x° = 3 ⋅ 18 = 54

2x° = 2 ⋅ 18 = 36

The two acute angles measure 36° and 54°.


Chapter 5

20. (3x + 2)° + x° = 90° 4x + 2 = 90

4x = 88

x = 22

(3x + 2)° = 3 ⋅ 22 + 2 = 68


21. (11x − 2)° + (6x + 7)° = 90° 17x + 5 = 90

17x = 85

x = 5

(11x − 2)° = 11 ⋅ 5 − 2 = 55 − 2 = 53

(6x + 7)° = 6 ⋅ 5 + 7 = 30 + 7 = 37


22. (19x − 1)° + (13x − 5)° = 90° 32x − 6 = 90

32x = 96

x = 3

(19x − 1)° = 19 ⋅ 3 − 1 = 57 − 1 = 56

(13x − 5)° = 13 ⋅ 3 − 5 = 39 − 5 = 34


23. x° + 5x° = 90° 6x = 90

x = 15

5x° = 5 ⋅ 15 = 75


24. x° + 8x° = 90° 9x = 90

x = 10

8x° = 8 ⋅ 10 = 80


25. x° + [ 3(x + 8) ] ° = 90° x + 3x + 24 = 90

4x = 66

x = 16.5

3(x + 8)° = 3(16.5 + 8) = 3 ⋅ 24.5 = 73.5

The two acute angles measure 16.5° and 73.5°.

26. x° + [2(x − 12)]° = 90° x + 2x − 24 = 90

3x = 114

x = 38

[2(x − 12)]° = 2(38 − 12) = 2 ⋅ 26 = 52


27. The sum of the measures of the interior angles of the triangle is 180°, not 360°. 115° + 39° + m∠1 = 180°

154° + m∠1 = 180° m∠1 = 26°

28. The measure of the exterior angle should be equal to the sum of the measures of the two nonadjacent interior angles.

m∠1 = 80° + 50° = 130°

29. m∠1 = 90° − 40° = 50°

30. m∠2 = 180° − 50° = 130°

31. m∠3 = m∠1 = 50° 32. m∠4 = m∠2 = 130°

33. m∠5 = 90° − 50° = 40°

34. m∠6 = 180° − 40° = 140°

35. m∠7 = 90°

36. m∠8 = 180° − 40° = 140°

37. None of the sides and none of the angles of the triangle are equal, therefore it is a scalene triangle. When you measure the three angles of the triangle, you fi nd that they are all acute. Therefore the triangle is acute. So, the triangle is an acute scalene triangle.

38. The following sets of angle measures could form a triangle:

B. 96° + 74° + 10° = 180° D. 101° + 41° + 38° = 180° E. 90° + 45° + 45° = 180° F. 84° + 62° + 34° = 180°

39. 2 ⋅ 6 + x = 20 2x + 6 = 20

12 + x = 20 2x = 14

x = 8 x = 7

You could make another bend 6 inches from the fi rst bend and leave the last side 8 inches long, or you could make another bend 7 inches from the fi rst bend and then the last side will also be 7 inches long.

40. Sample answer:

3

12GO TEAM!

When a triangular pennant is on a stick, the two angles on the top edge of the pennant (∠1 and ∠2) should have measures such that their sum is equal to the measure of the angle formed by the stick and the bottom edge of the pennant (∠3).


Chapter 5

41. Given △ABC is a right triangle. A

C B

Prove ∠A and ∠B are complementary.

STATEMENTS REASONS

1. △ABC is a right triangle. 1. Given

2. ∠C is a right angle. 2. Given (marked in diagram)

3. m∠C = 90° 3. Defi nition of a right angle

4. m∠A + m∠B + m∠C = 180° 4. Triangle Sum Theorem (Thm. 5.1)

5. m∠A + m∠B + 90° = 180° 5. Substitution Property of Equality

6. m∠A + m∠B = 90° 6. Subtraction Property of Equality

7. ∠A and ∠B are complementary.

7. Defi nition of complementary angles

42. Given △ABC, exterior ∠BCD

A C D

B Prove m∠A + m∠B = m∠BCD

STATEMENTS REASONS

1. △ABC, exterior ∠BCD 1. Given

2. m∠A + m∠B + m∠BCA = 180°

2. Triangle Sum Theorem (Thm. 5.1)

3. ∠BCA and ∠BCD form a linear pair.

3. Defi nition of linear pair

4. m∠BCA + m∠BCD = 180° 4. Linear Pair Postulate (Post 2.8)

5. m∠A + m∠B + m∠BCA = m∠BCA + m∠BCD

5. Transitive Property of Equality

6. m∠A + m∠B = m∠BCD 6. Subtraction Property of Equality

43. It is possible to draw an obtuse isosceles triangle. Sample answer:

An obtuse equilateral triangle is not possible, because when two sides form an obtuse angle the third side that connects them must be longer than the other two.

44. It is possible to draw a right isosceles triangle. It will have angle measurements of 45°, 45°, and 90°.

A right equilateral triangle is not possible, because the hypotenuse must be longer than either leg in a right triangle.

45. a. AB + AB + BC = Perimeter

x + x + 2x − 4 = 32

4x − 4 = 32

4x = 36

x = 9

AB + AB + BC = Perimeter

x + 2x − 4 + 2x − 4 = 32

5x − 8 = 32

5x = 40

x = 8

b. AB + AB + BC = Perimeter

x + x + 2x − 4 = 12

4x − 4 = 12

4x = 16

x = 4

AB + AB + BC = Perimeter

x + 2x − 4 + 2x − 4 = 12

5x − 8 = 12

5x = 20

x = 4

There is one value for x, x = 4.

46. a. The triangle appears to have three congruent sides and three congruent angles. So, the triangle is equiangular, acute, equilateral, and isosceles.

b. The triangle appears to have two congruent sides, no right angles, and no obtuse angle. So, the triangle is acute and isosceles.

c. The triangle appears to have one obtuse angle and no congruent sides. So, the triangle is obtuse and scalene.

d. The triangle appears to have a right angle and no congruent sides. So, the triangle is right and scalene.


Chapter 5

47. Exterior angle = Sum of the two nonadjacent interior angles

A. 100° =? 62° + 38° B. 81° =

? 57° + 24°

100° = 100° ✓ 81° = 81° ✓

F. 149° =? 101° + 48°

149° = 149° ✓

48. no; According to the Exterior Angle Theorem (Thm. 5.2), the measure of an exterior angle in a triangle is always equal to the sum of the measures of the two nonadjacent interior angles.

49. By the Alternate Interior Angles Theorem (Thm. 3.2), x = 43. Use the Exterior Angle theorem (Thm. 5.2) to fi nd y.

75° = y° + 43° 32 = y

So, x = 43 and y = 32.

50. By the Corresponding Angle Theorem (Thm. 3.1), x = 118. Use the Exterior Angle Theorem (Thm. 5.2) to fi nd the value of y.

118° = y° + 22° 96 = y

So, x = 118 and y = 96.

51. The sum of the measures of the two acute angles of a right triangle is 90°.

y° + 25° = 90° y = 65

Use the Exterior Angle Theorem (Thm. 5.2) to fi nd the value of x.

x° = 65° + 20° = 85° So, x = 85 and y = 65.

52. The sum of the measures of the two acute angles of a right triangle is 90°.

x° + 64° = 90° x = 26

By the Alternate Interior Angles Theorem (Thm. 3.2), the unmarked angle of the triangle containing y° measures 26°. Because both triangles contain a right angle and an angle measuring 26°, y° must be 64°. So, x = 26 and y = 64.

53. Given ⃖ ''⃗ AB ) ⃖ ''⃗ CD

Prove m∠1 + m∠2 + m∠3 = 180°

A

B

D

C E1

2

3 4 5

STATEMENTS REASONS

1. ⃖ ''⃗ AB ) ⃖ ''⃗ CD 1. Given (marked in diagram)

2. ∠ACD and ∠5 form a linear pair.

2. Defi nition of linear pair

3. m∠ACD + ∠5 = 180° 3. Linear Pair Postulate (Post. 2.8)

4. m∠3 + m∠4 = m∠ACD 4. Angle Addition Postulate (Post. 1.4)

5. m∠3 + m∠4 + m∠5= 180°

5. Substitution Property of Equality

6. m∠1 ≅ m∠5 6. Corresponding Angle Theorem (Thm. 3.1)

7. m∠2 ≅ m∠4 7. Alternate Interior Angles Theorem (Thm 3.2)

8. m∠1 = m∠5, m∠2 = m∠4

8. Defi nition of congruent angles

9. m∠3 + m∠2 + m∠1= 180°


Maintaining Mathematical Profi ciency 54. (5x − 27)° = (3x + 1)° 2x − 27 = 1

2x = 28

x = 14

3(14) + 1 = 42 + 1 = 43

So, m∠KHL = 43°.

55. m∠ABC = m∠GHK

(6x + 2)° = (5x − 27)° + (3x + 1)° 6x + 2 = 8x − 26

−2x = −28

x = 14

6 ⋅ 14 + 2 = 86

So, m∠ABC = 86°.

56. 5y − 8 = 3y

2y = 8

y = 4

3y = 3 ⋅ 4 = 12

So, GH = 12.


Chapter 5

57. 3z + 6 = 8z − 9

−5z + 6 = −9

−5z = −15

z = 3

3z + 6 = 3 ⋅ 3 + 6 = 9 + 6 = 15

So, BC = 15.

5.2 Explorations (p. 239) 1. translation, refl ection, rotation; A rigid motion maps

each part of a fi gure to a corresponding part of its image. Because rigid motions preserve length and angle measure, corresponding parts of congruent fi gures are congruent. In congruent triangles, this means that the corresponding sides and corresponding angles are congruent, which is suffi cient to say that the triangles are congruent.

2. a. Sample answer: Refl ect △ABC in the x-axis and translate 3 units right.

b. Sample answer: Rotate △ABC 180° about the origin.

c. Sample answer: Rotate △ABC 270° counterclockwise about the origin and translate 3 units down.

d. Sample answer: Refl ect △ABC in the line y = x.

3. Look at the orientation of the original triangle and decide which rigid motion or composition of rigid motions will result in the same orientation as the second triangle. Then, if necessary, use a translation to move the fi rst triangle so that it coincides with the second.

4. Sample answer: Refl ect △ABC in the y-axis and translate 3 units right and 2 units down.

x

y4

−2

4−2

A B

C

DE

F

5.2 Monitoring Progress (pp. 241–242) 1. corresponding sides: — AB ≅ — CD , — BG ≅ — DE , — GH ≅ — EF ,

— AH ≅ — CF

corresponding angles: ∠A ≅ ∠C, ∠B ≅ ∠D, ∠G ≅ ∠E, ∠H ≅ ∠F

2. 4x + 5 = 105

4x = 100

x = 25

3. From the diagram, — PS ≅ — RQ , — PT ≅ — RT , and — ST ≅ — QT . Also, by the Vertical Angles Congruence Theorem (Thm. 2.6), ∠PTS ≅ ∠RTQ. From the diagram, — PS ) — RQ , and ∠P ≅ ∠R and ∠S ≅ ∠Q by the Alternate Interior Angles Theorem (Thm. 3.2). Because all corresponding parts are congruent △PTS ≅ △RTQ.

4. ∠NSR ≅ ∠NDC and ∠CND ≅ ∠RNS, so by the Third Angles Theorem (Thm. 5.4), ∠SRN ≅ ∠DCN. So, ∠DCN = 75°.

5. The additional information that is needed to conclude that △NDC ≅ △NSR is that — CD ≅ — RS or — DN ≅ — SN .


Vocabulary and Core Concept Check 1. To show that two triangles are congruent, you need to show

that all corresponding parts are congruent. If two triangles have the same side lengths and angle measures, then they must be the same size and shape.

2. “Is △JLK ≅ △STR?” is different. Because corresponding angles and sides are not congruent, △JLK is not congruent to △STR. for the other three questions, corresponding angles are congruent and corresponding sides are congruent. So, the triangles are congruent.

Monitoring Progress and Modeling with Mathematics 3. corresponding sides: — AB ≅ — DE , — BC ≅ — EF , — AC ≅ — DF corresponding angles: ∠A ≅ ∠D, ∠B ≅ ∠E, ∠C ≅ ∠F

Sample answer: △CBA ≅ △FED

4. corresponding sides: — GH ≅ — QR , — HJ ≅ — RS , — JK ≅ — ST , GK ≅ — QT corresponding angles: ∠G ≅ ∠Q, ∠H ≅ ∠R, ∠J ≅ ∠S, ∠K ≅ ∠TSample answer: GHJK ≅ QRST

5. ∠N ≅ ∠Y, so m∠Y = 124°.

6. ∠X ≅ ∠M, so m∠M = 33°.

7. ∠Z ≅ ∠L and m∠L = 180° − 124° − 33° = 23°. So, m∠Z = 23°.

8. — MN ≅ — XY , so XY = 8.

9. 135 = 10x + 65

70 = 10x

7 = x

4y − 4 = 28

4y = 32

y = 8

So, x = 7 and y = 8.


Chapter 5

10. m∠N = 180° − (142° + 24°) = 180° − 166° = 14° m∠N = m∠U

14° = (2x − 50)° 64 = 2x

x = 32

NP = US

2x − y = 13

2(32) − y = 13

64 − y = 13

−y = −51

y = 51

So, x = 32 and y = 51.

11. — VZ ≅ — KJ , — ZY ≅ — JN , — YX ≅ — NM , — XW ≅ — ML , — VW ≅ — KL

∠V ≅ ∠K, ∠Z ≅ ∠J, ∠Y ≅ ∠N, ∠X ≅ ∠M, ∠W ≅ ∠L

Because all corresponding parts of the polygons are congruent, VZYXW ≅ KJNML.

12. From the diagram — WX ≅ — YZ and — XY ≅ — ZW . By the Refl exive Property of Congruence (Thm. 2.1), — WY ≅ — WY . Also from the diagram, ∠X ≅ ∠Z. Then, from the markings in the diagram, — XY ) — ZW and — XW ) — ZY . You can conclude that ∠XYW ≅ ∠ZWY and ∠XWY ≅ ∠ZYW by the Alternate Interior Angles Theorem (Thm. 3.2). Because all corresponding parts are congruent, △WXY ≅ △YZW.

13. ∠L ≅ ∠Z and ∠N ≅ ∠Y, so by the Third Angles Theorem (Thm. 5.4), ∠1 ≅ ∠M. By the Triangle Sum Theorem (Thm. 5.1), m∠1 = 180° − 90° − 70° = 20°.

14. ∠B ≅ ∠Q and ∠A ≅ ∠S, so by the Third Angles Theorem (Thm. 5.4), ∠C ≅ ∠1. By the Triangle Sum Theorem (Thm. 5.1), m∠1 = 180° − 45° − 80° = 55°.

15. Given — AB ) — DC , — AB ≅ — DC , A B

E

CD

E is the midpoint of — AC and — BD .

Prove △AEB ≅ △CED

STATEMENTS REASONS

1. — AB ) — DC , E is the midpoint of — AC and — BD .

1. Given

2. ∠AEB ≅ ∠CED 2. Vertical Angles Congruence Theorem (Thm. 2.6)

3. ∠BAE ≅ ∠DCE, ∠ABE ≅ ∠CDE

3. Alternate Interior Angles Theorem (Thm. 3.2)

4. — AE ≅ — EC , — BE ≅ — DE 4. Defi nition of midpoint

5. △AEB ≅ △CED 5. All corresponding parts are congruent.

16. Prove △ABG ≅ △DCF B

G D

C

E

FA

STATEMENTS REASONS

1. — AB ≅ — DC , — AF ≅ — DG , — BE ≅ — CE ≅ — EF ≅ — EG , ∠B ≅ ∠C, ∠A ≅ ∠D

1. Given (marked in diagram)

2. ∠BGA ≅ ∠CFD 2. Third Angles Theorem (Thm. 5.4)

3. AF + FG = AG, DG + FG = DF, BE + EG = BG, CE + EF = CF

3. Segment Addition Postulate (Post. 1.2)

4. AF = DG, BE = CE = EF = EG

4. Defi nition of congruent segments

5. DG + FG = AG,BE + EG = CF


6. DF = AG,BG = CF


7. — DF ≅ — AG , — BG ≅ — CF


8. △ABG ≅ △DCF 8. All corresponding parts are congruent.

17. The congruence statement should be used to ensure that corresponding parts are matched up correctly.

∠S ≅ ∠Y

m∠S = m∠Y

m∠S = 90° − 42° = 48°

18. In order to conclude that triangles are congruent, the sides must also be congruent; △MNP is not congruent to △RSP because the corresponding sides are not congruent.


Chapter 5

19. Given ∠A ≅ ∠D and ∠B ≅ ∠E

Prove ∠C ≅ ∠F

A

B

C D

E

F

STATEMENTS REASONS

1. ∠A ≅ ∠D, ∠B ≅ ∠E 1. Given

2. m∠A = m∠D, m∠B = m∠E


3. m∠A + m∠B + m∠C= 180°, m∠D + m∠E+ m∠F = 180°

3. Triangle Sum Theorem (Thm. 5.1)

4. m∠D + m∠E + m∠F= m∠A + m∠B + m∠C


5. m∠D + m∠E + m∠F= m∠D + m∠E + m∠C


6. m∠F = m∠C 6. Subtraction Property of Equality

7. ∠F = ∠C 7. Defi nition of congruent angles

8. ∠C = ∠F 8. Symmetric Property

20. Sample answer:

The Transitive Property of Triangle Congruence (Thm. 5.3) guarantees that all the triangles are congruent.

