1. Congruent Figures Solve each equation.
GEOMETRY LESSON 4-1
(For help, go to page 24.)
Solve each equation.
1. x + 6 = x = 33
3. 5x = x + 10 = 2x
5. For the triangle at the right, use the Triangle Angle-Sum Theorem to find the value of y.
4-1

2. 1. Subtract 6 from both sides: x = 19
Congruent Figures
GEOMETRY LESSON 4-1
Solutions
1. Subtract 6 from both sides: x = 19
2. Combine like terms: x + 20 = 33; subtract 20 from both sides: x = 13
3. Divide both sides by 5: x = 108
4. Subtract x from both sides of x + 10 = 2x: 10 = x, or x = 10
5. y = 180; combine like terms: y = 180; subtract 130 from both sides: y = 50
4-1

3. ABC QTJ. List the congruent corresponding parts.
Congruent Figures
GEOMETRY LESSON 4-1
ABC QTJ. List the congruent corresponding parts.
Sides: AB QT BC TJ AC QJ
Angles: A Q B T C J
List the corresponding vertices in the same order.
List the corresponding sides in the same order.
4-1

4. XYZ KLM, mY = 67, and mM = 48. Find mX.
Congruent Figures
GEOMETRY LESSON 4-1
XYZ KLM, mY = 67, and mM = 48. Find mX.
Use the Triangle Angle-Sum Theorem and the definition of congruent polygons to find mX.
mX + mY + mZ = 180 Triangle Angle-Sum Theorem
mZ = mM Corresponding angles of congruent triangles that are congruent
mZ = 48 Substitute 48 for mM.
mX = 180 Substitute.
mX = 180 Simplify.
mX = 65 Subtract 115 from each side.
4-1

5. Can you conclude that ABC CDE in the figure below?
Congruent Figures
GEOMETRY LESSON 4-1
Can you conclude that ABC CDE in the figure below?
List corresponding vertices in the same order.
If ABC CDE, then BAC DCE.
The diagram above shows BAC DEC, not DCE.
Corresponding angles are not necessarily congruent, therefore you cannot conclude that ABC CDE.
4-1

6. Show how you can conclude that CNG DNG. List statements and reasons.
Congruent Figures
GEOMETRY LESSON 4-1
Show how you can conclude that CNG DNG. List statements and reasons.
Congruent triangles have three congruent corresponding sides and three congruent corresponding angles.
Examine the diagram, and list the congruent corresponding parts for CNG and DNG.
a. CG DG Given
b. CN DN Given
c. GN GN Reflexive Property of Congruence
d. C D Given
e. CNG DNG Right angles are congruent.
f. CGN DGN If two angles of one triangle are congruent to two angles of another triangle, then the third angles are congruent. (Theorem 4-1.)
g. CNG DNG Definition of triangles
4-1

7. Congruent Figures Pages 182-185 Exercises 1. CAB DAB; C D; ABC ABD;
GEOMETRY LESSON 4-1
Pages Exercises
1. CAB DAB;
C D;
ABC ABD;
AC AD; AB AB
CB DB
2. GEF JHI;
GFE JIH;
EGF HJI;
GE JH; EF HI
FG IJ
3. BK
4. CM
5. ML
6. B
7. C
8. J
9. KJB
CLM
JBK
MCL
13. E, K, G, N
14. PO SI; OL ID;
LY DE; PY SE
P S; O I;
L D; Y E
in.
in.
19. 77
4-1

8. If two of a are to two of another , the third
Congruent Figures
GEOMETRY LESSON 4-1
in.
in.
22. 75
24. Yes; RTK UTK, R U (Given)
RKT UKT
If two of a are to two of another , the third
are . TR TU, RK UK (Given) TK TK (Reflexive Prop. of ) TRK TUK (Def. of )
25. No; the corr. sides are not .
26. No; corr. sides are not necessarily .
27. Yes; all corr. sides and
are .
28. a. Given
b. If || lines, then alt. int are .
c. Given
d. If 2 of one are to two of another , then 3rd are .
e. Reflexive Prop. of
f. Given
g. Def. of s s s s s s s s s s
4-1

