1. Bisectors in Triangles
GEOMETRY LESSON 5-2
Pages 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 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.
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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.
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3. Bisectors in Triangles
GEOMETRY LESSON 5-2
33. a.
b. The bisectors intersect at the same point.
c. Check students’ work
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 =
36. C(3, 0), D(3, 0);
AC = BC = 3, AD = BD =
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 =
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4. Bisectors in Triangles
GEOMETRY LESSON 5-2
43. (continued) 5 2 26 3 4 25
39. C , , D(5, 3);
AC = BC = , AD = BD =
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.
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5. 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
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6. Bisectors in Triangles
GEOMETRY LESSON 5-2
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
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7. 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
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8. 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 = , AC = CB =
65. C 0, ; AB = , AC = BC =
66. C , 5 ; AB = , AC = CB = 13
3 5 11 7
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