Bisectors in Triangles GEOMETRY LESSON 5-2 Pages 251-254 Exercises 1. AC is the bis. of BD 2. 15 3. 18 4. 8 5. The set of points equidistant from H and S is the bis. of HS. 6. x = 12; JK = 17; JM =17 7. y = 3; ST = 15; TU = 15 8. HL is the bis. of JHG because a point on HL is equidistant from J and G. 9. y = 9; m FHL = 54; m KHL = 54 10. 27 11. Point E is on the bisector of KHF. 12. 5 13. 10 14. 10 15. Isosceles; it has 2 sides 16. equidistant; RT = RZ 17. A point on the bis. of a segment if and only if it is equidistant from the endpoints of the segment. 5-2

Bisectors in Triangles GEOMETRY LESSON 5-2 18. 12 19. 4 20. 4 21. 16 22. 5 23. 10 24. 7 25. 14 26. Isosceles: CS = CT 26. (continued) and CT = CY by the Bis. Thm. 27. Answers may vary. Sample: the student needs to know that QS bisects PR. 28. No; A is not equidistant from the sides of X. 29. Yes; AX bis. TXR. 30. Yes; A is equidistant from the sides of X. 31. The pitcher’s plate 32. a. b. The bisectors intersect at the same point. c. Check student's work. 5-2

Bisectors in Triangles GEOMETRY LESSON 5-2 33. a. b. The bisectors intersect at the same point. c. Check students’ work 34-39. Answers may vary. Samples are given. 34. C(0, 2), D(1, 2); AC = BC = 2, AD = BD = 5 35. C(3, 2), D(3, 0); AC = BC = 3, AD = BD = 13 36. C(3, 0), D(3, 0); AC = BC = 3, AD = BD = 3 2 37. C(0, 0), D(1, 1); AC = BC = 3, AD = BD = 5 38. C(2, 2), D(4, 3); AC = BC = 5, AD = BD = 10 5-2

Bisectors in Triangles GEOMETRY LESSON 5-2 43. (continued) 5 2 26 3 4 25 39. C , , D(5, 3); AC = BC = , AD = BD = 13 40. a. : y = – x + ; x = 10 b. (10, 5) c. CA = CB = 5 d. C is equidist. from OA and OB. 41. bisector; right; Reflexive; SAS; CPCTC 42. PQ; BAQ; CPCTC; bisector 43. Answers may vary. Sample: proof of the Conv. of the Bis. Thm. 5-2

Bisectors in Triangles GEOMETRY LESSON 5-2 44. x = 3 45. y = –(x – 2) 46. y = – x + 4 47. Line is equidistant from points A, B, and C if it is to the plane determined 47. (continued) by A, B, and C and if it goes through the point that is the intersection of the bisectors of the sides of ABC. 48. BP AB and PC AC, thus ABP and ACP are rt. . Since AP bisects BAC, BAP CAP. AP AP by the Reflexive Prop. of . Thus ABP ACP by AAS and PB PC by CPCTC. Therefore, PB = PC. 1 2 s 5-2

Bisectors in Triangles GEOMETRY LESSON 5-2 49. 1. SP QP; SR QR 1. Given 2. QPS and QRS are rt. 2. Def. of 3. QPS QRS 3. All rt. are . 4. SP = SR 4. Given 5. QS QS 5. Refl. Prop. of 6. QPS QRS 6. HL 7. PQS RQS 7. CPCTC 8. QS bisects PQR. 8. Def. of bis s s 5-2

Bisectors in Triangles GEOMETRY LESSON 5-2 50. D 51. H 52. D 53. [2] Since MK MR, MK KV, and MR RV, the Bisector Thm. states that MV is the bisector of KVR. [1] partially correct logical argument 54. [4] MK MR. By the Reflexive Prop. of , MV MV. It is given that MKV and 54. [4] (continued) MRV are rt. . By HL, MKV MRV. By CPCTC, KV RV. By the Converse of the Bisector Thm., points M and V lie on the bisector, so MV is the bisector of KR [3] appropriate steps with one logical error OR one incorrect reason statement [2] two logical errors OR two incorrect reasons statements [1] proved but failed to reach desired conclusion s s 5-2

Bisectors in Triangles GEOMETRY LESSON 5-2 97 2 17 55. 8 56. 4 57. 6 58. Reflexive Prop. of = 59. Div. Prop. of = 60. Add. Prop. of = 61. Distr. Prop. 62. Subst. or Transitive Prop. of = 63. Trans. Prop. of 64. C 3, ; AB = 3 5, AC = CB = 65. C 0, ; AB = 97, AC = BC = 66. C , 5 ; AB = 17, AC = CB = 13 3 5 11 7 5-2