# Using Variables - Mrs. Powers' Class

Bisectors in Triangles GEOMETRY GEOMETRY LESSON LESSON 5-2 5-2 Pages 251-254 Exercises 1. AC is the BD bis. of 7. y = 3; ST = 15; TU = 15 12. 5 13. 10 2. 3. 15 18 8. HL is the bis. of JHG because a point on HL is equidistant from J and G. 4. 8 5. The set of points equidistant from H and S is the bis. of HS. 9. y = 9; m FHL = 54; m KHL = 54 x = 12; JK = 17; JM =17 11. Point E is on the

bisector of KHF. 6. 10. 27 5-2 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. Bisectors in Triangles GEOMETRY GEOMETRY LESSON LESSON 5-2 5-2 18. 12 19. 4 20. 4 21. 16 26. (continued) and CT = CY by the Bis. Thm. 31. The pitchers plate 32. a. 27. Answers may vary. Sample: the student needs to know that QS bisects PR. 22. 5 23. 10 28. No; A is not equidistant from the sides of X. 24. 7 29. Yes; AX bis.

TXR. 25. 14 26. Isosceles: CS = CT 30. Yes; A is equidistant from the sides of X. 5-2 b. The bisectors intersect at the same point. c. Check student's work. Bisectors in Triangles GEOMETRY GEOMETRY LESSON LESSON 5-2 5-2 33. a. 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 b. The bisectors intersect at the same point. 37. C(0, 0), D(1, 1); AC = BC = 3, AD = BD = c. Check students work 38. C(2, 2), D(4, 3);

AC = BC = 5, AD = BD = 5-2 2 5 10 Bisectors in Triangles GEOMETRY GEOMETRY LESSON LESSON 5-2 5-2 5 5 39. C 2 , 2 , D(5, 3); 26 AC = BC = , AD = BD = 2 40. a. 43. (continued) 13 : y = 3 x + 25 ; x = 10 4 5 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 GEOMETRY LESSON

LESSON 5-2 5-2 44. x = 3 45. y = (x 2) 46. y = 1 x + 4 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. 2 47. 48. BP AB and PC AC, thus ABP and ACP are rt. s . Since AP bisects BAC, BAP Line is equidistant from points A, B, and C if it is to the plane determined CAP. AP AP by the Reflexive Prop. of . Thus ABP ACP by AAS and PB PC by CPCTC. Therefore, PB = PC. 5-2 Bisectors in Triangles GEOMETRY GEOMETRY LESSON LESSON 5-2 5-2 49. 1. SP QP; SR 2. QPS and 3. QPS QR QRS are rt.

QRS 1. Given s 2. Def. of 3. All rt. s are 4. SP = SR 4. Given 5. QS 5. Refl. Prop. of QS 6. QPS QRS 6. HL 7. PQS RQS 7. CPCTC 8. QS bisects PQR. 8. Def. of 5-2 bis .

Bisectors in Triangles GEOMETRY GEOMETRY LESSON LESSON 5-2 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. 54. [4] (continued) MRV are rt. s . 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 [1] partially correct logical argument [2] two logical errors OR two incorrect reasons statements 54. [4] MK MR. By the Reflexive Prop. of , MV MV. It is given that MKV and [1] proved s but failed to reach desired conclusion 5-2 Bisectors in Triangles GEOMETRY GEOMETRY LESSON LESSON 5-2 5-2

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

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