21. corresponding sides: — JK ≅ — XY , — KL ≅ — YZ , — JL ≅ — XZ

corresponding angles: ∠J ≅ ∠X, ∠K ≅ ∠Y, ∠L ≅ ∠Z

22. a. They are congruent because corresponding parts of congruent fi gures are congruent.

b. They are congruent because they are both supplementary to congruent angles.

c. ∠GEB is also a right angle, and all right angles are congruent.

d. yes; From parts (a)–(c), you know that — BE ≅ — DE , ∠ABE ≅ ∠CDE, ∠GBE ≅ ∠GDE, and ∠GEB ≅ ∠GED. Also, — GE ≅ — GE by the Refl exive Property of Congruence (Thm. 2.1), ∠BGE ≅ ∠DGE by the Third Angles Theorem (Thm. 5.4), and — BG ≅ — DG from the diagram markings. So, △BEG ≅ △DEG because all corresponding parts are congruent.

23. m∠L + m∠M + m∠N = 180° 40° + 90° + m∠N = 180° 130° + m∠N = 180° m∠N = 50°

m∠N = m∠R m∠L = m∠P

50° = (2x + 4y)° 40° = (17x − y)°

So, a system of equations is 2x + 4y = 5017x − y = 40

.

Solve the second equation for y to get y = 17x − 40. Substitute this for y in the fi rst equation and solve for x.

2x + 4(17x − 40) = 50

2x + 68x − 160 = 50

70x − 160 = 50

70x = 210

x = 3

Substitute 3 for x in the second equation and solve for y.

y = 17 ⋅ 3 − 40

y = 51 − 40 = 11

So, x = 3 and y = 11.

24. m∠S + m∠T + m∠U = 180° 130° + 28° + (4x + y)° = 180° 158 + 4x + y = 180

4x + y = 22

m∠Y = m∠T

(8x − 6y)° = 28°

So, a system of equations is 4x + y = 228x − 6y = 28

.

Solve the fi rst equation for y to get y = 22 − 4x. Substitute this for y in the second equation and solve for x.

8x − 6(22 − 4x) = 28

8x − 132 + 24x = 28

32x − 132 = 28

32x = 160

x = 5

Substitute 5 for x in the fi rst equation and solve for y.

4(5) + y = 22

20 + y = 22

y = 2

So, x = 5 and y = 2.

25. A rigid motion maps each part of a fi gure to a corresponding part of its image. Because rigid motions preserve length and angle measure, corresponding parts of congruent fi gures are congruent, which means that the corresponding sides and corresponding angles are congruent.


Chapter 5

Maintaining Mathematical Profi ciency 26. ∠Z ≅ ∠W 27. ∠N ≅ ∠T, — RS ≅ — PQ

28. ∠J ≅ ∠M, — JK ≅ — KM (K is the midpoint), and m∠LKM = 90°.

29. ∠D ≅ ∠H, — DF ) — GH , — DE ≅ — HI , ∠DFE ≅ ∠HGI



c. BC ≈ 1.95 units, m∠B ≈ 98.79°, m∠C ≈ 41.21° d. Check students’ work. If two sides and the included angle

of a triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent.

2. Two triangles can be proved congruent if two pairs of corresponding sides and corresponding included angles are congruent.

3. Start with two triangles so that two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle. Then show that one triangle can be translated until it coincides with the other triangle by a composition of rigid motions.

5.3 Monitoring Progress (pp. 247–248)Given ABCD is a square, — AB ≅ — BC ≅ — CD ≅ — AD , ∠A, ∠B, ∠C,

∠D are right angles. R, S, T, and U are midpoints of the sides ABCD. — RT ⊥ — SU and — SV ≅ — VU .

CB S

DA

VT

U

R

1. Prove △SVR ≅ △UVR

STATEMENTS REASONS

1. — SV ≅ — VU , — RT ⊥ — SU 1. Given

2. — VR ≅ — VR 2. Refl exive Property of Congruence (Thm. 2.1)

3. ∠SVR and ∠UVR are right angles.

3. Defi nition of perpendicular lines

4. ∠SVR ≅ ∠UVR 4. Right Angles Congruence Theorem (Thm. 2.3)

5. △SVR ≅ △UVR 5. SAS Congruence Theorem (Thm. 5.5)

2. STATEMENTS REASONS

1. ∠B and ∠D are right angles, — BA ≅ — DA ≅ — DC ≅ — BC , R is the midpoint of — BA , U is the midpoint of — DA , T is the midpoint of — DC ,and S is the midpoint of — BC .

1. Given

2. ∠B ≅ ∠D 2. Right Angles Congruence Theorem (Thm. 2.3)

3. BA = DA = DC = BC 3. Defi nition of congruent segments

4. BA = BR + RA, DA = DU + UA, DC = DT + TC, BC = BS + SC


5. BR + RA = DU + UA = DT + TC = BS + SC


6. — BR ≅ — RA , — DU ≅ — UA , — DT ≅ — TC , — BS ≅ — SC

6. Defi nition of midpoint

7. BR = RA, DU = UA,DT = TC, BS = SC


8. BR + BR = DU + DU = DT + DT = BS + BS


9. 2 ⋅ BR = 2 ⋅ DU= 2 ⋅ DT = 2 ⋅ BS

9. Distributive Property

10. BR = DU = DT = BS 10. Division Property of Equality

11. — BR ≅ — DU ≅ — DT ≅ — BS 11. Defi nition of congruent segments

12. △BSR ≅ △DUT 12. SAS Congruence Theorem (Thm. 5.5)

3. Given — DA ≅ — DG and

GRA

D

∠ADR ≅ ∠GDR

Prove △DRA ≅ △DRG

STATEMENTS REASONS

1. — DA ≅ — DG , ∠ADR ≅ ∠GDR

1. Given

2. — DR ≅ — DR 2. Refl exive Property of Congruence (Thm. 2.1)

3. △DRA ≅ △DRG 3. SAS Congruence Theorem (Thm. 5.5)


Chapter 5


Vocabulary and Core Concept Check 1. An included angle is an angle formed by two adjacent

consecutive sides of a triangle.

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

Monitoring Progress and Modeling with Mathematics 3. ∠JKL is the included angle between — JK and — KL .

4. ∠PKL is the included angle between — PK and — LK .

5. ∠KLP is the included angle between — LP and — LK .

6. ∠LJK is the included angle between — JL and — JK .

7. ∠KLJ is the included angle between — KL and — JL .

8. ∠KPL is the included angle between — KP and — PL .

9. no; The congruent angles are not the included angle.

10. yes; Two pairs of sides and the included angles are congruent.

11. no; One of the congruent angles is not the included angle.


13. yes; Two pairs of sides and the included angles are congruent.

14. no; ∠NKM and ∠KML are congruent by the Alternate Interior Angles Theorem (Thm. 3.2) but they are not the included angles.

15. Given — PQ bisects ∠SPT,

Q

T

P

S — SP ≅ — TP

Prove △SPQ ≅ △TPQ

STATEMENTS REASONS

1. — SP ≅ — TP , — PQ bisects ∠SPT.

1. Given

2. — PQ ≅ — PQ 2. Refl exive Property of Congruence (Thm. 2.1)

3. ∠SPQ ≅ ∠TPQ 3. Defi nition of angle bisector

4. △SPQ ≅ △TPQ 4. SAS Congruence Theorem (Thm. 5.5)

16. Given — AB ≅ — CD , — AB ) — CD A D

CB

1

2

Prove △ABC ≅ △CDA

STATEMENTS REASONS

1. — AB ≅ — CD , — AB ) — CD 1. Given

2. — AC ≅ — AC 2. Refl exive Property of Congruence (Thm. 2.1)

3. ∠1 ≅ ∠2 3. Alternate Interior Angle Congruence Theorem (Thm. 3.2)

4. △ABC ≅ △CDA 4. SAS Congruence Theorem (Thm. 5.5)

17. Given C is the midpoint

A

D

EC

B

of — AE and — BD .

Prove △ABC ≅ △EDC

STATEMENTS REASONS

1. C is the midpoint of — AE and — BD .1. Given

2. ∠ACB ≅ ∠ECD 2. Vertical Angles Congruence Theorem (Thm. 2.6)

3. — AC ≅ — EC , — BC ≅ — DC

3. Defi nition of midpoint

4. △ABC ≅ △EDC 4. SAS Congruence Theorem (Thm. 5.5)

18. Given — PT ≅ — RT , — QT ≅ — ST P Q

RS

T Prove △PQT ≅ △RST

STATEMENTS REASONS

1. — PT ≅ — RT , — QT ≅ — ST

1. Given

2. ∠PTQ ≅ ∠RTS 2. Vertical Angles Congruence Theorem (Thm. 2.6)

3. △PQT ≅ △RST 3. SAS Congruence Theorem (Thm. 5.5)


Chapter 5

19. △SRT ≅ △URT; — RT ≅ — RT by the Refl exive Property of Congruence (Thm. 2.1). Also, because all points on a circle are the same distance from the center, — RS ≅ — RU . It is given that ∠SRT ≅ ∠URT. So, △SRT and △URT are congruent by the SAS Congruence Theorem (Thm. 5.5).

20. △BAD ≅ △DCB; Because the sides of the square are congruent, — BA ≅ — DC and — AD ≅ — CB . Also, because the angles of the square are congruent, ∠A ≅ ∠C. So, △BAD and △DCB are congruent by the SAS Congruence Theorem (Thm. 5.5).

21. △STU ≅ △UVR; Because the sides of the pentagon are congruent, — ST ≅ — UV and — TU ≅ — VR . Also, because the angles of the pentagon are congruent ∠T ≅ ∠V. So, △STU and △UVR are congruent by the SAS Congruence Theorem (Thm. 5.5).

22. △NMK ≅ △NLK; Because all points on a circle are the same distance from the center, — NM ≅ — NL and — KM ≅ — KL .Because — MK ⊥ — MN and — KL ⊥ — NL , ∠M and ∠L are right angles by defi nition of perpendicular lines, which means that ∠M ≅ ∠L by the Right Angles Congruence Theorem (Thm. 2.3). So, △NMK and △NLK are congruent by the SAS Congruence Theorem (Thm. 5.3.)

23. Construct side — DE so that it is congruent to — AC . Construct ∠D, with vertex D and side '''⃗ DE so that it is congruent to ∠A. Construct — DF so that it is congruent to — AB . Draw △DFE. By the SAS Congruence Theorem (Thm. 5.5), △ABC ≅ △DFE.

A

B E

F

D

C

24. Construct side — DE so that it is congruent to — AC . Construct ∠D, with vertex D and side '''⃗ DE so that it is congruent to ∠A. Construct — DF so that it is congruent to — AB . Draw △DFE. By the SAS Congruence Theorem (Thm. 5.5), △ABC ≅ △DFE.

A B DF

EC

25. △XYZ and △WYZ are congruent so either the expressions for — XZ and — WZ or the expressions for — XY and — WY should be set equal to each other because they are corresponding sides.

5x − 5 = 3x + 9

2x − 5 = 9

2x = 14

x = 7

26. In order to prove △ABC ≅ △DBC, you will need to know that ∠ACB ≅ ∠DCB.

27. Given △ABC, △BCD, and △CDE are isosceles triangles

and ∠B ≅ ∠D.

Prove △ABC ≅ △CDE

A C E

DB

Because △ABC, △BCD, and △CDE are isosceles triangles, you know that — AB ≅ — BC , — BC ≅ — CD , and — CD ≅ — DE . So, by the Transitive Property of Congruence (Thm. 2.1), — AB ≅ — CD and — BC ≅ — DE . It is given that ∠B ≅ ∠D, So △ABC ≅ △CDE by the SAS Congruence Theorem (Thm. 5.5).

28. SSS, ASA, AAS, and SAS all correspond to congruence theorems.

Counterexample for SSA

Counterexample for AAA

29. Prove △ABC ≅ △DEC

STATEMENTS REASONS

1. — AC ≅ — DC , — BC ≅ — EC

1. Given (marked in diagram)

2. ∠ACB ≅ ∠DCE 2. Vertical Angles Congruence Theorem (Thm. 2.6)

3. △ABC ≅ △DEC 3. SAS Congruence Theorem (Thm. 5.5)

AC = CD BC = CE

4y − 6 = 2x + 6 3y + 1 = 4x

4y = 2x + 12 3 ( 1 — 2 x + 3 ) + 1 = 4x

y = 1 — 2 x + 3 1.5x + 9 + 1 = 4x

1.5x + 10 = 4x

10 = 2.5x

x = 4

y = 1 — 2 ⋅ 4 + 3 = 2 + 3 = 5

So, x = 4 and y = 5.

30. no; When you construct — AB and — AC , you have to construct them at an angle that is congruent to ∠A. Otherwise, when you construct an angle congruent to ∠C, you might not get a third segment that is congruent to — BC .


Chapter 5

31. Given A refl ection in line K A″

MA′

K

AP

k

mmaps point A to A′, a refl ection in line m maps A′ to A″, and m∠MPK = x°.

Prove The angle of rotation is 2x°.

STATEMENTS REASONS

1. A refl ection in line k maps point A to A′, a refl ection in line m maps A′ to A″, and m∠MPK = x°.

1. Given

2. Line k is the perpendicular bisector of — AA′ , and line m is the perpendicular bisector of — A′A″ .

2. Defi nition of refl ection

3. — AK ≅ — KA′ , ∠AKP and ∠A′KP are right angles, — A′M ≅ — MA″ , and ∠A′MP and ∠A″MP are right angles.

3. Defi nition of perpendicular bisector

4. ∠AKP ≅ ∠A′KP, ∠A′MP ≅ ∠A″MP

4. Right Angles Congruence Theorem (Thm. 2.3)

5. — KP ≅ — KP 5. Refl exive Property of Congruence (Thm. 2.1)

6. △AKP ≅ △A′KP, △A′MP ≅ △A″MP

6. SAS Congruence Theorem (Thm. 5.5)

7. — AP ≅ — A′P , — A′P ≅ — A″P , ∠APK ≅ ∠A′PK, ∠A′PM ≅ ∠A″PM

7. Corresponding parts of congruent triangles are congruent.

8. — AP ≅ — A″P 8. Transitive Property of Congruence (Thm. 2.1)

9. m∠APK = m∠A′PK, m∠A′PM = m∠A″PM


10. m∠MPK = m∠A′PK + m∠A′PM, m∠APA″ = m∠APK + m∠A′PK + m∠A′PM + m∠A″PM

10. Angle Addition Postulate (Post. 1.4)

11. m∠APA″ = m∠A′PK + m∠A′PK + m∠A′PM + m∠A′PM


STATEMENTS REASONS

12. m∠APA″ = 2(m∠A′PK + m∠A′PM)

12. Distributive Property

13. m∠APA″ = 2(m∠MPK) 13. Substitution Property of Equality

14. m∠APA″ = 2(x°) = 2x° 14. Substitution Property of Equality

15. Point A is rotated about point P, and the angle of rotation is 2x°.

15. Defi nition of rotation

Maintaining Mathematical Profi ciency 32. Two sides are equivalent and one angle is a right angle.

So, the triangle is a right isosceles triangle.

33. Two sides are equivalent and one angle is obtuse. So, the triangle is an obtuse isosceles triangle.

34. All sides are congruent, and therefore all angles are congruent. So, the triangle is a equiangular equilateral triangle.

35. None of the sides are congruent and one angle is obtuse. So, the triangle is an obtuse scalene triangle.

5.4 Explorations (p. 251) 1. a. Check students’ work. b. Check students’ work. c. Because all points on a circle are the same distance from

the center, — AB ≅ — AC .

d. ∠B ≅ ∠C e. Check students’ work. If two angles of a triangle are congruent, then the angles

opposite them are congruent. f. If two angles of a triangle are congruent, then the sides

opposite them are congruent. The converse is true.

2. In an isosceles triangle, two sides are congruent and the angles opposite them are congruent.

3. Draw the angle bisector of the included angle between the congruent sides to divide the given isosceles triangle into two triangles. Use the SAS Congruence Theorem (Thm. 5.5) to show that these two triangles are congruent. Then, use properties of congruent triangles to show that the two angles opposite the shared sides are congruent.

For the converse, draw the angle bisector of the angle that is not congruent to the other two. This divides the given triangle into two triangles that have two pairs of corresponding congruent angles. The third pair of angles are congruent by the Third Angles Theorem (Thm. 5.4). Also, the angle bisector is congruent to itself by the Refl exive Property of Congruence (Thm. 2.1). So, the triangles are congruent, and the sides opposite the congruent angles in the original triangle are congruent.


Chapter 5

5.4 Monitoring Progress (pp. 253–255) 1. If — HG ≅ HK, then ∠HKG ≅ ∠G.

2. If ∠KHJ ≅ ∠KJH, then — KJ ≅ — KH .

3. The triangle is equiangular; therefore, by Corollary to the Converse of the Base Angles Theorem (Cor. 5.3) the length of — ST is 5 units.

4. The triangle on the right is an equilateral triangle. Each angle has a measure of 60°, therefore, x = 60. The triangle on the left is an isosceles triangle. Both base angles are 90° − 60° = 30°. So, y = 180 − 2 ⋅ 30 = 120.

