Isosceles and Equilateral Triangles GEOMETRY LESSON 4-5 Pages 213-216 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

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 17. 150; 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 22. 140 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

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. 4-3. 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

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

Isosceles and Equilateral Triangles GEOMETRY LESSON 4-5 43. a. 5 b. 44. 0 < measure of base < 45 45. 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 = 60 so m PAB = 60. Since PAB and QAB are compl., m QAB = 30. QAB is isosc. so m AQB = 120. [1] one part correct 4-5

Isosceles and Equilateral Triangles GEOMETRY LESSON 4-5 50. RC = GV; RC GV by CPCTC since RTC GHV by ASA. 51. AAS 52. SSS 53. 24 sides 4-5