9. 29. A and H; B and G; C and E; D and F
Congruent Figures
GEOMETRY LESSON 4-1
29. A and H; B and G; C and E; D and F
30. x = 15; t = 2
31. 5
32. m A = m D = 20
33. m B = m E = 21
34. BC = EF = 8
35. AC = DF = 19
36. Answers may vary. Sample: It is important that PACH OLDE for the patch to completely fill the hole.
37. Answers may vary. Sample: She could arrange them in a neat pile and pull out the ones of like sizes.
JYB XCH
BCE ADE
TPK TRK
JLM NRZ;
JLM ZRN
42. Answers may vary. Sample: The die is a mold that is used to make items that are all the same size.
43. Answers may vary. Sample: TKR MJL: TK MJ; TR ML; KR JL; TKR MJL;
TRK MLJ; KTR JML
4-1

10. b. If || lines, then alt. int. are .
Congruent Figures
GEOMETRY LESSON 4-1
44. a. Given
b. If || lines, then alt. int. are .
c. If || lines, then alt. int. are .
d. Vertical are .
e. Given
f. Given
g. Def. of segment bisector
h. Def. of
45. Answers may vary. Sample: Since the sum of the of a is 180, and if 2 of one are the same as 2 of a second , then their sum subtracted from 180 has to be the same.
46. KL = 4; LM = 3; KM = 5
47. 2; either (3, 1) or (3, –7) s
48. a. 15 b.
4-1

11. 53. Answers may vary. Sample: 54.
Congruent Figures
GEOMETRY LESSON 4-1
51. 40
52. 72
53. Answers may vary. Sample:
54.
        
56. RS = PQ        
        
58. 12
59. AB GH
4-1

12. In Exercises 1 and 2, quadrilateral WASH quadrilateral NOTE.
Congruent Figures
GEOMETRY LESSON 4-1
In Exercises 1 and 2, quadrilateral WASH quadrilateral NOTE.
1. List the congruent corresponding parts.
2. mO = mT = 90 and mH = 36. Find mN.
3. Write a statement of triangle congruence. 4. Write a statement of triangle congruence. 5. Explain your reasoning in Exercise 4 above.
WA NO, AS OT, SH TE, WH NE;
W N, A O, S T, H E
144
Sample: DFH ZPR
Sample: ABD CDB
Sample: Two pairs of corresponding sides and two pairs of corresponding angles are given. C A because all right angles are congruent. BD BD by the Reflexive Property of ABD CDB by the definition of congruent triangles.
4-1

13. Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
(For help, go to Lesson 2-5.)
What can you conclude from each diagram?
4-2

14. Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
Solutions
1. According to the tick marks on the sides, AB DE. According to the tick marks on the angles, C F.
2. The two triangles share a side, so PR PR. According to the tick marks on the angles, QPR SRP and Q S.
3. According to the tick marks on the sides, TO NV. The tick marks on the angles show that M S. Since MO || VS, by the Alternate Interior Angles Theorem MON SVT. Since OV OV by the Reflexive Property, you can use the Segment Addition Property to show TV NO.
4-2

15. Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
Given: M is the midpoint of XY, AX AY
Prove: AMX AMY
Write a paragraph proof.
Copy the diagram. Mark the congruent sides.
You are given that M is the midpoint of XY, and AX AY. Midpoint M implies MX MY. AM AM by the Reflexive Property of Congruence, so AMX AMY by the SSS Postulate.
4-2

16. Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
AD BC. What other information do you need to prove ADC BCD?
It is given that AD BC. Also, DC CD by the Reflexive Property of Congruence. You now have two pairs of corresponding congruent sides. Therefore:
Solution 1: If you know AC BD, you can prove ADC BCD
by SSS.
Solution 2: If you know ADC BCD, you can prove ADC BCD by SAS.
4-2

17. Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
Given: RSG RSH, SG SH
From the information given, can you prove RSG RSH? Explain.
Copy the diagram. Mark what is given on the diagram.
It is given that RSG RSH and SG SH.
RS RS by the Reflexive Property of Congruence.
Two pairs of corresponding sides and their included angles are congruent, so RSG RSH by the SAS Postulate.
4-2

18. Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
1. SSS
2. cannot be proved
3. SAS
4. SSS
5. Yes; OB OB by Refl. Prop.; BOP
BOR since all rt. are ; OP OR (Given); the are by SAS. s
Pages Exercises
6. Yes; AC DB (Given); AE CE and BE DE
(Def. of midpt.); AEB
CED (vert. are )
AEB CED by SAS.
7. a. Given
b. Reflexive
c. JKM
d. LMK
8. WV, VU
9. W s
U, V
11. WU X
13. XZ, YZ
14. LG MN
T V or RS WU
16. DC CB
4-2

19. Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
17. additional information not needed
18. Yes; ACB EFD by SAS.
19. Yes; PVQ STR by SSS.
20. No; need YVW
ZVW or YW ZW.
21. Yes; NMO LOM by SAS.
ANG RWT; SAS
KLJ MON; SSS
24. Not possible; need
H P or DY TK.
JEF SVF or
JEF SFV; SSS
BRT BRS; SSS
PQR NMO; SAS
28. No; even though the are , the sides may not be. s
29. No; you would need
H K or GI JL.
30. yes; SAS
31.
4-2

20. Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
32.
33. a. Vertical are .
b. Given
c. Def. of midpt.
d. Given
e. Def. of midpt.
f. SAS s
34. Answers may vary. Sample:
35. a–b. Answers may vary. Sample:
a. wallpaper designs; ironwork on a bridge; highway warning signs
35. (continued)
b produce a well-balanced, symmetric appearance. In construction, enhance designs. Highway warning signs are more easily identified if they are .
ISP PSO; ISP OSP by SAS.
37. IP PO; ISP
OSP by SSS. s
4-2

21. Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
38. Yes; ADB CBD by SAS; ADB
DBC because if || lines, then alt. int. are .
39. Yes; ABC CDA by SAS; DAC
ACB because if || lines, then alt. int. are .
40. No; ABCD could be a square with side 5 and EFGH could be a polygon with side 5 but no rt. . s
FG || KL (Given)
2. GFK FKL (If || lines, then alt. int. are .)
3. FG KL (Given)
4. FK FK (Reflexive Prop. of )
5. FGK KLF (SAS) s
42. AE and BD bisect each other, so AC CE and BC CD. ACB DCE  because vert. are .
ACB ECD by SAS. s
4-2

22. Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
43.
44. AM MB because M is the midpt. of AB.
B AMC because all right are
. CM DB is given.
AMC MBD by SAS. s
45. D
46. G
47. C
48. [2] a. AB AB; Reflexive Prop. of
b. No; AB is not a corr. side.
[1] one part correct E
50. AB
51. FG C
53. The product of the slopes of two lines is –1 if and only if the lines are .
54. If x = 2, then 2x = 4. If 2x = 4, then x = 2.
55. If 2x = 6, then x = 3. The statement and the converse are both true.
4-2

23. Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
56. If x2 = 9, then x = 3. The statement is true but the converse is false.
4-2

24. Triangle Congruence by SSS and SAS
GEOMETRY LESSON 4-2
1. In VGB, which sides include B?
2. In STN, which angle is included between NS and TN?
3. Which triangles can you prove congruent?
Tell whether you would use the SSS or SAS Postulate.
4. What other information do you need to prove
DWO DWG?
5. Can you prove SED BUT from the information given? Explain.
BG and BV N
APB XPY; SAS
If you know DO DG, the triangles are by SSS; if you know DWO DWG, they are by SAS.
No; corresponding angles are not between corresponding sides.
4-2

25. Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
(For help, go to the Lesson 4-2.)
In JHK, which side is included between the given pair of angles?
1. J and H 2. H and K
In NLM, which angle is included between the given pair of sides?
3. LN and LM 4. NM and LN
Give a reason to justify each statement.
5. PR PR 6. A D
4-3

26. Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
Solutions
1. JH
2. HK
3. L
4. N
5. By the Reflexive Property of Congruence, a segment is congruent to itself.
6. By the Triangle Angle-Sum Theorem, the sum of the angles of any triangle is 180. If mC = mF = x and mB = mE = y, then mA = 180 – x – y = mD, so A = D.
4-3

27. Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
Suppose that F is congruent to C and I is not congruent to C. Name the triangles that are congruent by the ASA Postulate.
The diagram shows N A D and FN CA GD.
If F C, then F C G
Therefore, FNI CAT GDO by ASA.
4-3

28. Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
Write a paragraph proof.
Given: A B, AP BP
Prove: APX BPY
It is given that A B and AP BP.
APX BPY by the Vertical Angles Theorem.
Because two pairs of corresponding angles and their included sides are congruent, APX BPY by ASA.
4-3

29. Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
Write a Plan for Proof that uses AAS.
Given: B D, AB || CD
Prove: ABC CDA
Because AB || CD, BAC DCA by the Alternate Interior Angles Theorem.
Then ABC CDA if a pair of corresponding sides are congruent.
By the Reflexive Property, AC AC so ABC CDA by AAS.
4-3

30. Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
Write a two-column proof that uses AAS.
Given: B D, AB || CD
Prove: ABC CDA
Statements Reasons
1. B D, AB || CD 1. Given
2. BAC DCA 2. If lines are ||, then alternate interior angles are .
3. AC CA 3. Reflexive Property of Congruence
4. ABC CDA 4. AAS Theorem
4-3

31. Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
Pages Exercises
1. PQR VXW
2. ACB EFD
3. RS
4. N and O
5. yes
6. not possible
7. yes
8. a. Reflexive
b. ASA
9. AAS
10. ASA
11. not possible
FDE GHI; DFE HGI
13. a. UWV
b. UW
c. right
d. Reflexive
B D
15. MU UN
16. PQ QS
WZV WZY
4-3

32. Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
18. a. Vert. are .
b. Given
c. TQ QR
d. AAS
PMO NMO; ASA
UTS RST; AAS
ZVY WVY; AAS
TUX DEO; AAS
23. The are not because no sides are . s
TXU ODE; ASA
25. The are not because the are not included .
26. Yes; if 2 of a are to 2 of another , then the 3rd are So, an AAS proof can be rewritten as an ASA proof.
27. a. SRP
b. PR
c. alt. int.
d. PR
e. Reflexive
28. a. Given
b. Def. of bis.
c. Given
d. Reflexive Prop. of
e. AAS
4-3

33. Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
29. a. Def. of
b. All right are .
c. QTP STR
d. Def. of midpt.
e. AAS
30. s
34. Yes; by ASA since
BDH FDH by def. of bis. and DH DH by the Reflexive Prop. of .
35. Answers may vary. Sample:
36. a. Check students’ work.
b. Answers may vary; most likely ASA.
31. Yes; by AAS since
MON QOP.
32. Yes; by AAS since
FGJ HJG because when lines are ||, then alt. int. are and GJ GJ by the Reflexive Prop. of .
33. Yes; by ASA, since
EAB DBC because || lines have
corr. . s
4-3

34. Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
AEB CED,
BEC DEA,
ABC CDA,
BAD DCB
AEB CED,
BEC DEA,
ABC CDA,
ABD DCA,
BAD DCB,
ABD DCB,
CBA DAB,
BCD ADC
39. They are bisectors; ASA.
44. [2] a. RPQ SPQ, RQP SQP (Def. of bisector)
b. ASA
[1] one part correct
40.
41.
42. D
43. F 13 20
4-3

35. Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
45. [4] a. Def. of midpt.
b. Yes; JLM KGM because they are alt. int. of || lines, and LMJ GMK because vertical are . So the are by ASA.
c. Yes; if two of one are to of another , the third are .
[3] incorrect for part b or c, but otherwise correct
[2] correct conclusions but incomplete explanations for parts b and c
[1] at least one part correct
ONL MLN; SAS
47. not possible
48. AC and CB
49. If corr. are , then the lines are ||.
photos
photos
52. 60% more paper s s s s s s s s
4-3

36. Triangle Congruence by ASA and AAS
GEOMETRY LESSON 4-3
1. Which side is included between R and F in FTR?
2. Which angles in STU include US?
Tell whether you can prove the triangles congruent by ASA or AAS. If you can, state a triangle congruence and the postulate or theorem you used. If not, write not possible. RF
S and U
GHI PQR AAS
not possible
ABX ACX AAS
4-3

37. Using Congruent Triangles: CPCTC
GEOMETRY LESSON 4-4
(For help, go to the Lesson 4-1.)
In the diagram, JRC HVG.
1. List the congruent corresponding angles.
2. List the congruent corresponding sides.
You are given that TIC LOK.
3. List the congruent corresponding angles.
4. List the congruent corresponding sides.
4-4

38. Using Congruent Triangles: CPCTC
GEOMETRY LESSON 4-4
Solutions
1. In the triangle congruence statement, the corresponding vertices are listed in the same order. So, J H, R V, and C G.
2. In the triangle congruence statement, the corresponding vertices are listed in the same order. So, JR HV, RC VG, and JC HG.
3. In the triangle congruence statement, the corresponding vertices are listed in the same order. So, T L, I O, and C K.
4. In the triangle congruence statement, the corresponding vertices are listed in the same order. So, TI LO, IC OK, and TC LK.
4-4

39. Using Congruent Triangles: CPCTC
GEOMETRY LESSON 4-4
What other congruence statements can you prove from the diagram, in which SL SR, and 1 2 are given? SC SC by the Reflexive Property of Congruence, and LSC RSC by SAS. 3 4 by corresponding parts of congruent triangles are congruent.
When two triangles are congruent,
you can form congruence statements about three pairs of corresponding angles and three pairs of corresponding sides. List the congruence statements.
4-4