5. From Example 4, you know that — PT ≅ — QT and ∠1 ≅ ∠2. It is stated that — PS ≅ — QR and ∠QPS ≅ ∠PQR. By defi nition of congruent segments and angles, m∠1 = m∠2 and m∠QPS ≅ m∠PQR. Also, by the Angle Addition Postulate (Post 1.4), m∠1 + m∠TPS + m∠QPS and m∠2 + m∠TQR + m∠PQR. By substituting m∠1 for m∠2 and m∠QPS for m∠PQR, you get m∠1 + m∠TQR = m∠QPS. Then, by the Transitive Property of Equality, m∠1 + m∠TPS = m∠1 + m∠TQR. So, by the Subtraction Property of Equality, m∠TPS = m∠TQR. Because ∠TPS = ∠TQR by defi nition, you can conclude that △PTS ≅ △QTR by the SAS Congruence Theorem (Thm. 5.5).


Vocabulary and Core Concept Check 1. The vertex angle is the angle formed by the congruent sides,

or legs, of an isosceles triangle.

2. The base angles of an isosceles triangle are opposite the congruent sides, and they are congruent by the Base Angles Theorem (Thm. 5.6).

Monitoring Progress and Modeling with Mathematics 3. If — AE ≅ — DE , then ∠D ≅ ∠A by the Base Angles Theorem

(Thm. 5.6).

4. If — AB ≅ — EB , then ∠AEB ≅ ∠A by the Base Angles Theorem (Thm. 5.6).

5. If ∠D ≅ ∠CED, then — EC ≅ — DC , by the Converse of the Base Angles Theorem (Thm. 5.7).

6. If ∠EBC ≅ ∠ECB, then — EC ≅ — EB , by the Converse of the Base Angles Theorem (Thm. 5.7).

7. △ABC is an equiangular triangle and, therefore, an equilateral triangle, So, x = 12.

8. △MLN is an equiangular triangle and, therefore, an equilateral triangle, So, x = 16.

9. △RST is an equilateral triangle and, therefore, an equiangular triangle, So, x = 60.

10. △DEF is an equilateral triangle and, therefore, each angle is 3x° = 60°, or x = 20.

11. The pennant is an isosceles triangle. So, x = 79 and y = 180 − 2 ⋅ 79 = 180 − 158 = 22.

12. Sample answer:

7 cm ReuseReduceRecycle

7 cm

7 cm

13. The triangle in the center is equilateral; therefore, each angle measures 60°. So, x = 60. Because the top and bottom lines are parallel, the alternate interior angles have a measure of 60°. By the Base Angles Theorem (Thm. 5.6), y = 60.

14. The vertex angle of the isosceles triangle measures 180° − 40° = 140°. To fi nd the base angles:

2x° + 140° = 180° 2x = 40

x = 20

To fi nd y:

y° + x° = 90° y + 20 = 90

y = 70

So, x = 20 and y = 70.

15. The triangle on the left is equiangular and, therefore, equilateral.

8y = 40

y = 5

The triangle on the right is isosceles and the vertex angle measures 180° − 60° = 120°. So, by the Base Angles

Theorem (Thm. 5.6), 180° − 120° —— 2 = 60° —

2 = 30°. So, x = 30

and y = 5.


Chapter 5

16. By the Converse of Base Angles Theorem (Thm. 5.7):

3x − 5 = y + 12

3x − 5 − 12 = y

y = 3x − 17

The triangle on the right is equiangular and, therefore, equilateral.

3x − 5 = 5y − 4

3x − 5 = 5(3x − 17) − 4

3x − 5 = 15x − 85 − 4

3x − 5 = 15x − 89

−12x = − 84

x = 7

y = 3x − 17

y = 3 ⋅ 7 − 17

y = 21 − 17 = 4

So, x = 7 and y = 4.

17. Draw a segment with length 3 inches. Draw an arc with center at one endpoint and radius 3 inches. Draw an arc with center at the other endpoint and radius 3 inches. Connect the intersection of the arcs with two segments to form an equilateral triangle.

3 in.

18. Draw a segment with length 1.25 inches. Draw an arc with center at one endpoint and radius 1.25 inches. Draw an arc with center at the other endpoint and radius 1.25 inches. Connect the intersection of the arcs with two segments to form an equilateral triangle.

1.25 in.

19. When two angles of a triangle are congruent, the sides opposite the angles are congruent; Because ∠A ≅ ∠C, — AB ≅ — BC . So BC = 5.

20. a. Because △ABD and △CBD are congruent and equilateral, you know that — AB ≅ — CB . So △ABC is isosceles.

b. Because △ABC is isosceles, ∠BAE ≅ ∠BCE by the Base Angels Theorem (Thm. 5.6).

c. By the Refl exive Property of Congruence (Thm. 2.1), — BE ≅ — BE . Because △ABD and △CBD are congruent and equilateral, and also equiangular by the Corollary to the Base Angels Theorem (Cor. 5.2), you can conclude that ∠ABE ≅ ∠CBE. Also, — AB ≅ — CB as explained in part (a). So, by the SAS Congruence Theorem (Thm 5.5), △ABE and △CBE.

d. m∠ABE + m∠CBE = m∠ABC

Angle Addition Postulate (Post. 1.4)

m∠ABE = 60°, m∠CBE = 60° Defi nition of equiangular triangle

m∠ABC = 60° + 60° = 120° Substitution Property of Equality

m∠ABC + m∠BAE + m∠BCE = 180° Triangle Sum Theorem (Thm. 5.1)

120° + m∠BAE + m∠BCE = 180° Substitution Property of Equality

m∠BAE + m∠BCE = 60° Subtraction Property of Equality

m∠BAE = m∠BCE

Corresponding parts of congruent triangles are congruent.

m∠BAE + m∠BCE = 60° Substitution Property of Equality

2m∠BAE = 60° Simplify.

m∠BAE = 30° Division Property of Equality

21. a. Each edge is made out of the same number of sides of the original equilateral triangle.

b. The areas of the fi rst four triangles in the pattern are 1 square unit, 4 square units, 9 square units, and 16 square units.

c. Triangle 1 has an area of 12 = 1, Triangle 2 has an area of 22 = 4, Triangle 3 has an area of 32 = 9, and so on. So, by deductive reasoning, you can predict that Triangle n has an area of n2.

Seventh triangle: 72 = 49 square units.

22. A, C; If the base of isosceles △XYZ is — YZ , then the legs are — XY and — XZ and — XY ≅ — XZ and ∠Y ≅ ∠Z.


Chapter 5

23. x + 4 = 4x + 1

4 = 3x + 1

3 = 3x

1 = x

Perimeter = 7 + 1 + 4 + 4 + 1 = 17 inches

24. 2x − 3 = x + 5

x = 8

21 − x = x + 5

16 = 2x

8 = x

21 − x = 2x − 3

24 = 3x

8 = x

Perimeter = (2 ⋅ 8 − 3) + (8 + 5) + (21 − 8)

= 13 + 13 + 13 = 39 inches

25. By the Refl exive Property of Congruence (Thm. 2.1), the yellow triangle and the yellow-orange triangle share a congruent side. Because the triangles are all isosceles, by the Transitive Property of Congruence (Thm. 2.1), the yellow-orange triangle and the orange triangle share a side that is congruent to the one shared by the yellow triangle and the yellow-orange triangle. This reasoning can be continued around the wheel, so the legs of the isosceles triangles are all congruent. Because you are given that the vertex angles are all congruent, you can conclude that the yellow triangle is congruent to the purple triangle by the SAS Congruence Theorem (Thm. 5.5).

26. 180° − 30° = 150° 150

— 2 = 75

The measures of the base angles are each 75°.

27.

purple

yellow

red-purple

blue-purple

yellow-orange

yellow-green

blue red

green orange

blue-green

red-orange

The three sides of the triangle are congruent. So, the triangle is an equiangular equilateral triangle.

28. Every fourth color is a triad: Yellow-green, blue-purple, red-orange; green, purple, orange; and blue-green, red-purple, yellow-orange.

29. no; The two sides that are congruent can form an obtuse angle or a right angle.

30. no; The sum of the angles of a triangle is always 180°. So, if all three angles are congruent, then they will always be

180° — 3 = 60°.

31. 3t = 5t − 12 or 3t = t + 20 or 5t − 12 = t + 20

−2t = −12 2t = 20 4t = 32

t = 6 t = 10 t = 8

The values of t could be 6, 8, and 10.

32.

x°

The base angles are (180 − x)° and the vertex angle is 180° − 2(180 − x)° = (2x − 180)°.

x°

The vertex angle is (180 − x)° and the base angles are

180° − (180 − x)° —— 2 = ( x —

2 ) °.

33. If the base angles are x°, then the vertex angle is (180 − 2x)°, or [2(90 − x)]°. Because 2(90 − x) is divisible by 2, the vertex angle is even when the angles are whole numbers.

34. a. ∠XVY, ∠UXV; ∠WUX ≅ ∠XVY because they are both vertex angles of congruent isosceles triangles. Also, m∠UXV + m∠VXY = m∠UXY by the Angle AdditionPostulate (Post. 1.4), and m∠UXY = m∠WUX + m∠UWXby the Exterior Angle Theorem (Thm. 5.2). So, by the Transitive Property of Equality, m∠UXV + m∠VXY = m∠WUX + m∠UWX. Also, m∠UWX = m∠VXY because they are base angles of congruent isosceles triangles. By substituting m∠UWX for m∠VXY, you get m∠UXV + m∠UWX = m∠WUX + m∠UWX. By the Subtraction Property of Equality, m∠UXV = m∠WUX, so ∠UXV ≅ ∠WUX.

b. Because the triangles are congruent isosceles triangles and from part (a), — UX ≅ — VY ≅ — UW ≅ — VX , then — WX ≅ — XY ≅ — UV . So, the distance between points U and V is 8 meters.

35. a. 2.1 mi; by the Exterior Angle Theorem (Thm. 5.2), m∠L = 70° − 35° = 35°. Because m∠SRL = 35°= m∠RLS, by defi nition of congruent angles, ∠SRL ≅ ∠RLS. So, by the Converse of the Base Angles Theorem (Thm. 5.7), — RS ≅ — SL . So, SL = RS = 2.1 miles.

b. Find the point on the shore line that has an angle of 45° from the boat. Then, measure the distance that the boat travels until the angle is 90°. That distance is the same as the distance between the boat and the shore line because the triangle formed is an isosceles right triangle.

36. no; The sum of the angle measures of a very large spherical triangle will be greater than 180°, but for smaller spherical triangles, the sum will be closer to 180°.


Chapter 5

37. Given △ABC is equilateral.

B

A

C

Prove △ABC is equiangular.

STATEMENTS REASONS

1. △ABC is equilateral.

1. Given

2. — AB ≅ — AC , — AB ≅ — BC , — AC ≅ — BC

2. Defi nition of equilateral triangle

3. ∠B ≅ ∠C, ∠A ≅ ∠C, ∠A ≅ ∠B

3. Base Angles Theorem (Thm. 5.6)

4. △ABC is equiangular.

4. Defi nition of equiangular triangle

38. a. By the markings, — AE ≅ — DE , — AB ≅ — DC , and ∠BAE ≅ ∠CDE. So, △ABE ≅ △DCE by SAS Congruence Theorem (Thm. 5.5).

b. △AED and △BEC are isosceles triangles.

c. ∠EDA, ∠BCA, and ∠CBD are all congruent to ∠EAD.

39. Given △ABC is equiangular.

B

A

C

Prove △ABC is equilateral.

STATEMENTS REASONS

1. △ABC is equiangular.

1. Given

2. ∠B ≅ ∠C, ∠A ≅ ∠C, ∠A ≅ ∠B


3. — AB ≅ — AC , — AB ≅ — BC , — AC ≅ — BC

3. Converse of the Base Angles Theorem (Thm. 5.7)

4. △ABC is equilateral.


40. no; The distance between point T(0, 6) and a point on y = x is equal to the distance between U(6, 0) and y = x (using the same point). So any point V on y = x will be the same distance from T and U. Therefore, TV = VU and △TVU is an isosceles triangle. Unless V(3, 3) is the third point, in which case, T, V, and U are collinear and perpendicular to y = x.

41. Given △ABC is equilateral.

B C

A

D

EF

∠CAD ≅ ∠ABE ≅ ∠BCF

Prove △DEF is equilateral.

STATEMENTS REASONS

1. △ABC is equilateral. ∠CAD ≅ ∠ABE ≅ ∠BCF

1. Given

2. △ABC is equiangular. 2. Corollary to the Base Angles Theorem (Cor. 5.2)

3. ∠ABC ≅ ∠BCA ≅ ∠BAC


4. m∠CAD = m∠ABE = m∠BCF, m∠ABC = m∠BCA = m∠BAC


5. m∠ABC = m∠ABE + m∠EBC, m∠BCA = m∠BCF + m∠ACF, m∠BAC = m∠CAD + m∠BAD


6. m∠ABE + m∠EBC = m∠BCF + m∠ACF = m∠CAD + m∠BAD


7. m∠ABE + m∠EBC = m∠ABE + m∠ACF = m∠ABE + m∠BAD


8. m∠EBC = m∠ACF = m∠BAD

8. Subtraction Property of Equality

9. ∠EBC ≅ ∠ACF ≅ ∠BAD


10. ∠FEB ≅ ∠DFC ≅ ∠EDA

10. Third Angles Theorem (Thm. 5.4)

11. ∠FEB and ∠FED are supplementary, ∠DFC and ∠EFD are supplementary, and ∠EDA and ∠FDE are supplementary.

11. Linear Pair Postulate (Post. 2.8)

12. ∠FED ≅ ∠EFD ≅ ∠FDE

12. Congruent Supplements Theorem (Thm. 2.4)

13. △DEF is equiangular. 13. Defi nition of equiangular triangle

14. △DEF is equilateral. 14. Corollary to the Converse of the Base Angles Theorem (Cor. 5.3)


Chapter 5

Maintaining Mathematical Profi ciency 42. Refl exive Property of Congruence (Theorem 2.1): — SE ≅ — SE

43. Symmetric Property of Congruence (Theorem 2.1):If — JK ≅ — RS , then — RS ≅ — JK .

44. Transitive Property of Congruence (Theorem 2.1):If — EF ≅ — PQ , and — PQ ≅ — UV , then — EF ≅ — UV .

5.1–5.4 What Did You Learn? (p. 259) 1. You are given a diagram of the triangle made from the

segments that connect each person to the other two, along with the length of each segment; the people are all standing on the same stage (plane), so the points are coplanar; You are asked to classify the triangle by its sides and by measuring its angles.

2. There is a pair of congruent triangles, so all pairs of corresponding sides and angles are congruent. By the Triangle Sum Theorem (Thm. 5.1), the three angles in △LMN have measures that add up to 180°. You are given two measures, so you can fi nd the third using this theorem. The measure of ∠P is equal to the measure of its corresponding angle, ∠L. The measure of ∠R is equal to the measure of its corresponding angle, ∠N. Once you write this system of equations, you can solve for the values of the variables.

3. Sample answer: a large triangle made up of 9 small triangles, a hexagon, a parallelogram

5.1–5.4 Quiz (p. 260) 1. x° = 30° + 80° = 110°; The exterior angle measures 110°.

2. (5x + 2)° + 6x° = 90° 11x = 88

x = 8

y° = 180° − (5x + 2)° = 180° − (5 ⋅ 8 + 2)° = 180° − 42° = 138° The exterior angle measures 138°.

3. 29° + (12x + 26)° = (15x + 34)° 55 + 12x = 15x + 34

21 = 3x

7 = x

(15x + 34)° = (15 ⋅ 7 + 34)° = (105 + 34)° = 139° The exterior angle measures 139°.

4. corresponding angles: ∠C ≅ ∠F, ∠A ≅ ∠D, ∠B ≅ ∠E

corresponding sides: — CA ≅ — FD , — AB ≅ — DE , — CB ≅ — FE

Sample answer: △CAB ≅ △FDE

5. corresponding angles: ∠Q ≅ ∠W, ∠R ≅ ∠X, ∠S ≅ ∠Y, ∠T ≅ ∠Z

corresponding sides: — QR ≅ — WX , — RS ≅ — XY , — ST ≅ — YZ , — QT ≅ — WZ

Sample answer: RSTQ ≅ XYZW


7. yes; △GHF ≅ △KHJ by the SAS Congruence Theorem (Thm. 5.5).

STATEMENTS REASONS

1. — GH ≅ — HK , — FH ≅ — HJ

1. Given

2. ∠GHF ≅ ∠KHJ 2. Vertical Angles Congruence Theorem (Thm. 2.6)

3. △GHF ≅ △KHJ 3. SAS Congruence Theorem (Thm. 5.5)

8. yes; △LMP ≅ △NMP by the SAS Congruence Theorem (Thm. 5.5).

STATEMENTS REASONS

1. — LM ≅ — NM ,∠LMP ≅ ∠NMP

1. Given

2. — MP ≅ — MP 2. Refl exive Property of Congruence (Thm. 2.1)

3. △LMP ≅ △NMP 3. SAS Congruence Theorem (Thm. 5.5)

9. If — VW ≅ — WX , then ∠VXW ≅ ∠XVW by the Base Angles Theorem (Thm. 5.6).

10. If — XZ ≅ — XY , then ∠XYZ ≅ ∠XZY by the Base Angles Theorem (Thm. 5.6).

11. If ∠ZVX ≅ ∠ZXV, then — XZ ≅ — VZ , by the Converse of the Base Angles Theorem (Thm. 5.7).

12. If ∠XYZ ≅ ∠ZXY, then — XZ ≅ — YZ , by the Converse of the Base Angles Theorem (Thm. 5.7).


Chapter 5

13. DE = QR

5y − 7 = 38

5y = 45

y = 9

m∠S = m∠F

m∠F = 180° − (123° + 29°) = 28° 2x + 2 = 28

2x = 26

x = 13

So, x = 13 and y = 9.

