1. Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
Pages Exercises
1. M
2. DF
3. XY 4. 5. 6. 7. 8. 9.
10. a. Given
b. Reflexive Prop. of
c. Given
d. AAS
e. CPCTC
LQP PML; HL
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2. Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
RST UTS; SSS
QDA UAD; SAS
QPT RUS; AAS
15. TD RO if TDI
ROE by AAS. TID
REO if TEI
RIE. TEI RIE by SSS.
16. AE DE if AEB
DEC by AAS. AB DC and A D since they are corr. parts of
ABC and DCB, which are by HL.
QET QEU by SAS if QT QU. QT and QU are corr. parts of QTB and QUB which are by ASA.
ADC EDG by ASA if A E. A and E are corr. parts in ADB and
EDF, which are by SAS.
19–22.  Answers may vary. Samples are given.
19.
20.
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3. Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
21. a. b.
22. a.
ACE BCD by ASA; AC BC, A
B (Given) C C (Reflexive Prop. of )
ACE BCD (ASA)
WYX ZXY by HL; WY YX, ZX YX, WX
ZY (Given) WYX and ZXY are rt. (Def. of ) XY XY (Reflexive Prop. of )
WYX ZXY (HL)
25. m 1 = 56; m 2 = 56; m 3 = 34; m 4 = 90; m 5 = 22; m 6 = 34; m 7 = 34; m 8 = 68; m 9 = 112
ABC FCG; ASA
27. a. Given
b. Reflexive Prop. of     
c. Given
d. ETI s
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4. Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
27. (continued)
e. IRE
f. CPCTC
g. Given
h. All rt. are .
i. TDI
j. ROE   
k. CPCTC
28. a. Given
b. Def. of
c. Def. of rt.
d. Given
e. BC BC
f. Reflexive
g. HL
h. CPCTC
28. (continued)
i. DEC
j. Vert. are .
k. AAS
l. AE DE
m. CPCTC s s
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5. Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
29–30. Proofs may vary. Samples are given.
29. It is given that 1
2 and Since QB QB by the Reflexive Prop. of
, QTB QUB by ASA. So QT QU by CPCTC. Since QE
QE by the Reflexive Prop. of , then QET QEU by SAS.
AD ED (Given)
2. D is the midpt. of BF. (Given)
3. FD DB (Def. of midpt.)
4. FDE ADB   (Vert. are .)
5. FDE BDA   (SAS)
6. E A (CPCTC) s
30. (continued)
7. GDE CDA   (Vert. are .)
ADC EDG  (ASA)
31. a. AD BC; AB DC; AE EC; DE EB
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6. Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
31. (continued)
b. Use DB DB (refl.) and alt. int. to show ADB
CBD (ASA). AB
DC and AD BC (CPCTC). AEB CED (ASA) and AED CEB (ASA). Then AE EC and DE EB (CPCTC). s
AC EC; CB CD (Given)
2. C C (Reflexive Prop. of )
3. ACD ECB  (SAS)
4. A E (CPCTC)
33. PQ RQ and PQT
RQT by Def. of bisector. QT QT so
PQT RQT by SAS. P R by CPCTC. QT bisects
VQS so VQT
SQT and PQT and RQT are both rt So VQP
SQR since they are compl. of PQV
RQS by ASA so QV QS by CPCTC.
34. C
35. F s
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7. Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
36. A
37. [2] a. HBC HED
b. HB HE by CPCTC if HBC
HED by ASA. Since BDC
CED by AAS, then DBC
CED by CPCTC and CHB
DHE because vertical are .
[1] one part correct s
38. [4] a. HL b.
c. x = 30. In ADC m A + m ADC + m ACD = Substituting, 90 + x + x + x = Solving, x = 30.
d. 120; it is suppl. to a 60°
e. 6 m; DC = 2(AD)
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8. Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7 1 3
38. (continued)
[3] 4 parts answered correctly
[2] 3 parts answered correctly
[1] 2 parts answered correctly
39. a. right b.
c. Reflexive
d. HL
40.
41.
42. y + 6 = (x – 2)
43. y – 5 = 1(x – 0)
44. y – 6 = –2(x + 3)
45. y – 0 = – (x – 0)
46–48. Eqs. may vary, depending on pt. chosen.
46. y – 4 = 2(x – 1)
47. y + 5 = (x – 3)
48. y + 3 = – (x + 4) 5 3 5 6 1 2
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