1. Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
Pages Exercises
1. a. RS
b. RS
c. Given
d. Def. of bisector
e. Reflexive Prop. of
f. AAS
2. a. KM
b. KM
c. By construction
2. (continued)
d. Def. of segment bisector
e. Reflexive Prop. of
f. SSS 
g. CPCTC
3. VX; Conv. of the Isosc.
Thm.
4. UW; Conv. of the Isosc.
5. VY; VT = VX (Ex. 3) and UT = YX (Ex. 4), so VU = VY by the Subtr. Prop. of =.
6. Answers may vary. Sample: VUY; opp. sides are .
7. x = 80; y = 40
8. x = 40; y = 70
9. x = 38; y = 4 s
4-5

2. Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
10. x = 4 ; y = 60
11. x = 36; y = 36
12. x = 92; y = 7
13. 64
14. 2
15. 42
16. 35
; 15
18. 24, 48, 72, 96, 120 1 2
19. a.
30, 30, 120
b. 5; 30, 60, 90, 120, 150
c. Check students’ work.
20. 70
21. 50
23. 6
24. x = 60; y = 30
25. x = 64; y = 71
26. x = 30; y = 120
27. Two sides of a are
if and only if the opp. those sides are .
28. 80, 80, 20; 80, 50, 50 s
4-5

3. Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
29. a. isosc.
b. 900 ft; 1100 ft
c. The tower is the bis. of the base of each .
30. No; the can be positioned in ways such that the base is not on the bottom.
31. 45; they are = and have sum 90. s
32. Answers may vary. Sample: Corollary to Thm. 4-3: Since XY YZ, X Z by Thm YZ ZX, so Y
X by Thm. 4-3 also. By the Trans. Prop., Y
Z, so X Y
Z. Corollary to Thm. 4-4: Since X Z, XY YZ by Thm. 4-4.
Y X, so YZ ZX by Thm. 4-4 also. By the Trans. Prop. XY ZX, so XY YZ ZX.
33. a. Given
b. A D
c. Given
d. ABE DCE
34. m = 36; n = 27
35. m = 60; n = 30
36. m = 20; n = 45
37. (0, 0), (4, 4), (–4, 0),
(0, –4), (8, 4), (4, 8)
38. (5, 0); (0, 5); (–5, 5);
(5, –5); (0, 10); (10, 0)
4-5

4. Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
39. (5, 3); (2, 6); (2, 9); (8, 3); (–1, 6); (5, 0)
40. a. 25
b. 40; 40; 100
c. Obtuse isosc. ; 2 of the are and one is obtuse. s
41. AC CB and ACD
DCB are given. CD CD by the Refl. Prop. of
, so ACD BCD by SAS. So AD DB by CPCTC, and CD bisects AB. Also ADC
BDC by CPCTC, m ADC + m BDC = 180 by Add. Post., so m ADC = m BDC = 90 by the Subst. Prop. So CD is the bis. of AB.
42. The bis. of the base of an isosc. is the bis. of the vertex ; given isosc. ABC with bis. CD, ADC
BDC and AD DB by def. of bis. Since CD CD by Refl. Prop., ACD BCD by SAS. So ACD
BCD by CPCTC, and CD bisects ACB.
4-5

5. Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
43. a. 5 b.
44. 0 < measure of base < 45
< measure of base
< 90
46. C
47. G
48. D
49. [2] a. 60; since m PAB = m PBA, and m PAB + m PBA = 120, m PAB = 60.
b. 120; m APB = so m PAB = Since PAB and QAB are compl., m QAB = QAB is isosc. so m AQB = 120.
[1] one part correct
4-5

6. Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
50. RC = GV; RC GV by CPCTC since
RTC GHV by ASA.
51. AAS
52. SSS
sides
4-5