14. 5x − 1 = 24

5x = 25

x = 5

6y = 120

y = 20

So, x = 5 and y = 20.

15. x = 4 [ (90 − x) − 5 ] x = 360 − 4x − 20

5x = 340

x = 68

90° − 68° = 22° The acute angles are 68° and 22°.

16. a. Triangle 1 has a right angle. So, it is a right triangle. It appears that triangle 2 has an obtuse angle. So, it is an obtuse triangle. It appears that triangle 3 has three nonequivalent acute angles. So, it is an acute triangle.

Triangle 4 has three congruent angles. So, it is an equiangular triangle.

b. Triangle 4 has three congruent sides. So, it is an equilateral triangle. It appears that triangle 5 has three noncongruent sides. So, it is a scalene triangle.

Triangle 6 has two congruent sides. So, it is an isosceles triangle.

c. yes;

7

A

BC D F

E

8

STATEMENTS REASONS

1. — AB ≅ — DE , — BC ≅ — EF , ∠B ≅ ∠E

1. Given

2. △ABC ≅ △DEF 2. SAS Congruence Theorem (Thm. 5.5)



c. AB = 2 because — AB has one endpoint at the origin and one endpoint on a circle with a radius of 2 units.

AC = 3 because — AC has one endpoint at the origin and one endpoint on a circle with a radius of 3 units. BC = 4 because it was created that way.

d. m∠A = 104.43°, m∠B = 46.61°, m∠C = 28.96° e. Check students’ work; If two triangles have three pairs

of congruent sides, then they will have three pairs of congruent angles.

2. When the corresponding sides of two triangles are congruent, the corresponding angles are also congruent.

3. Use rigid transformations to map triangles.

5.5 Monitoring Progress (pp. 263–265) 1. yes; From the diagram markings, — DF ≅ — HJ , — FG ≅ — JK , and — DG ≅ — HK . So, △DFG ≅ △HJK by the SSS Congruence

Theorem (Thm. 5.8).

2. no; — AB corresponds with — CD , but they are not the same measure. In order for two triangles to be congruent, all pairs of corresponding sides must be congruent.

3. yes; From the diagram markings, — QP ≅ — RS , — PT ≅ — ST , and — QT ≅ — RT . So, △QPT ≅ △RST by the SSS Congruence Theorem (Thm. 5.8).

4. not stable; This square is not stable because there are many possible quadrilaterals with the given side lengths.

5. stable; The diagonal support in this fi gure forms triangles with fi xed side lengths. By the SSS Congruence Theorem (Thm. 5.8), these triangles cannot change shape, so the fi gure is stable.

6. not stable; The diagonal support in this fi gure forms a triangle and a quadrilateral. The triangle would be stable, but the quadrilateral is not because there are many possible quadrilaterals with the given side lengths.

7. Redraw △ABC and △DCB.

C

B A

B

C D


Chapter 5

8. Given — AC ≅ — DB , ∠ABC and ∠DCB are right angles.

Prove △ABC ≅ △DCB

STATEMENTS REASONS

1. — AC ≅ — DB , ∠ABC and ∠DCB are right angles.

1. Given

2. — CB ≅ — BC 2. Refl exive Property of Congruence (Thm. 2.1)

3. △ABC and △DCB are right triangles.

3. Defi nition of a right triangle

4. △ABC ≅ △DCB 4. HL Congruence Theorem (Thm. 5.9)


Vocabulary and Core Concept Check 1. The side opposite the right angle is called the hypotenuse of

the right triangle.

2. Three of the triangles are confi rmed as right triangles. The second triangle from the left is the only triangle with legs that are not the legs of a right triangle.

Monitoring Progress and Modeling with Mathematics 3. yes; — AB ≅ — DB , — BC ≅ — BE , — AC ≅ — DE

4. no; You cannot tell for sure from the diagram whether — PS and — RS are congruent.

5. yes; ∠B and ∠E are right angles, — AB ≅ — FE , — AC ≅ — FD

6. no; the hypotenuses are not marked as congruent.

7. no; You are given that — RS ≅ — PQ , — ST ≅ — QT , and — RT ≅ — PT .So, it should say △RST ≅ △PQT by the SSS Congruence Theorem (Thm. 5.8).

8. yes; You are given that — AB ≅ — CD and — AD ≅ — CB . Also — BD ≅ — BD by the Refl exive Property of Congruence (Thm. 2.1). So, △ABD ≅ △CDB by the SSS Congruence Theorem (Thm. 5.8).

9. yes; You are given that — EF ≅ — GF and — DE ≅ — DG . Also — DF ≅ — DF by the Refl exive Property of Congruence (Thm. 2.1). So, △DEF ≅ △DGF by the SSS Congruence Theorem (Thm. 5.8).

10. no; You are given that — JK ≅ — KL ≅ — LM ≅ — MJ . Also, — JL ≅ — JL by the Refl exive Property of Congruence (Thm. 2.1). So, it should say △JKL ≅ △LMJ or △JKL ≅ △JML by the SSS Congruence Theorem (Thm. 5.8).

11. yes; The diagonal supports in this fi gure form triangles with fi xed side lengths. By the SSS Congruence Theorem (Thm. 5.8), these triangles cannot change shape, so the fi gure is stable.

12. no; The support in this fi gure forms two quadrilaterals, which are not stable because there are many possible quadrilaterals with the given side lengths.

13. Given — AC ≅ — DB , A

D

B

D

A

C

— AB ⊥ — AD , — CD ⊥ — AD

Prove △BAD ≅ △CDA

STATEMENTS REASONS

1. — AC ≅ — DB , — AB ⊥ — AD , — CD ⊥ — AD

1. Given

2. — AD ≅ — AD 2. Refl exive Property of Congruence (Thm. 2.1)

3. ∠BAD and ∠CDA are right angles.


4. △BAD and △CDA are right triangles.


5. △BAD ≅ △CDA 5. HL Congruence Theorem (Thm. 5.9)

14. Given G is the midpoint of — EH , E

G

F

H

G

I

— FG ≅ — GI , ∠E and ∠H are right angles.

Prove △EFG ≅ △HIG


Chapter 5

STATEMENTS REASONS

1. G is the midpoint of — EH , — FG ≅ — GI , ∠E and ∠H are right angles.

1. Given

2. — EG ≅ — HG 2. Defi nition of midpoint

3. △EFG and △HIG are right triangles.


4. △EFG ≅ △HIG 4. HL Congruence Theorem (Thm. 5.9)

15. Given — LM ≅ — JK , — MJ ≅ — KL

MJ

K L

Prove △LMJ ≅ △JKL

STATEMENTS REASONS

1. — LM ≅ — JK , — MJ ≅ — KL 1. Given

2. — JL ≅ — JL 2. Refl exive Property of Congruence (Thm. 2.1)

3. △LMJ ≅ △JKL 3. SSS Congruence Theorem (Thm. 5.8)

16. Given — WX ≅ — VZ ,

ZV

W X

Y — WY ≅ — VY , — YZ ≅ — YX

Prove △VWX ≅ △WVZ

STATEMENTS REASONS

1. — WX ≅ — VZ , — WY ≅ — VY , — YZ ≅ — YX

1. Given

2. — WV ≅ — WV 2. Refl exive Property of Congruence (Thm. 2.1)

3. WY = VY, YZ = YX 3. Defi nition of congruent segments

4. WZ = WY + YZ, VX = VY + YX


5. VX = WY + YZ 5. Substitution Property of Equality

6. VX = WZ 6. Transitive Property of Equality

7. — VX ≅ — WZ 7. Defi nition of congruent segments

8. △VWX ≅ △WVZ 8. SSS Congruence Theorem (Thm. 5.8)

17. Construct a side that is congruent to — QS . Open your compass to the length of — QR . Use this length to draw an arc. Draw an arc with radius RS. Complete the triangle. By the SSS Congruence Theorem (Thm. 5.8), the two triangles are congruent.

Q

R

S

18. Construct a side that is congruent to — QS . Open your compass to the length of — QR . Use this length to draw an arc. Draw an arc with radius RS. Complete the triangle. By the SSS Congruence Theorem (Thm. 5.8), the two triangles are congruent.

Q

R

S

19. The order of the points in the congruence statement should refl ect the corresponding sides and angles.

△TUV ≅ △ZYX by the SSS Congruence Theorem (Thm. 5.8).

20. When you substitute 1 — 4 for x, KL ≠ LM.

6x = 4x + 4 2x + 1 = 3x − 1

2x = 4 1 = x − 1

x = 2 2 = x

21. no; The sides of a triangle do not have to be congruent to each other, but each side of one triangle must be congruent to the corresponding side of the other triangle.


Chapter 5

22. Given — HF ≅ — FS ≅ — ST ≅ — TH ; — FT ≅ — SH ;∠H, ∠F, ∠S, and ∠T are right angles.

Prove △HFS ≅ △FST ≅ △STH

S

F

H

T

STATEMENTS REASONS

1. — HF ≅ — FS ≅ — ST ≅ — TH ; — FT ≅ — SH ; ∠H, ∠F, ∠S, and ∠T are right angles.

1. Given

2. — SH ≅ — SH 2. Refl exive Property of Congruence (Thm. 2.1)

3. △HFS, △FST, and △STH are right triangles.


4. △HFS ≅ △FST ≅ △STH

4. HL Congruence Theorem (Thm. 5.9)

23. a. You need to know that the hypotenuses are congruent: — JL ≅ — ML .

b. SAS Congruence Theorem (Thm. 5.5); By defi nition of midpoint, — JK ≅ — MK . Also, — LK ≅ — LK , by the Refl exive Property of Congruence (Thm. 2.1), and ∠JKL ≅ ∠MKL by the Right Angles Congruence Theorem (Thm. 2.3).

24. a. To use the SSS Congruence Theorem (Thm. 5.8) to prove △ABC ≅ △CDE, you need to know — AB ≅ — CD .

b. To use the HL Congruence Theorem (Thm. 5.9) to prove △ABC ≅ △CDE, you need to know that ∠B and ∠D are right angles.

25. AB = √———

[ 4 − (−2) ] 2 + (−2 − (−2))2

= √——

(4 + 2)2 + (−2 + 2)2 = √—

62 = 6

BC = √——

(4 − 4)2 + [ 6 − (−2) ] 2 = √—

(6 + 2)2 = √—

82 = 8

AC = √———

[ 4 − (−2) ] 2 + [ 6 − (−2) ] 2 = √

—— (4 + 2)2 + (6 + 2)2 = √

— 62 + 82

= √—

36 + 64 = √—

100 = 10

DE = √——

(5 − 5)2 + (1 − 7)2 = √—

(−6)2 = √—

36 = 6

EF = √——

(13 − 5)2 + (1 − 1)2 = √—

82 = 8

DF = √——

(13 − 5)2 + (1 − 7)2 = √——

(8)2 + (−6)2

= √—

64 + 36 = √—

100 = 10

AB = DE, BC = EF, and AC = DF, so △ABC ≅ △DEF.

26. AB = √———

[ 3 − (−2) ] 2 + (−3 − 1)2 = √——

(3 + 2)2 + (−4)2

= √—

52 + 16 = √—

25 + 16 = √—

41 ≈ 6.4

BC = √——

(7 − 3)2 + [ 5 − (−3) ] 2 = √——

(4)2 + (5 + 3)2

= √—

16 + 82 = √—

16 + 64 = √—

80 ≈ 8.9

AC = √——

[ 7 − (−2) ] 2 + (5 − 1)2 = √——

(7 + 2)2 + (4)2

= √—

92 + 42 = √—

81 + 16 = √—

97 ≈ 9.8

DE = √——

(8 − 3)2 + (2 − 6)2 = √——

(5)2 + (−4)2

= √—

25 + 16 = √—

41 ≈ 6.4

EF = √——

(10 − 8)2 + (11 − 2)2 = √—

22 + 92

= √—

4 + 81 = √—

85 ≈ 9.2

DF = √——

(10 − 3)2 + (11 − 6)2 = √—

(7)2 + (5)2

= √—

49 + 25 = √—

74 ≈ 8.6

The triangles are not congruent.

27. AB = √——

(6 − 0)2 + (5 − 0)2 = √—

62 + 52

= √—

36 + 25 = √—

61 ≈ 7.8

BC = √——

(9 − 6)2 + (0 − 5)2 = √—

32 + (−5)2

= √—

9 + 25 = √—

34 ≈ 5.8

AC = √——

(9 − 0)2 + (0 − 0)2 = √—

92 = √—

9 = 9

DE = √———

(6 − 0)2 + [ −6 − (−1) ] 2 + √——

62 + (−6 + 1)2

= √—

36 + (−5)2 = √—

36 + 25 = √—

61 ≈ 7.8

EF = √———

(9 − 6)2 + [ −1 − (−6) ] 2 = √——

32 + (−1 + 6)2

= √—

9 + (−5)2 = √—

9 + 25 = √—

34 ≈ 5.8

DF = √———

(9 − 0)2 + [ −1 − (−1) ] 2 = √——

92 + (−1 + 1)2

= √—

81 = 9

AB = DE, BC = EF, and AC = DE, so △ABC ≅ △DEF.

28. AB = √———

[ −5 − (−5) ] 2 + (2 − 7)2 = √——

(−5 + 5)2 + (−5)2

= √—

25 = 5

BC = √——

[ 0 − (−5) ] 2 + (2 − 2)2 = √—

52 = √—

25 = 5

AC = √——

[ 0 − (−5) ] 2 + (2 − 7)2 = √—

52 + (−5)2

= √—

25 + 25 = √—

50 ≈ 7.1

DE = √——

(0 − 0)2 + (1 − 6)2 = √—

(−5)2 = √—

25 = 5

EF = √——

(4 − 0)2 + (1 − 1)2 = √—

42 = √—

16 = 4

DF = √——

(4 − 0)2 + (1 − 6)2 = √—

42 + (−5)2

= √—

16 + 25 = √—

41 ≈ 6.4

The triangles are not congruent.

29. yes; You could use the string to measure the length of each side on one triangle and compare it to the length of the corresponding side of the other triangle to determine whether SSS Congruence Theorem (Thm. 5.8) applies.

30. SAS Congruence Theorem (Thm. 5.5), HL Congruence Theorem (Thm. 5.9), SSS Congruence Theorem (Thm. 5.8)


Chapter 5

31. both; — JL ≅ — JL by the Refl exive Property of Congruence (Thm. 2.1), and the other two pairs of sides are marked as congruent. So, the SSS Congruence Theorem (Thm. 5.8) can be used. Also, because ∠M and ∠K are right angles, they are both right triangles, and the legs and hypotenuses are congruent. So, the HL Congruence Theorem (Thm. 5.9) can be used.

32. yes; They would have to be formed from circles that were at the same angle with each other. So, all corresponding parts would be congruent.

33. Sample answer:

L

M N

S

T U

34. Sample answer:

L

M N

S

T U

35. a. — BD ≅ — BD by the Refl exive Property of Congruence (Thm. 2.1). It is given that — AB ≅ — CB and that ∠ADB and ∠CDB are right angles. So, △ABC and △CBD are right triangles and are congruent by the HL Congruence Theorem (Thm. 5.9).

b. yes; Because — AB ≅ — CB ≅ — CE ≅ — FE , — BD ≅ — EG , and they are all right triangles, it can be shown that △ABD ≅ △CBD ≅ △CEG ≅ △FEG by the HL Congruence Theorem (Thm. 5.9).

36. 5x = 3x + 10 5x − 2 = 4x + 3

2x = 10 x = 5

x = 5

AB = 5 ⋅ 5 = 25

AC = 5 ⋅ 5 − 2 = 25 − 2 = 23

BD = 4 ⋅ 5 + 3 = 23

CD = 3 ⋅ 5 + 10 = 25

For x = 5, — AB ≅ — CD , — AC ≅ — BD , and — BC ≅ — CB (Refl exive Property of Congruence, Thm. 2.1). Then, by the SSS Congruence Theorem (Thm. 5.8), △ABC ≅ △DCB.

5x = 4x + 3

x = 3

AB = 5 ⋅ 3 = 15

AC = 5 ⋅ 3 − 2 = 15 − 2 = 13

BD = 4 ⋅ 3 + 3 = 15

CD = 3 ⋅ 3 + 10 = 19

For x = 3, — AC ≅ — CD , so the triangles are not congruent.

5x − 2 = 3x + 10

2x = 12

x = 6

AB = 5 ⋅ 6 = 30

AC = 5 ⋅ 6 − 2 = 30 − 2 = 28

BD = 4 ⋅ 6 + 3 = 27

CD = 3 ⋅ 6 + 10 = 28

For x = 6, — AB ≅ — BD , so the triangles are not congruent.

x = 5 is the only solution that yields △ABC ≅ △DCB.

Maintaining Mathematical Profi ciency 37. — DF corresponds to — AC . So, — AC ≅ — DF .

38. — BC corresponds to — EF . So, — EF ≅ — BC .

39. ∠E corresponds to ∠B. So, ∠B ≅ ∠E.

40. ∠C corresponds to ∠F. So, ∠F ≅ ∠C.



c. The side that △ABC and △ABD share is — AB . So, — AB ≅ — AB . Also, — BC of △ABC is congruent to — BD of △ABD. The nonincluded angle that these two triangles share is ∠BAC.

d. no; The third pair of sides are not congruent.

e. no; These two triangles provide a counterexample for SSA. They have two pairs of congruent sides and a pair of nonincluded congruent angles, but the triangles are not congruent.


Chapter 5

2. Possible congruence theorem

Valid or not valid?

SSS Valid

SSA Not valid

SAS Valid

AAS Valid

ASA Valid

AAA Not valid

Sample answer: A counterexample for SSA is given in Exploration 1. A counterexample for AAA is shown here.

x

y

456

−4

5−5 ABC′

B′ A′

C

In this example, each pair of corresponding angles is congruent, but the corresponding sides are not congruent.

3. In order to determine that two triangles are congruent, one of the following must be true.

All three pairs of corresponding sides are congruent (SSS).

Two pairs of corresponding sides and the pair of included angles are congruent (SAS).

Two pairs of corresponding angles and the pair of included sides are congruent (ASA).

Two pairs of corresponding angles and one pair of nonincluded sides are congruent (AAS).

The hypotenuses and one pair of corresponding legs of two right triangles are congruent (HL).

4. yes; Sample answer: A

DB C

In the diagram, △ABD ≅ △ACD by theHL Congruence Theorem (Thm. 5.9), the SSS Congruence Theorem (Thm. 5.8), and the SAS Congruence Theorem (Thm. 5.5).

5.6 Monitoring Progress (pp. 271–273) 1. yes; By the Alternate Interior Angles Theorem (Thm. 3.2),

∠1 ≅ ∠3 and ∠4 ≅ ∠2, and by the Refl exive Property of Congruence (Thm. 2.1), — WY ≅ — WY . So, △WXY ≅ △YZW by the ASA Congruence Theorem (Thm. 5.10).

2. Given — AB ⊥ — AD , — DE ⊥ — AD , E

DA

B

C

— AC ≅ — DC

Prove △ABC ≅ △DEC

STATEMENTS REASONS

1. — AB ⊥ — AD , — DE ⊥ — AD , — AC ≅ — DC

1. Given

2. ∠BAC and ∠EDC are right angles.


3. ∠BAC ≅ ∠EDC 3. Right Angles Congruence Theorem (Thm. 2.3)

4. ∠ACB ≅ ∠DCE 4. Vertical Angles Congruence Theorem (Thm. 2.6)

5. △ABC ≅ △DEC 5. ASA Congruence Theorem (Thm. 5.10)

3. Given ∠S ≅ ∠U, — RS ≅ — VU U

V

T

S

R

Prove △RST ≅ △VUT

STATEMENTS REASONS

1. ∠S ≅ ∠U, — RS ≅ — VU

1. Given

2. ∠RTS ≅ ∠VTU 2. Vertical Angles Congruence Theorem (Thm. 2.6)

3. △RST ≅ △VUT 3. AAS Congruence Theorem (Thm. 5.11)


Vocabulary and Core Concept Check 1. Both theorems are used to prove that two triangles are

congruent, and both require two pairs of corresponding angles to be congruent. In order to use the AAS Congruence Theorem (Thm. 5.11), one pair of corresponding nonincluded sides must also be congruent. In order to use the ASA Congruent Theorem (Thm. 5.10), the pair of corresponding included sides must be congruent.

2. You need to know that one pair of corresponding sides are congruent.


Chapter 5

Modeling Progress and Modeling with Mathematics 3. yes; △ABC ≅ △QRS by the AAS Congruence Theorem

(Thm. 5.11).

4. yes; △ABC ≅ △DBC by the AAS Congruence Theorem (Thm. 5.11).

5. no; Two sides and a nonincluded angle are not suffi cient to determine congruence.

6. yes; △RSV ≅ △UTV by the ASA Congruence Theorem (Thm. 5.10).

7. Given — GH ≅ — MN , ∠G ≅ ∠M, ∠F ≅ ∠L

8. Given — FG ≅ — LM , ∠G ≅ ∠M, ∠F ≅ ∠L

9. yes; △ABC ≅ △DEF by the ASA Congruence Theorem (Thm. 5.10).

10. no; The congruence statements follow the pattern SSA, which is not suffi cient to conclude that the triangles are congruent.

11. no; — AC and — DE are not corresponding sides.

12. yes; △ABC ≅ △DEF by the AAS Congruence Theorem (Thm. 5.11).

13. Construct a side that is congruent to — DF . Construct an angle that is congruent to ∠D and a second angle that is congruent to ∠F. Complete the triangle. By the ASA Congruence Theorem (Thm. 5.10), △DEF ≅ △ACB.

14. Construct a side that is congruent to — JK . Construct an angle that is congruent to ∠J and construct a second angle that is congruent to ∠K. Complete the triangle. By the ASA Congruence Theorem (Thm. 5.10), △JKL ≅ △BAC.

15. In the congruence statement, the vertices should be in corresponding order; △JKL ≅ △FGH by the ASA Congruence Theorem (Thm. 5.10).

16. The diagram shows ∠Q ≅ ∠V, ∠R ≅ ∠W, and — QR ≅ — VW . Because — QR and — VW are included sides, △QRS ≅ △VWX by the ASA Congruence Theorem (Thm. 5.10).

17. Given M is the midpoint of — NL , — NL ⊥ — NQ , — NL ⊥ — MP , — QM ) — PL

Prove △NQM ≅ △MPL

N M

Q

L

P

STATEMENTS REASONS

1. M is the midpoint of — NL , — NL ⊥ — NQ , — NL ⊥ — MP , — QM ) — PL

1. Given

2. ∠QNM and ∠PML are right angles.


3. ∠QNM ≅ ∠PML 3. Right Angles Congruence Theorem (Thm. 2.3)

4. ∠QMN ≅ ∠PLM 4. Corresponding Angles Theorem (Thm. 3.1)

5. — NM ≅ — ML 5. Defi nition of midpoint

6. △NQM ≅ △MPL 6. ASA Congruence Theorem (Thm. 5.10)

18. Given — AJ ≅ — KC , B

KJA C

∠BJK ≅ ∠BKJ, ∠A ≅ ∠C

Prove △ABK ≅ △CBJ

STATEMENTS REASONS

1. — AJ ≅ — KC , ∠BJK ≅ ∠BKJ, ∠A ≅ ∠C

1. Given

2. AJ = KC 2. Defi nition of congruent segments

3. JC = JK + KC,AK = AJ + JK


4. AK = KC + JK 4. Substitution Property of Equality

5. AK = JK + KC 5. Commutative Property of Addition

6. AK = JC 6. Transitive Property of Equality

7. — AK ≅ — JC 7. Defi nition of congruent segments

8. △ABK ≅ △CBJ 8. ASA Congruence Theorem (Thm. 5.10)


Chapter 5

19. Given — VW ≅ — UW , Z XY

U

W

V∠X ≅ ∠Z

Prove △XWV ≅ △ZWU

STATEMENTS REASONS

1. — VW ≅ — UW , ∠X ≅ ∠Z

1. Given

2. ∠W ≅ ∠W 2. Refl exive Property of Congruence (Thm. 2.2)

3. △XWV ≅ △ZWU 3. ASA Congruence Theorem (Thm. 5.11)

20. Given ∠NKM ≅ ∠LMK,

MK

L N

∠L ≅ ∠N

Prove △NMK ≅ △LKM

STATEMENTS REASONS

1. ∠NKM ≅ ∠LMK, ∠L ≅ ∠N

1. Given

2. — KM ≅ — MK 2. Refl exive Property of Congruence (Thm. 2.1)

3. △NMK ≅ △LKM 3. AAS Congruence Theorem (Thm. 5.11)

21. You are given two right triangles, so the triangles have congruent right angles by the Right Angles Congruence Theorem (Thm. 2.3). Because another pair of angles and a pair of corresponding nonincluded sides (the hypotenuses) are congruent, the triangles are congruent by the AAS Congruence Theorem (Thm. 5.11).

22. You are given two right triangles, so the triangles have congruent right angles by the Right Angles Congruence Theorem (Thm. 2.3). Because both pairs of legs are congruent, and the congruent right angles are the included angles, the triangles are congruent by the SAS Congruence Theorem (Thm. 5.5).

23. You are given two right triangles, so the triangles have congruent right angles by the Right Angles Congruence Theorem (Thm. 2.3). There is also another pair of congruent corresponding angles and a pair of congruent corresponding sides. If the pair of congruent sides is the included side, then the triangles are congruent by the ASA Congruence Theorem (Thm. 5.10). If the pair of congruent sides is a nonincluded pair, then the triangles are congruent by the AAS Congruence Theorem (Thm. 5.11).

24. D; To prove △JKL ≅ △MNL using ASA Congruence Theorem (Thm. 5.10), you know that ∠L ≅ ∠L and — LM ≅ — LJ . So, the additional information needed is ∠LKJ ≅ ∠LNM.

25. By fi nding the values of x and y in each triangle and solving for the angle measurements, △ABC ≅ △DBC by the ASA Congruence Theorem (Thm. 5.10) or the AAS Congruence Theorem (Thm. 5.11). Both triangles have a common side, — BC ≅ — BC .

△ABC: (2x − 8)° + (5x + 10)° + (8x − 32)° = 180° 15x − 30 = 180

15x = 210

x = 14

m∠CAB = 2 ⋅ 14 − 8 = 28 − 8 = 20° m∠ABC = 8 ⋅ 14 − 32 = 80° m∠BCA = 5 ⋅ 14 + 10 = 80°

△DBC: (y − 6)° + (3y + 2)° + (4y − 24)° = 180° 8y − 28 = 180

8y = 208

y = 26

m∠DBC = 4 ⋅ 26 − 24 = 80° m∠BCD = 3 ⋅ 26 + 2 = 80° m∠CDB = 26 − 6 = 20°

∠BAC ≅ ∠CDB, — BC ≅ — BC , ∠BCA ≅ ∠BCD △ABC ≅ △DBC by the ASA Congruence Theorem

(Thm 5.10).

∠BAC ≅ ∠CDB, — BC ≅ — BC , ∠BCA ≅ ∠BCD △ABC ≅ △DBC by the AAS Congruence Theorem

(Thm 5.11).

∠BAC ≅ ∠BDC, — BC ≅ — BC , ∠ABC ≅ ∠DBC △ABC ≅ △DBC by the AAS Congruence Theorem

(Thm 5.11).

26. C; ∠TSV ≅ ∠VUW (given), — ST ≅ — UV (given), and ∠STV ≅ ∠UVW because they are right angles.


Chapter 5

27. Given ∠B ≅ ∠C A

CDB

Prove — AB ≅ — AC

STATEMENTS REASONS

1. Draw — AD , the angle bisector of ∠ABC.

1. Construction of angle bisector

2. ∠CAD ≅ ∠BAD 2. Defi nition of angle bisector

3. ∠B ≅ ∠C 3. Given

4. — AD ≅ — AD 4. Refl exive Property of Congruence (Thm. 2.1)

5. △ABD ≅ △ACD 5. AAS Congruence Theorem (Thm. 5.11)

6. — AB ≅ — AC 6. Corresponding parts of congruent triangles are congruent.

28. yes; If you are using the AAS Congruence Theorem (Thm. 5.11), you have enough information to conclude that two pairs of angles are congruent and one pair of non-included sides is congruent. By the Third Angles Theorem (Thm. 5.4), the other pair of angles is congruent. Now, choose the third pair of angles, the pair of congruent sides, and the other pair of angles so that the pair of congruent sides are included sides. So, you can use the ASA Congruence Theorem (Thm. 5.10) to show that the triangles are congruent.

29. a. Given ∠CDB ≅ ∠ADB, DB ⊥ AC

Prove △ABD ≅ △CBD

STATEMENTS REASONS

1. ∠CDB ≅ ∠ADB, DB ⊥ AC

1. Given

2. ∠ABD ≅ ∠CBD are right angles.


3. ∠ABD ≅ ∠CBD 3. Right Angles Congruence Theorem (Thm. 2.3)

4. — BD ≅ — BD 4. Refl exive Property of Congruence (Thm. 2.1)

5. △ABD ≅ △CBD 5. ASA Congruence Theorem (Thm. 5.10)

b. Because △ABD ≅ △CBD and corresponding parts of congruent triangles are congruent, you can conclude that — AD ≅ — CD , which means that △ACD is isosceles by defi nition.

c. no; For instance, because △ACD is isosceles, the girl sees her toes at the bottom of the mirror. This remains true as she moves backwards because △ACD remains isosceles.

30. △PTS ≅ △RTQ, △PTQ ≅ △RTS, △SQR ≅ △QSP, △SRP ≅ △QPR; Because ∠PTS ≅ ∠RTQ by the Vertical Angles Congruence Theorem (Thm. 2.6), △PTS ≅ △RTQ by the SAS Congruence Theorem (Thm. 5.5); Because ∠PTQ ≅ ∠RTS by the Vertical Angles Congruence Theorem (Thm. 2.6), △PTQ ≅ △RTS by the SAS Congruence Theorem (Thm. 5.5); Transversal — SQ intersects parallel sides — PS and — QR to create congruent alternate interior angles ∠SQR and ∠QSP, and transversal — SQ also intersects parallel sides — PQ and — SR to create congruent alternate interior angles ∠QSR and ∠SQP. Also, — SQ ≅ — SQ by the Refl exive Property of Congruence (Thm. 2.1). So, △SQR ≅ △QSP by the ASA Congruence Theorem (Thm. 5.10); Transversal — PR intersects parallel sides — PS and — QR to create congruent alternate interior angles ∠SPR and ∠QRP, and transversal — PR also intersects parallel sides — PQ and — SR to create congruent alternate interior angles ∠QPR and ∠SRP. Also, — PR ≅ — PR by the Refl exive Property of Congruence (Thm. 2.1). So, △SRP ≅ △QPR by the ASA Congruence Theorem (Thm. 5.10).

31. Two triangles with congruent B

E

A

D F

C

corresponding angles can be similar triangles without being congruent. This is based on dilating a triangle. The angles are equal and corresponding sides are proportional.

32. yes; Both triangles will always have three vertices. So, each vertex of one triangle can be mapped to one vertex of the other triangle so that each vertex is only used once, and adjacent vertices of one triangle must be adjacent in the other.


Chapter 5

33. a. The combinations that will make △TUV ≅ △XYZ by the ASA Congruence Theorem (Thm. 5.10) are:

∠T ≅ ∠X, — TU ≅ — XY , ∠U ≅ ∠Y

∠U ≅ ∠Y, — UV ≅ — YZ , ∠V ≅ ∠Z

∠T ≅ ∠X, — TV ≅ — XZ , ∠V ≅ ∠Z

The combinations that will make △TUV ≅ △XYZ by the AAS Congruence Theorem (Thm. 5.11) are:

— TU ≅ — XY , ∠T ≅ ∠X, ∠V ≅ ∠Z

— UV ≅ — YZ , ∠U ≅ ∠Y, ∠T ≅ ∠X

— TV ≅ — XZ , ∠T ≅ ∠X, ∠U ≅ ∠Y

— TU ≅ — XY , ∠U ≅ ∠Y, ∠V ≅ ∠Z

— UV ≅ — YZ , ∠V ≅ ∠Z, ∠T ≅ ∠X

— TV ≅ — XZ , ∠V ≅ ∠Z, ∠U ≅ ∠Y

The combination that will make △TUV ≅ △XYZ by the SSS Congruence Theorem (Thm. 5.8) is:

— TU ≅ — XY , — UV ≅ — YZ , — TV ≅ — XZ

The combinations that will make △TUV ≅ △XYZ by the SAS Congruence Theorem (Thm. 5.5) are:

— TU ≅ — XY , ∠T ≅ ∠X, — TV ≅ — XZ

— UV ≅ — YZ , ∠U ≅ ∠Y, — TU ≅ — XY

— UV ≅ — YZ , ∠V ≅ ∠Z, — TV ≅ — XZ

b. There are 20 ways to choose 3 items from 6 items ( 6C3 = 20 ) . There are 13 combinations that provide enough information. So, the probability of choosing at random enough information to prove that the triangles are congruent is 13

— 20 , or 65%.

Maintaining Mathematical Profi ciency

34. Midpoint = ( 1 + 5 — 2 , 0 + 4 —

2 ) = ( 6 —

2 , 4 —

2 ) = (3, 2)

35. Midpoint = ( −2 + 4 — 2 , 3 + (−1) —

2 ) = ( 2 —

2 , 2 —

2 ) = (1, 1)

36. Midpoint = ( −5 + 2 — 2 , −7 + (−4) —

2 ) = ( −

3 —

2 , −

11 —

2 )

37. Draw a segment. Label a point D

D E

F

on the segment. Draw an arc with center A, and label the intersection points B and C. Using the same radius, draw an arc with center D. Label the point of intersection of the arc and the segment as E. Draw an arc with radius BC with center E. Label the intersection F. Draw '''⃗ DF . So, ∠FDE ≅ ∠A.

38. Draw a segment. Label a point D

D E

Fon the segment. Draw an arc with center B, and label the intersection points A and C. Using the same radius, draw an arc with center D. Label the point of intersection of the arc and the segment as E. Draw an arc with radius AC with center E. Label the intersection F. Draw '''⃗ DF . So, ∠FDE ≅ ∠B.

5.7 Explorations (p. 277) 1. a. The surveyor can measure — DE , which will have the same

measure as the distance across the river ( — AB ). Because △ABC ≅ △DEC by the ASA Congruence Theorem (Thm. 5.10), the corresponding parts of the two triangles are also congruent.

b. Given ∠A and ∠D B

AC

E

D

are right angles. — AC ≅ — CD

Prove AB = DE

STATEMENTS REASONS

1. — AC ≅ — CD , ∠A and ∠D are right angles.

1. Given

2. ∠A ≅ ∠D 2. Right Angles Congruence Theorem (Thm. 2.3)

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

4. △BCA ≅ △ECD 4. ASA Congruence Theorem (Thm. 5.10)

5. — AB ≅ — DE 5. Corresponding parts of congruent triangles are congruent.

6. AB = DE 6. Defi nition of congruent segments

c. By creating a triangle on land that is congruent to a triangle that crosses the river, you can fi nd the distance across the river by measuring the distance of the corresponding congruent segment on land.

2. a. The offi cer’s height stays the same, he is standing perpendicular to the ground the whole time, and he tipped his hat the same angle in both directions. So, △DEF ≅ △DEG by the ASA Congruence Theorem (Thm. 5.10). Because corresponding parts of the two triangles are also congruent, — EG ≅ — EF . By the defi nition of congruent segments, EG equals EF, which is the width of the river.


Chapter 5

b. Given ∠DEG is a right angle.∠DEF is a right angle.∠EDG ≅ ∠EDF

Prove EG = EF

D

E

GF

STATEMENTS REASONS

1. ∠EDG ≅ ∠EDF, ∠DEG and ∠DEF are right angles.

1. Given

2. ∠DEG ≅ ∠DEF 2. Right Angles Congruence Theorem (Thm. 2.3)

3. — DE ≅ — DE 3. Refl exive Property of Congruence (Thm. 2.1)

4. △FDE ≅ △GDE 4. ASA Congruence Theorem (Thm. 5.10)

5. — EG ≅ — EF 5. Corresponding parts of congruent triangles are congruent.

6. EG = EF 6. Defi nition of congruent segments

c. By standing perpendicular to the ground and using the tip of your hat to gaze at two different points in such a way that the direction of your gaze makes the same angle with your body both times, you can create two congruent triangles, which ensures that you are the same distance from both points.

3. By creating a triangle that is congruent to a triangle with an unknown side length or angle measure, you can measure the created triangle and use it to fi nd the unknown measure indirectly.

4. You do not actually measure the side length or angle measure you are trying to fi nd. You measure the side length or angle measure of a triangle that is congruent to the one you are trying to fi nd.

5.7 Monitoring Progress (pp. 278–280) 1. All three pairs of sides are congruent. So, by the SSS

Congruence Theorem (Thm. 5.8), △ABD ≅ △CBD. Because corresponding parts of congruent triangles are congruent, ∠A ≅ ∠C.

2. no; As long as the rest of the steps are followed correctly, △LKM will still be congruent to △PNM. So, the corresponding parts will still be congruent, and you will still be able to fi nd NP by measuring — LK .

3. You know that — TU ≅ — QP and — UP ≅ — PU . You need to show that — PT ≅ — UQ to prove that the triangles are congruent by the SSS Congruence Theorem (Thm. 5.8). △QSP ≅ △TRU by the HL Congruence Theorem (Thm. 5.9). So △USQ ≅ △PRT by the SAS Congruence Theorem (Thm. 5.5) and — PT ≅ — UQ .

4. Segments that can be assumed congruent are — AC and — AB .


Vocabulary and Core Concept Check 1. Corresponding parts of congruent triangles are congruent.

2. Sample answer: Finding the width of a body of water, a distance across a valley or canyon, or any inaccessible distance can require indirect measurement.

Monitoring Progress and Modeling with Mathematics 3. All three pairs of sides are congruent. So, by the SSS

Congruence Theorem (Thm. 5.8), △ABC ≅ △DBC. Because corresponding parts of congruent triangles are congruent, ∠A ≅ ∠D.

4. Two pairs of sides and the pair of included angles are congruent. So, by the SAS Congruence Theorem (Thm. 5.5), △QPR ≅ △TPS. Because corresponding parts of congruent triangles are congruent, ∠Q ≅ ∠T.

5. The hypotenuses and one pair of legs of two right triangles are congruent. So, by the HL Congruence Theorem (Thm. 5.9), △JMK ≅ △LMK. Because corresponding parts of congruent triangles are congruent, — JM ≅ — LM .

6. ∠BAD ≅ ∠ABC by the Alternate Interior Angles Theorem (Thm. 3.3). From the diagram, ∠B ≅ ∠C. — AD ≅ — AD by the Refl exive Property of Congruence (Thm. 2.1). So, by the AAS Congruence Theorem (Thm. 5.11), △ACD ≅ △DBA. Because corresponding parts of congruent triangles are congruent, — AC ≅ — DB .

7. From the diagram, ∠JHN ≅ ∠KGL, ∠N ≅ ∠L, and — JN ≅ — KL . So, by the AAS Congruence Theorem (Thm. 5.11), △JNH ≅ △KLG. Because corresponding parts of congruent triangles are congruent, — GK ≅ — HJ .

8. From the diagram, ∠Q ≅ ∠W ≅ ∠RVT ≅ ∠T and — VW ≅ — RT . So, by the AAS Congruence Theorem (Thm. 5.11), △QVW ≅ △VRT. Because corresponding parts of congruent triangles are congruent, — QW ≅ — VT .

9. Use the AAS Congruence Theorem (Thm 5.11) to prove that △FHG ≅ △GKF. Then, state that ∠FGK ≅ ∠GFH. Use the Congruent Complements Theorem (Thm. 2.5) to prove that ∠1 ≅ ∠2.

10. Use the AAS Congruence Theorem (Thm 5.11) to prove that △ABE ≅ △DCE. Then, state that — BE ≅ — CE because corresponding parts of congruent triangles are congruent. Use the Base Angles Theorem (Thm. 5.6) to prove that ∠1 ≅ ∠2.


Chapter 5

11. Use the ASA Congruence Theorem (Thm 5.10) to prove that △STR ≅ △QTP. Then, state that — PT ≅ — RT because corresponding parts of congruent triangles are congruent. Use the SAS Congruence Theorem (Thm. 5.5) to prove that △STP ≅ △QTR. So, ∠1 ≅ ∠2.

12. Use the ASA Congruence Theorem (Thm 5.10) to prove that △ABE ≅ △CBE. Then, state that — AE ≅ — CE and ∠BCE ≅ ∠BAE because corresponding parts of congruent triangles are congruent. Use the Congruent Complements Theorem (Thm. 2.5) to state that ∠ECD ≅ ∠EAF. So, you can use the SAS Congruence Theorem (Thm. 5.5) to prove that △ECD ≅ △EAF, so ∠1 ≅ ∠2.

13. Given — AP ≅ — BP and — AQ ≅ — BQ P

BA

Q

M Prove ∠AMP and ∠BMP are right angles.

STATEMENTS REASONS

1. — AP ≅ — BP , — AQ ≅ — BQ 1. Given


3. △APQ ≅ △BPQ 3. SSS Congruence Theorem (Thm. 5.8)

4. ∠APQ ≅ ∠BPQ 4. Corresponding parts of congruent triangles are congruent.

5. — PM ≅ — PM 5. Refl exive Property of Congruence (Thm. 2.1)

6. △APM ≅ △BPM 6. SAS Congruence Theorem (Thm. 5.5)

7. ∠AMP ≅ ∠BMP 7. Corresponding parts of congruent triangles are congruent.

8. ∠AMP and ∠BMP form a linear pair.

8. Defi nition of a linear pair

9. — MP ⊥ — AB 9. Linear Pair Perpendicular Theorem (Thm. 3.10)

10. ∠AMP and ∠BMP are right angles.


14. Given — PA ≅ — PB , — QA ≅ — QB

PBA

Q

Prove ∠QPA and ∠QPB are right angles.

STATEMENTS REASONS

1. — PA ≅ — PB , — QA ≅ — QB 1. Given


3. △APQ ≅ △BPQ 3. SSS Congruence Theorem (Thm. 5.8)

4. ∠QPA ≅ ∠QPB 4. Corresponding parts of congruent triangles are congruent.

5. ∠QPA and ∠QPB form a linear pair.

5. Defi nition of a linear pair

6. — PQ ⊥ — AB 6. Linear Pair Perpendicular Theorem (Thm. 3.10)

7. ∠QPA and ∠QPB are right angles.



Chapter 5

15. Given — FG ≅ — GJ ≅ — HG ≅ — GK , F H

NL

M

KJ

G — JM ≅ — LM ≅ — KM ≅ — NM

Prove — FL ≅ — HN

STATEMENTS REASONS

1. — FG ≅ — GJ ≅ — HG ≅ — GK , — JM ≅ — LM ≅ — KM ≅ — NM

1. Given

2. ∠FGJ ≅ ∠HGK, ∠JML ≅ ∠KMN

2. Vertical Angles Congruence Theorem (Thm. 2.6)

3. △FGJ ≅ △HGK,△JML ≅ △KMN

3. SAS Congruence Theorem (Thm. 5.5)

4. ∠F ≅ ∠H, ∠L ≅ ∠N

4. Corresponding parts of congruent triangles are congruent.

5. FG = GJ = HG = GK


6. HJ = HG + GJ, FK = FG + GK


7. FK = HG + GJ 7. Substitution Property of Equality

8. FK = HJ 8. Transitive Property of Equality

9. — FK ≅ — HJ 9. Defi nition of congruent segments

10. △HJN ≅ △FKL 10. AAS Congruence Theorem (Thm. 5.11)

11. — FL ≅ — HN 11. Corresponding parts of congruent triangles are congruent.

16. Given ∠PRU ≅ ∠QVS, Q

V

Y

W

X

R

P

TUS

∠PUR ≅ ∠QSV≅ ∠RUX ≅ ∠VSY, — RS ≅ — VU

Prove △PUX ≅ △QSY

STATEMENTS REASONS

1. ∠PRU ≅ ∠QVS, ∠PUR ≅ ∠QSV ≅ ∠RUX ≅ ∠VSY, — RS ≅ — VU

1. Given

2. RS = VU 2. Defi nition of congruent segments

3. RU = RS + SU, VS = VU + SU


4. VS = RS + SU 4. Substitution Property of Equality

5. RU = VS 5. Transitive Property of Equality

6. — RU ≅ — VS 6. Defi nition of congruent segments

7. △PUR ≅ △QSV 7. ASA Congruence Theorem (Thm. 5.10)

8. ∠P ≅ ∠Q, — PU ≅ — QS 8. Corresponding parts of congruent triangles are congruent.

9. m∠PUR = m∠QSV = m∠RUX = m∠VSY


10. m∠PUX = m∠PUR + m∠RUX, m∠QSY = m∠QSV + m∠VSY


11. m∠QSY = m∠PUR + m∠RUX


12. m∠PUX = m∠QSY 12. Transitive Property of Equality

13. ∠PUX ≅ ∠QSY 13. Defi nition of congruent angles

14. △PUX ≅ △QSY 14. ASA Congruence Theorem (Thm. 5.10)


Chapter 5

17. Because — AC ⊥ — BC and — ED ⊥ — BD , ∠ACB and ∠EDB are congruent right angles. Because B is the midpoint of — CD , — BC ≅ — BD . The vertical angles ∠ABC and ∠EBD are congruent. So, △ABC ≅ △EBD by the ASA Congruence Theorem (Thm. 5.10). Then, because corresponding parts of congruent triangles are congruent, — AC ≅ — ED . So, you can fi nd the distance AC across the canyon by measuring — ED .

18. a. Because the base of the red triangle is twice the base of the purple triangle, the red triangle has an area twice the area of the purple triangle.

b. Because the base of the orange triangle is twice the base of the purple triangle and the height of the orange triangle is twice the height of the purple triangle, the area of the orange triangle is four times the area of the purple triangle.

19. Given — AD ) — BC E is the midpoint of — AC .

Prove △AEB ≅ △CED

A B

CD

E

STATEMENTS REASONS

1. — AD ) — BC , E is the midpoint of — AC .

1. Given

2. — AE ≅ — CE 2. Defi nition of midpoint

3. ∠AED ≅ ∠BEC,∠AEB ≅ ∠CED

3. Vertical Angles Congruence Theorem (Thm. 2.6)

4. ∠DAE ≅ ∠BCE 4. Alternate Interior Angles Theorem (Thm. 3.2)

5. △DAE ≅ △BCE 5. ASA Congruence Theorem (Thm. 5.10)

6. — DE ≅ — BE 6. Corresponding parts of congruent triangles are congruent.

7. △AEB ≅ △CED 7. SAS Congruence Theorem (Thm. 5.5)

20. From the Internet, the distance form Miami to Bermuda is about 1035 miles. The distance from Bermuda to San Juan is about 956 miles. The distance form San Juan to Miami is about 1034 miles. The perimeter of the triangle created by these locations is about 1034 + 956 + 1034 = 3025 miles. To fi nd the area of this triangle, which is close to an isosceles triangle, let the distance between Miami and Bermuda be the same as the distance between Miami and San Juan. Let h be the height of the triangle.

B

S

J

478 miles

478 miles1035 miles

1035 miles

hM

h2 = 10352 − 4782

h2 = 1,071,225 − 228,484

h2 = 842,741

h = √—

842,741 ≈ 918

The approximate height of the triangle is 918 miles.

Area ≈ 1 — 2 ⋅ 956 ⋅ 918 = 438,804 mi2

The area of the triangle is about 439,000 square miles.

Sample answer: Three cities with approximate distances as the three cities in the Bermuda Triangle are Erie, PA, to Orlando, FL (970 miles); Erie, PA, to Oklahoma City, OK (1056 miles); and Oklahoma City, OK, to Orlando, FL (1064 miles).

Erie, PA

Orlando, FL

Oklahoma City, OK 970 miles

1056 miles

1064 miles

21. yes; You can show that WXYZ is a rectangle. This means that the opposite sides are congruent. Because △WZY and △YXW share a hypotenuse, the two triangles have congruent hypotenuses and corresponding legs, which allows you to use the HL Congruence Theorem (Thm 5.9) to prove that the triangles are congruent.

22. a. If two triangles have the same perimeter, then they are congruent is false. The converse is true: If two triangles are congruent, then their perimeters are the same.

b. If two triangles are congruent, then their areas are the same is true.


Chapter 5

23. — AC ≅ — GJ (given), ∠A ≅ ∠G because they are right angles, and — AB ≅ — GH (given). So, △ABC ≅ △GHJ by the SAS Congruence Theorem (Thm 5.5).

∠C ≅ ∠Q (given), — CA ≅ — QN (given), and ∠A ≅ ∠N because they are right angles. So, △ABC ≅ △NPQ by the ASA Congruence Theorem (Thm. 5.10).

Maintaining Mathematical Profi ciency

24. AB = √——

[ 4 − (−1) ] 2 + (1 − 1)2 = √—

(4 + 1)2

= √—

52 = √—

25 = 5

BC = √——

(4 − 4)2 + (−2 − 1)2 = √—

(−3)2 = √—

9 = 3

CD = √———

[ (−1) − 4 ] 2 + [ −2 − (−2) ] 2 = √—

(−1 − 4)2

= √—

(−5)2 = √—

25 = 5

AD = √———

[ (−1) − (−1) ] 2 + (−2 − 1)2

= √——

(−1 + 1)2 + (−3)2 = √—

(−3)2 = √—

9 = 3

Perimeter = AB + BC + CD + AD

= 5 + 3 + 5 + 3 = 16 units

25. JK = √———

[ (−2) − (−5) ] 2 + (1 − 3)2

= √——

(−2 + 5)2 + (−2)2

= √—

(3)2 + 4 = √—

9 + 4 = √—

13 ≈ 3.6

KL = √——

[ 3 − (−2) ] 2 + (4 − 1)2 = √——

(3 + 2)2 + 32

= √—

25 + 9 = √—

34 ≈ 5.8

JL = √——

[ 3 − (−5) ] 2 + (4 − 3)2

= √——

(3 + 5)2 + 12 = √—

82 + 1 = √—

65 = 8.1

Perimeter = JK + KL + JL

≈ 3.6 + 5.8 + 8.1 = 17.5 units

5.8 Explorations (p. 283) 1. a. Check students’ work. b. Check students’ work. c. Check students’ work. d. Using the Distance Formula, AC = √

— 9 + y2 and

AB = √—

9 + y2 . — AC ≅ — BC , so △ABC is an isosceles triangle.

2. a. Check students’ work.


c. AB = ∣ 6 − 0 ∣ = ∣ 6 ∣ = 6

BC = √——

(6 − 3)2 + (0 − 3)2 = √—

32 + (−3)2

= √—

9 + 9 = √—

18 = 3 √—

2

AC = √——

(0 − 3)2 + (0 − 3)2 = √——

(−3)2 + (−3)2

= √—

9 + 9 = √—

18 = 3 √—

2

m — AC = 3 − 0 — 3 − 0

= 3 — 3 = 1

m — BC = 3 − 0 — 3 − 6

= 3 — −3

= −1

Because AC = BC = 3 √—

2 and m — AC = 1 and m — BC = −1, AC ⊥ BC and △ABC is a right isosceles triangle.

d. Refl ect the triangle in the x-axis, so that △ABC is still an isosceles right triangle. So, C(3, −3).

x

y4

23

1

−3−4

−2

4 5 6321−2−1

A B

C

90°

e. If C lies on the line x = 3, then the coordinates are C(3, y). Because △ABC is an isosceles triangle,

m — AC = y — 3 , and m — BC = y —

−3 .

△ABC is a right triangle, so it must have a right angle. Because — AC and — BC are the congruent legs of △ABC, ∠A and ∠B are the congruent base angles by the Base Angles Theorem (Thm. 5.6). The vertex angle, ∠C, must be the right angle, which means that — AC ⊥ — BC by defi nition of perpendicular lines. By the Slopes of Perpendicular Lines Theorem (Thm. 3.14),

y — 3 ⋅ y —

−3 = −1 and y = ±3.

So, the coordinates of C must be (3, 3) or (3, −3).

3. You can position the fi gure in a coordinate plane and then use deductive reasoning to show that what you are trying to prove must be true based on the coordinates of the fi gure.

4. AB = √——

(6 − 0)2 + (0 − 0)2 = √—

62 = √—

36 = 6

AC = √——

(3 − 0)2 + (3 √—

3 − 0)2 = √——

32 + ( 3 √—

3 ) 2 = √

— 9 + 9 ⋅ 3 = √

— 9 + 27 = √

— 36 = 6

BC = √——

(6 − 3)2 + ( 0 − 3 √—

3 ) 2 = √——

32 + ( 3 √—

3 ) 2 = √

— 9 + 9 ⋅ 3 = √

— 9 + 27 = √

— 36 = 6

△ABC with vertices A(0, 0), B(6, 0), C ( 3, 3 √—

3 ) has side lengths AB = 6, AC = 6, and BC = 6. By the defi nition of congruent sides, △ABC is an equilateral triangle.

5.8 Monitoring Progress (pp. 284–286) 1. Another way of placing the rectangle in Example 1(a) that

is convenient for fi nding side lengths would be to place the width (k) on the x-axis.

x

y

(0, 0)

(0, h)

(k, 0)

(k, h)

h

k


Chapter 5

2. A square with vertices (0, 0), (m, 0), and (0, m) will have a fourth vertex at (m, m).

x

y

(0, m) (m, m)

(m, 0)(0, 0)

3. Use the Distance Formula to fi nd GO and GH in order to show they are congruent. State that ∠OGJ ≅ ∠HGJ by the defi nition of angle bisector and — GJ ≅ — GJ by the Refl exive Property of Congruence (Thm. 2.1). Then use the SAS Congruence Theorem (Thm. 5.5) to show that △GJO ≅ △GJH.

4.

x

y

O(0, 0) J(m, 0)

H(m, n)

Side lengths of △OHJ:

OH = √——

(m − 0)2 + (n − 0)2 = √—

m2 + n2

OJ = √——

(m − 0)2 + (0 − 0)2 = √—

m2 = m

HJ = √——

(m − m)2 + (0 − n)2 = √—

(−n)2 = √—

n2 = n

Midpoints of each side:

Midpoint of — OH = ( m + 0 — 2 , n + 0 —

2 ) = ( m —

2 , n —

2 )

Midpoint of — OJ = ( m + 0 — 2 , 0 + 0 —

2 ) = ( m —

2 , 0 )

Midpoint of — HJ = ( m + m — 2 , n + 0 —

2 ) = ( 2m —

2 , n —

2 ) = ( m, n —

2 )

Slopes of each side:

Slope of — OH = n − 0 — m − 0

= n — m

Slope of — OJ = 0 − 0 — m − 0

= 0 — m

= 0

Slope of — HJ = 0 − n — m − m

= −n — 0 = undefi ned

— OJ ⊥ — HJ at angle J. Therefore, △OHJ is a right triangle.

5. Given Coordinates of △NPO are N(h, h), P(0, 2h), and O(0, 0). Coordinates of △NMO are N(h, h), M(2h, 0), and O(0, 0).

Prove △NPO ≅ △NMO

x

y

M(2h, 0)

N(h, h)

P(0, 2h)

O(0, 0)

NP = √——

(0 − h)2 + (2h − h)2 = √—

(−h)2 + h2

= √—

2h2 = h √—

2

PO = √——

(0 − 0)2 + (0 − 2h)2 = √——

02 + (−2h)2

= √—

4h2 = 2h

NM = √——

(2h − h)2 + (0 − h)2 = √—

h2 + (−h)2

= √—

2h2 = h √—

2

MO = √——

(0 − 2h)2 + (0 − 0)2 = √——

(−2h)2 + 02

= √—

4h2 = 2h

So, — PN ≅ — MN and — PO ≅ — MO , and by the Refl exive Property of Congruence (Thm. 2.1), — ON ≅ — ON . So, you can apply the SSS Congruence Theorem (Thm. 5.8) to conclude that △NPO ≅ △NMO.


Vocabulary and Core Concept Check 1. In a coordinate proof, you have to assign coordinates to

vertices and write expressions for side length and the slope of segments in order to show how sides are related; As with other types of proofs, you still have to use deductive reasoning and justify every conclusion with theorems, proofs, and properties of mathematics.

2. When the triangle is positioned as shown, you are using zeros in your expressions, so the side lengths are often the same as one of the coordinates.

Monitoring Progress and Modeling with Mathematics 3. Sample answer: Place the legs on the x- and y-axes.

x

y

2

3

1

2 31A(0, 0)

B(3, 0)

C(0, 2)

It is easy to fi nd the lengths of horizontal and vertical segments and distances from the origin.


Chapter 5

4. Sample answer: Place one side on the x-axis and one side on the y-axis.

x

y

2

3

1

2 31X(0, 0)

Y(3, 0)

W(0, 3) Z(3, 3)


5. Sample answer: Place the legs on the x- and y-axes.

x

y

S(0, 0) T(p, 0)

R(0, p)


6. Sample answer: Place the side with length 2m on the x-axis.

x

y

K(2m, 0)

L(n, p)

J(0, 0)

It is easy to fi nd the lengths of horizontal segments and distances from the origin.

7. Find the lengths of — OP , — PM , — MN , and — NO to show that — OP ≅ — PM and — MN ≅ — NO .

8. Find the coordinates of G using the Midpoint Formula. Use these coordinates and the Distance Formula to show that — OG ≅ — JG . Show that — HG ≅ — FG by the defi nition of midpoint, and ∠HGL ≅ ∠FGO by the Vertical Angles Congruence Theorem (Thm. 2.6). Then use the SAS Congruence Theorem (Thm. 5.5) to conclude that △GHJ ≅ △GFO.

9.

x

y

45678

23

1

4 5 6 7 8 9321O(0, 0)

D(9, 0)

C(0, 7)

CD = √——

(0 − 9)2 + (7 − 0)2 = √—

(−9)2 + 72

= √—

81 + 49 = √—

130 ≈ 11.4 The length of the hypotenuse is about 11.4 units.

10.

x

y

10

20

30

40

50

20 3010−10−20−30

C(30, 0)

A(0, 50)

B(−30, 0)

Use the Pythagorean Theorem c2 = a2 + b2, where the hypotenuse of the right triangle is one of the legs of the isosceles triangle.

AC2 = 502 + 302

AC2 = 2500 + 900 AC2 = 3400 AC = √

— 3400 ≈ 58.3

The length of one of the legs of the isosceles triangle is about 58.3 units.

11.

x

y

456

23

1

4 5 6321O(0, 0)

L(0, 4) M(5, 4)

N(5, 0)

NL = √——

(5 − 0)2 + (0 − 4)2 = √—

52 + (−4)2

= √—

25 + 16 = √—

41 ≈ 6.4

The length of the diagonal is about 6.4 units.


Chapter 5

12.

x

y

O(0, 0)

Z(0, n)

X(n, 0)

Y(n, n)

XZ = √——

(n − 0)2 + (0 − n)2 = √—

n2 + (−n)2

= √—

n2 + n2 = √—

2n2 = n √—

2

The length of the diagonal is n √—

2 units.

13.

x

y

A(0, 0) C(2h, 0)

B(h, h)

Side lengths of △ABC:

AB = √——

(h − 0)2 + (h − 0)2 = √—

h2 + h2 = √—

2h2 = h √—

2

BC = √——

(2h − h)2 + (0 − h)2 = √—

h2 + h2 = √—

2h2 = h √—

2

AC = √——

(2h − 0)2 + (0 − 0)2 = √—

4h2 = 2h

Slopes of each side of △ABC:

Slope of — AB = h − 0 — h − 0

= h — h = 1

Slope of — BC = 2h − h — 0 − h

= h — −h

= −1

Slope of — AC = 0 − 0 — 2h − 0

= 0 — 2h

= 0

Midpoints of each side of △ABC:

Midpoint of — AB = ( 0 + h — 2 , 0 + h —

2 ) = ( h —

2 , h —

2 )

Midpoint of — BC = ( h + 2h — 2 , h + 0 —

2 ) = ( 3h —

2 , h —

2 )

Midpoint of — AC = ( 2h + 0 — 2 , 0 + 0 —

2 ) = ( 2h —

2 , 0 ) = (h, 0)

Because m — AB ⋅ m — BC = −1, — AB ⊥ — BC by the Slopes of Perpendicular Lines Theorem (Thm. 3.14). So, ∠ABC is a right angle. — AB ≅ — BC because AB = BC. So, △ABC is a right isosceles triangle.

14.

x

y

D(0, n)

F(m, 0)

E(m, n)

Side lengths of △DEF:

DE = √——

(m − 0)2 + (n − n)2 = √—

m2 + 0 = √—

m2 = m

EF = √——

(m − m)2 + (0 − n)2 = √—

0 + (−n)2 = √—

n2 = n

DF = √——

(m − 0)2 + (0 − n)2 = √—

m2 + n2

Slopes of each side of △DEF:

Slope of — DE = n − n — m − 0

= 0 — m

= 0

Slope of — EF = 0 − n — m − m

= −n — 0 = undefi ned

Slope of — DF = 0 − n — m − 0

= − n —

m

Midpoints of each side of △DEF:

Midpoint of — DE = ( 0 + m — 2 , n + n —

2 ) = ( m —

2 , 2n —

2 ) = ( m —

2 , n )

Midpoint of — EF = ( m + m — 2 , n + 0 —

2 ) = ( 2m —

2 , n —

2 ) = ( m, n —

2 )

Midpoint of — DF = ( 0 + m — 2 , n + 0 —

2 ) = ( m —

2 , n —

2 )

Because m — DE = 0, — DE is horizontal. Because m — EF = undefi ned, — EF is vertical. So, — DE ⊥ — BC by Postulate 3.5, and ∠DEF is a right angle by the defi nition of perpendicular lines. Also, none of the sides have the same length. So, △DEF is a right scalene triangle.

15. The coordinates of the unlabeled vertex is N(h, k).

ON = √——

(h − 0)2 + (k − 0)2 = √—

h2 + k2

MN = √——

(2h − h)2 + (0 − k)2 = √—

h2 + (−k)2 = √—

h2 + k2

16. The coordinates of the vertices are O(0, 0), U(k , 0), R(k, k), S(k, 2k), and T(2k, 2k).

OT = √——

(2k − 0)2 + (2k − 0)2 = √——

(2k)2 + (2k)2

= √—

4k2 + 4k2 = √—

8k2 = 2k √—

2


Chapter 5

17. Find the lengths of — DE , — EC , and — DC of △DEC.

DE = √——

(2h − h)2 + (2k − 2k)2 = √—

(h)2 = h

EC = √——

(2h − h)2 + (2k − k)2 = √—

h2 + k2

DC = √——

(h − h)2 + (2k − k)2 = √—

k2 = k

Find the lengths of — BO , — OC , and — BC of △BOC.

BO = √——

(h − 0)2 + (0 − 0)2 = √—

(h)2 = h

OC = √——

(h − 0)2 + (k − 0)2 = √—

h2 + k2

BC = √——

(h − h)2 + (k − 0)2 = √—

k2 = k

Because DE = BO, EC = OC, and DC = BC, by the defi nition of congruent segments — DE ≅ — BO , — EC ≅ — OC , — DC ≅ — BC . By the SSS Congruence Theorem (Thm. 5.8), △DEC ≅ △BOC.

18. Because H is the midpoint of — DA , the coordinates of point H are

H ( 0 − 2h — 2 , 2k + 0 —

2 ) = H(−h, k).

Because G is the midpoint of — EA , the coordinates of point G are

G ( 0 + 2h — 2 , 2k + 0 —

2 ) = (h, k).

DG = √——

(−2h − h)2 + (0 − k)2 = √——

(−3h)2 + (−k)2

= √—

9h2 + k2

EH = √———

[2h − (−h)]2 + (0 − k)2 = √——

(3h)2 + (−k)2

= √—

9h2 + k2

Because DG = EH, by the defi nition of congruent segments, — DG ≅ — EH .

19. The triangle formed by your position (Y ), your cousins position (C), and the campsite (O) has the coordinates Y(500, 1200), C(1000, 0), and O(0, 0).

x

y

500

1000

500 1000O(0, 0)

C(1000, 0)

Y(500, 1200)

OY = √———

(500 − 0)2 + (1200 − 0)2

= √——

5002 + 12002

= √——

250,0000 + 1,440,000

= √—

1,690,000 = 1300

The distance between your position and the campsite is 1300 meters.

YC = √———

(1000 − 500)2 + (0 − 1200)2

= √——

(5002 + (−1200)2

= √——

250,000 + 1,440,000

= √—

1,690,000 = 1300

The distance between your position and your cousin’s position is 1300 meters.

OH = √——

(1000 − 0)2 + (0 − 0)2

= √—

(1000)2

= 1000

The distance between the campsite and your cousin’s position is 1000 meters.

Because — OY ≅ — YC , the triangle formed by your position, your cousin’s position, and the campsite is an isosceles triangle.


Chapter 5

20.

x

y

−4

42−2−4

P(0, 2)S(−2, 1)

R(1, −5)

Q(3, −4)

Side lengths:

PQ = √——

(3 − 0)2 + (−4 − 2)2 = √—

32 + (−6)2

= √—

9 + 36 = √—

45 = 3 √—

5

QR = √———

(1 − 3)2 + [ −5 − (−4) ] 2 = √——

(−2)2 + (−5 + 4)2

= √—

4 + (−1)2 = √—

4 + 1 = √—

5

RS = √———

(−2 − 1)2 + [ 1 − (−5) ] 2 = √——

(−3)2 + (1 + 5)2

= √—

9 + 36 = √—

45 = 3 √—

5

PS = √——

(−2 − 0)2 + (1 − 2)2 = √——

(−2)2 + (−1)2

= √—

4 + 1 = √—

5

Slopes of the sides:

Slope of — PQ = −4 − 2 — 3 − 0

= −6 — 3 = −2

Slope of — QR = −5 − (−4) — 1 − 3

= −5 + 4 — −2

= −1 — −2

= 1 — 2

Slope of — RS = 1 − (−5) — −2 − 1

= 1 + 5 — −3

= 6 — −3

= −2

Slope of — PS = 1 − 2 — −2 − 0

= −1 — −2

= 1 — 2

So, — PQ ≅ — SR and — SP ≅ — RQ , which shows that opposite sides are congruent. Also, m — PQ ⋅ m — SP = −1, m — PQ ⋅ m — RQ = − 1, m — SR ⋅ m — SP = −1, m — SR ⋅ m — RQ = −1.

So, — PQ ⊥ — SP , — PQ ⊥ — RQ , — SR ⊥ — SP , and — SR ⊥ — RQ by the Slopes of Perpendicular Lines Theorem (Thm. 3.14). So, by defi nition of perpendicular lines, ∠PSR, ∠SRQ, ∠RQP, and ∠QPS are right angles. So, the quadrilateral is a rectangle. The second friend is correct.

21. The endpoints of a segment with the origin as the midpoint are (x, y) and (−x, −y) because

M ( x + (−x) — 2 , y + (−y) —

2 ) = M ( 0 —

2 , 0 —

2 ) = M(0, 0).

22. B

x

y

(0, v)

(w, 0)

(0, −v)

(−w, 0)

Midpoint = ( −w + 0 — 2 , 0 + (−v) —

2 ) = ( −

w —

2 , −

v —

2 )

23. A

x

y

2

6

−2

2

(−h, k) (2h, k)

(2h, 0)(−h, 0)

24. Sample answer: It would be easy to prove the Base Angles Theorem (Thm. 5.6) with a coordinate proof. First, position the given isosceles triangle, △ABC, on the coordinate plane so that the base is on the x-axis, and one vertex is at the origin.

x

y

B(0, 0) D(k, 0) C(2k, 0)

A(k, m)

x = k

This is an isosceles triangle because BA = √—

k2 + m2 and CA = √

— k2 + m2 . Draw the line x = k that intersects △ABC

in point A(m, k) and the x-axis in the point (k, 0). Call this point D, — BD and — DC are congruent because BD = k and DC = k. — AD ≅ — AD by the Refl exive Property of Congruence (Thm. 2.1). Because — AD is vertical and — BC is horizontal, AD ⊥ BC is the Slopes of Perpendicular Lines Theorem (Thm. 3.14). So, ∠BDA and ∠CDA are congruent right angles. By the SAS Congruence Theorem (Thm. 5.5), △ABD ≅ △ACD. Because corresponding parts of congruent triangles are congruent, ∠B ≅ ∠C.

25. Sample answer: Refl ect the triangle in the y-axis and translate 5d units right and 5d units up.

(5d, −5d) → (−5d, −5d) → (0, 0) (0, −5d) → (0, −5d) → (5d, 0) (5d, 0) → (−5d, 0) → (0, 5d)

26. Diagonal — WU is horizontal, and diagonal — TV is vertical. So, by the Slopes of Perpendicular Lines Theorem (Thm. 3.14), — WU ⊥ — TV ; Change the coordinates to T(0, m), U(m, 0), V(0, −m), and W(−m, 0). These coordinates can be used for any square, and the diagonals are still horizontal and vertical. So, the diagonals are perpendicular for any square.


Chapter 5

27. a.

x

y

B(0, 0)

A(0, 2m)

C(2n, 0)

M(n, m)

Because M is the midpoint of — AC , the coordinates of M are M(n, m). Using the Distance Formula, AM = √

— n2 + m2 ,

BM = √—

n2 + m2 , and CM = √—

n2 + m2 . So, the midpoint of the hypotenuse of a right triangle is the same distance from each vertex of the triangle.

b.

x

y

T(m, 0)

R(0, m)

O(0, 0)S(−m, 0)

When any two congruent right isosceles triangles are positioned with the vertex opposite the hypotenuse on the origin and their legs on the axes as shown in the diagram, a triangle is formed and the hypotenuses of the original triangles make up two sides of the new triangle. SR = m √

— 2 and TR = m √

— 2 so these two sides are the

same length. So, by defi nition, △SRT is isosceles.

Maintaining Mathematical Profi ciency 28. m∠XYW = m∠WYZ

(3x − 7)° = (2x + 1)° x − 7 = 1

x = 8

29. m∠XYZ = (3x − 7)° + (2x + 1)° m∠XYZ = 5x − 6

From Exercise 28, you know that x = 8.

m∠XYZ = 5 ⋅ 8 − 6 = 34°

5.5 –5.8 What Did You Learn? (p. 289) 1. Given square ABCD with diagonal — BD , as shown, prove

△BAD ≅ △DCB; In this problem, the square could represent the baseball “diamond,” and then diagonal — BD would represent the distance from home plate to second base. So, you could use this problem to prove the equivalent of △HFS ≅ △STH. Then you could just redraw square ABCD with diagonal — AC this time, so that △CBA is the equivalent of △FST. It could easily be shown that the third triangle is congruent to the fi rst two.

2. The theorems on page 275 correspond to other triangle congruence theorems given. HA corresponds to AAS, LL corresponds to SAS, and AL corresponds to ASA or AAS.

3. Two sources that could be used to help solve Exercise 20 on page 282, are a map website and a website that calculates distances.

Chapter 5 Review (pp. 290–294) 1. The triangle is an acute isosceles triangle because it has two

equal sides and it appears that all angles are acute.

2. Because 46° + 86° = 132°, the measure of the exterior angle is 132°.

3. (9x + 9)° = 45° + 5x° 4x + 9 = 45

4x = 36

x = 9

The exterior angle:

(9x + 9)° = 9 ⋅ 9 + 9

= 81 + 9

= 90

The measure of the exterior angle is 90°.

4. 8x° + 7x° + 90° = 180° 15x + 90 = 180

15x = 90

x = 6

8x = 8 ⋅ 6 = 48

7x = 7 ⋅ 6 = 42

The measure of each acute angle is 42° and 48°.

5. (7x° + 6)° + (6x − 7)° + 90° = 180° 13x + 89 = 180

13x = 91

x = 7

7x + 6 = 7 ⋅ 7 + 6 = 49 + 6 = 55

6x − 7 = 6 ⋅ 7 − 7 = 35

The measure of each acute angle is 35° and 55°.

6. corresponding angles: ∠G ≅ ∠L, ∠H ≅ ∠M, ∠J ≅ ∠N, ∠K ≅ ∠P

corresponding sides: — GH ≅ — LM , — HJ ≅ — MN , — JK ≅ — NP , — GK ≅ — LP

Sample answer: Another congruence statement is KJHG ≅ PNML.


Chapter 5

7. m∠S = m∠T = 74° 90° + 74° + m∠V = 180° 164° + m∠V = 180° m∠V = 16°

8. no; There is enough information to prove two pairs of congruent sides and one pair of congruent angles, but the angle is not the included angle.

9. yes;

STATEMENTS REASONS

1. — WX ≅ — YZ , WZ ) YX 1. Given

2. — XZ ≅ — XZ 2. Refl exive Property of Congruence (Thm. 2.1)

3. ∠WXZ ≅ ∠YZX 3. Alternate Interior Angles Theorem (Thm. 3.2)

4. △WXZ ≅ △YZX 4. SAS Congruence Theorem (Thm. 5.5)

10. If — QP ≅ — QR , then ∠QRP ≅ ∠P.

11. If ∠TRV ≅ ∠TVR, then — TV ≅ — TR .

12. If — RQ ≅ — RS , then ∠RSQ ≅ ∠RQS.

13. If ∠SRV ≅ ∠SVR, then — SV ≅ — SR .

14. 8x° = 180° − 60° 8x = 120

x = 15

5y + 1 = 26

5y = 25

y = 5

So, x = 15 and y = 5.

15. no; There is only enough information to conclude that two pairs of sides are congruent.

16. yes;

STATEMENTS REASONS

1. — WX ≅ — YZ , ∠XWZ and ∠ZYX are right angles.

1. Given

2. — XZ ≅ — XZ 2. Refl exive Property of Congruence (Thm. 2.1)

3. △WXZ and △YZX are right triangles.


4. △WXZ ≅ △YZX 4. HL Congruence Theorem (Thm. 5.9)

17. yes;

STATEMENTS REASONS

1. ∠E ≅ ∠H, ∠F ≅ ∠J, — FG ≅ — JK

1. Given

2. △EFG ≅ △HJK 2. AAS Congruence Theorem (Thm. 5.11)

18. no; There is only enough information to conclude that one pair of angles and one pair of sides are congruent.

19. yes;

STATEMENTS REASONS

1. ∠PLN ≅ ∠MLN, ∠PNL ≅ ∠MNL

1. Given

2. — LN ≅ — LN 2. Refl exive Property of Congruence (Thm. 2.1)

3. △LPN ≅ △LMN 3. ASA Congruence Theorem (Thm. 5.10)

20. no; There is only enough information to conclude that one pair of angles and one pair of sides are congruent.

21. By the SAS Congruence Theorem (Thm. 5.5), △HJK ≅ △LMN. Because corresponding parts of congruent triangles are congruent, ∠K ≅ ∠N.

22. First, state that — QV ≅ — QV . then, use the SSS Congruence Theorem (Thm. 5.8) to prove that △QSV ≅ △QTV. Because corresponding parts of congruent triangles are congruent, ∠QSV ≅ ∠QTV. ∠QSV ≅ ∠1 and ∠QTV ≅ ∠2 by the Vertical Angles Congruence Theorem (Thm. 2.6). So, by the Transitive Property of Congruence (Thm. 2.2), ∠1 ≅ ∠2.

23. OP = √——

(h − 0)2 + (k − 0)2 = √—

h2 + k2

QP = √———

(h − h)2 + [ (k + j) − k ] 2 = √—

j 2 = j

QO = √——

(h − 0)2 + (k + j)2 = √——

h2 + (k + j)2

QR = √——

(h − 0)2 + (k − 0)2 = √—

h2 + k2

OR = √——

(0 − 0)2 + (0 − j)2 = √—

j 2 = j

So, — OP ≅ — QR and — OR ≅ — QP . Also, by the Refl exive Property of Congruence (Thm. 2.1), — QO ≅ — QO . So, you can apply the SSS Congruence Theorem (Thm. 5.8) to conclude that △OPQ ≅ △QRO.


Chapter 5

24. Place the base along the x-axis and the vertex of the legs on the y-axis.

x

y

C(p, 0)

B(0, k)

A(−p, 0)

25.

x

y

(0, k) (2k, k)

(2k, 0)(0, 0)

The fourth vertex of the rectangle has the vertex (2k, k).

Chapter 5 Test (p. 295) 1. Given — CA ≅ — CB ≅ — CD ≅ — CE A B

C

ED

Prove △ABC ≅ △EDC

STATEMENTS REASONS

1. — CA ≅ — CB ≅ — CD ≅ — CE 1. Given

2. ∠ACB ≅ ∠ECD 2. Vertical Angles Congruence Theorem (Thm. 5.5)

3. △ABC ≅ △EDC 3. SAS Congruence Theorem (Thm. 5.5)

2. Given — JK ) — ML ,

— MJ ) — KL J K

LM

Prove △MJK ≅ △KLM

STATEMENTS REASONS

1. — JK ) — ML , — MJ ) — KL

1. Given

2. — MK ≅ — KM 2. Refl exive Property of Congruence (Thm. 2.1)

3. ∠JKM ≅ ∠LMK, ∠JMK ≅ ∠LKM

3. Alternate Interior Angles Theorem (Thm. 3.2)

4. △MJK ≅ △KLM 4. ASA Congruence Theorem (Thm. 5.10)

3. Given — QR ≅ — RS , ∠P ≅ ∠T

P T

N

R

SQ Prove △SRP ≅ △QRT

STATEMENTS REASONS

1. — QR ≅ — RS , ∠P ≅ ∠T 1. Given

2. ∠R ≅ ∠R 2. Refl exive Property of Congruence (Thm. 2.2)

3. △SRP ≅ △QRT 3. AAS Congruence Theorem (Thm. 5.11)

4. (4x − 2)° + (3x + 8)° = 90° 7x + 6 = 90

7x = 84

x = 12

4x − 2 = 4 ⋅ 12 − 2 = 48 − 2 = 46

3x + 8 = 3 ⋅ 12 + 8 = 36 + 8 = 44

The measures of the acute angles are 44° and 46°.

5. no; By the Corollary to the Base Angles Theorem (Cor. 5.2), if a triangle is equilateral, then it is also equiangular.

6. no; The Third Angles Theorem (Thm. 5.4) can be used to prove that two triangles are equiangular, but AAA is not suffi cient to prove that the triangles are congruent. You need to know that at least one pair of corresponding sides are congruent.

7. First, use the HL Congruence Theorem (Thm. 5.9) to prove that △ACD ≅ △BED. Because corresponding parts of congruent triangles are congruent, — AD ≅ — BD . Then, use the Base Angles Theorem (Thm. 5.6) to prove that ∠1 ≅ ∠2.

8. Use the SSS Congruence Theorem (Thm. 5.8) to prove that △SVX ≅ △SZX. Use the Vertical Angles Congruence Theorem (Thm. 2.6) and the SAS Congruence Theorem (Thm. 5.5) to prove that △VXW ≅ △ZXY. Because corresponding parts of congruent triangles are congruent, ∠SZX ≅ ∠SVX and ∠W ≅ ∠Y. Use the Segment Addition Postulate (Post. 1.2) to show that — VY ≅ — ZW . Then, use the ASA Congruence Theorem (Thm. 5.10) to prove that △VYT ≅ △ZWR. Because corresponding parts of congruent triangles are congruent, ∠1 ≅ ∠2.

9. yes; HL Congruence Theorem (Thm. 5.9), ASA Congruence Theorem (Thm. 5.10), AAS Congruence Theorem (Thm. 5.11), SAS Congruence Theorem (Thm. 5.5)


Chapter 5

10. Use the Distance Formula to fi nd the lengths of — PQ and — ST .

PQ = √——

(21 − 3)2 + (30 − 30)2 = √—

182 = 18

ST = √——

(21 − 3)2 + (0 − 0)2 = √—

182 = 18

So, — PQ ≅ — ST . Also, the horizontal segments — PQ and — ST each have a slope of 0, which implies that they are parallel. So, — PS intersects — PQ and — ST to form congruent alternate interior angles, ∠P and ∠S. By the Vertical Angles Congruence Theorem (Thm. 2.6), ∠PRQ ≅ ∠SRT. So, by the AAS Congruence Theorem (Thm. 5.11), △PQR ≅ △STR.

11. a. The triangle shown is an isosceles triangle because it has two congruent sides.

b. By the Triangle Sum Theorem (Thm. 5.1):

m∠1 + m∠2 + m∠3 = 180° m∠1 + m∠2 + 40° = 180° m∠1 + m∠2 = 140° By the Base Angles Theorem (Thm. 5.6) ∠1 ≅ ∠2:

m∠2 + m∠2 = 140° 2m∠2 = 140° m∠2 = 70° m∠1 = 70° If m∠3 = 40°, then m∠1 = 70° and m∠2 = 70°.

Chapter 5 Standards Assessment (pp. 296–297) 1. no; the Exterior Angle Theorem (Thm. 5.2) follows from

the Triangle Sum Theorem (Thm. 5.1). Also, the Triangle Sum Theorem (Thm. 5.1) is used to prove the Exterior Angle Theorem (Thm. 5.2), so you cannot use the Exterior Angle Theorem (Thm. 5.2) to prove the Triangle Sum Theorem (Thm. 5.1).

2. By step 1, a line through point P intersects line m in point Q. By steps 2 and 3, — QA ≅ — QB ≅ — PC ≅ — PD and — AB ≅ — CD because congruent segments were drawn with the same compass setting. So, in step 4, you see that if — AB and — CD were drawn, then △AQB and △CPD would be congruent by the SSS Congruence Theorem (Thm. 5.8). Because corresponding parts of congruent triangles are congruent, ∠CBD ≅ ∠AQB, which means that ⃖ ''⃗ PD ) m by the Corresponding Angles Converse (Thm. 3.5).

3. a. Sample answer: Refl ect △JKL in the x-axis and then translate 4 units right.

b. yes; corresponding sides: — JK ≅ — XY , — KL ≅ — YZ , — JL ≅ — XZ ;corresponding angles: ∠J ≅ ∠X, ∠K ≅ ∠Y, ∠L ≅ ∠Z

4. C;

Slope = 5 − 0 — −2 − 8

= 5 — −10

= − 1 —

2

x = 2 — 5 ⋅ (10) + (−2) = 4 − 2 = 2

y = 2 — 5 ⋅ (−5) + 5 = −2 + 5 = 3

So, Q(2, 3).

5. a. In order to prove △ABC ≅ △DEF, show that — AB ≅ — DE .

AB = √——

(2 − 2)2 + (8 − 5)2 = √—

32 = √—

9 = 3

DE = √——

(8 − 5)2 + (2 − 2)2 = √—

32 = √—

9 = 3

The segments — AB and — DE have the same measure and are therefore congruent. Also, from the markings in the diagram, ∠B ≅ ∠E and — BC ≅ — EF . So, by the SAS Congruence Theorem (Thm. 5.5), △ABC ≅ △DEF.

b. Rotate △ABC 90° counterclockwise about the origin followed by a translation 13 units right.

6. yes; The coordinate rule for dilations is to multiply each coordinate of each point by the scale factor, which is 2 in this case. So, when you do this to the coordinates of point W, you get (2 ⋅ 0, 2 ⋅ 0) which is still (0, 0).

7. A, B, D

Figure A has 180° rotational symmetry. Figure B has 72° rotational symmetry. Figure D has 90° and 180° rotational symmetry.

8. To prove quadrilateral ABCD is a rectangle, opposite sides must be equal and right angles must be formed at the vertices.

Side lengths:

AB = √——

(4 − 2)2 + (7 − 5)2 = √—

22 + 22

= √—

4 + 4 = √—

8 = 2 √—

2

BC = √——

(4 − 7)2 + (7 − 4)2 = √—

(−3)2 + 32

= √—

9 + 9 = √—

18 = 3 √—

2

CD = √——

(7 − 5)2 + (4 − 2)2 = √—

22 + 22

= √—

4 + 4 = √—

8 = 2 √—

2

AD = √——

(5 − 2)2 + (2 − 5)2 = √——

(3)2 + (−3)2

= √—

9 + 9 = √—

18 = 3 √—

2

Slopes:

Slope of — AB = 7 − 5 — 4 − 2

= 2 — 2 = 1

Slope of — BC = 7 − 4 — 4 − 7

= 3 — −3

= −1

Slope of — CD = 4 − 2 — 7 − 5

= 2 — 2 = 1

Slope of — AD = 2 − 5 — 5 − 2

= −3 — 3 = −1

Sides — AD and — BC have the same measure and the same slope, as do — AB and — DC . So, by the Slopes of Parallel Lines Theorem (Thm. 3.13), — AD ) — BC and — AB ) — DC . Because the product of their slopes is −1, — AD ⊥ — AB , — AD ⊥ — DC , — BC ⊥ — AB , and — BC ⊥ — DC . So, ABCD is a rectangle.


Chapter 5

9. Prove △ABC is an equilateral triangle.

STATEMENTS REASONS

1. AB = AC, BA = BC 1. By construction

2. AC = BC 2. Transitive Property of Congruence (Thm. 2.1)

3. — AB ≅ — AC , — BA ≅ — BC , — AC ≅ — BC 3. Defi nition of congruent

segments

4. △ABC is an equilateral triangle.


hscc geom wsk 05 · 2015. 10. 18. · m∠ A + m∠B + m∠C = 180° m∠C = 180° − 32° = 148° m∠A + m∠B + 148 = 180° m∠A + m∠B = 32° The sum of the measures of the

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