40. Using Congruent Triangles: CPCTC
GEOMETRY LESSON 4-4
(continued)
SL SR Given
SC SC Reflexive Property of Congruence
CL CR Other congruence statement
Sides:
1 2 Given
3 4 Corresponding Parts of Congruent Triangles
CLS CRS Other congruence statement
Angles:
In the proof, three congruence statements are used, and one congruence statement is proven. That leaves two congruence statements remaining that also can be proved:
CLS CRS
CL CR
4-4

41. Using Congruent Triangles: CPCTC
GEOMETRY LESSON 4-4
The Given states that DEG and DEF are right angles. What conditions must hold for that to be true?
DEG and DEF are the angles the officer makes with the ground.
So the officer must stand perpendicular to the ground, and the ground must be level.
4-4

42. Using Congruent Triangles: CPCTC
GEOMETRY LESSON 4-4
Pages Exercises
1. PSQ SPR; SQ RP; PQ SR
2. AAS; ABC EBD;
A E; CB DB; DE CA by CPCTC
3. SAS; KLJ OMN;
K O; J N; KJ ON by CPCTC
4. SSS; HUG BUG;
H B; HUG
BUG; UGH
UGB by CPCTC
5. They are ; the are by AAS, so all corr. ext.
are also .
6. a. SSS
b. CPCTC
7. ABD CBD by ASA because BD BD by Reflexive Prop. of ; AB CB by CPCTC.
8. MOE REO by SSS because OE OE by Reflexive Prop. of ;
M R by CPCTC. s
9. SPT OPT by SAS because TP TP by Reflexive Prop. of
; S O by CPCTC.
PNK MNL by SAS because KNP
LNM by vert. are
; KP LM by CPCTC.
CYT RYP by AAS; CT RP by CPCTC. s
4-4

43. Using Congruent Triangles: CPCTC
GEOMETRY LESSON 4-4
ATM RMT by SAS because ATM
RMT by alt. int. are ; AMT
RTM by CPCTC.
13. Yes; ABD CBD by SSS so A C by CPCTC.
14. a. Given
b. Given
c. Reflexive Prop. of
d. AAS s
PKL QKL by def. of bisect, and KL KL by Reflexive Prop. of , so the are by SAS.
16. KL KL by Reflexive Prop. of ; PL LQ by Def. of bis.; KLP
KLQ by Def. of ; the
are by SAS.
KLP KLQ because all rt are ; KL KL by Reflexive Prop. of ; and PKL
QKL by def. of bisect; the are by ASA.
18. The are by SAS so the distance across the sinkhole is 26.5 yd by CPCTC.
19. a. Given
b. Def. of
c. All right are .
d. Given
e. Def. of segment bis.
f. Reflexive Prop. of
g. SAS
h. CPCTC s s s s s
4-4

44. Using Congruent Triangles: CPCTC
GEOMETRY LESSON 4-4
ABX ACX by SSS, so BAX
CAX by CPCTC.
Thus AX bisects
BAC by the def. of
bisector.
21. Prove ABE CDF by SAS since AE
FC by subtr.
22. Prove KJM QPM by ASA since P J and K Q by alt. int. are .
23. e or b, e or b, d, c, f, a s
24. BA BC is given; BD BD by the Reflexive Prop. of and since BD bisects ABC, ABD CBD by def. of an bisector; thus, ABD CBD by SAS; AD DC by CPCTC so BD bisects AC by def. of a bis.; ADB CDB by CPCTC and ADB and CDB are suppl.; thus, ADB and CDB are right and BD AC by def. of .
25. a. AP PB; AC BC
b. The diagram is constructed in such a way that the are by SSS. CPA CPB by CPCTC. Since these are and suppl., they are right . Thus, CP is to . s
4-4

45. Using Congruent Triangles: CPCTC
GEOMETRY LESSON 4-4
PR || MG; MP || GR (Given)
2. Draw PG. (2 pts. determine a line.)
3. RPG PGM and RGP GPM (If || lines, then alt. int. are .)
4. PGM GPR (ASA) A similar proof can be written if diagonal RM is drawn.
27. Since PGM GPR (or PMR GRM), then
PR MG and MP GR by CPCTC.
28. C
29. C s
30. D
31. B
32. C
33. [2] a. KBV KBT; yes; SAS
b. CPCTC
[1] one part correct
34. ASA
35. AAS
4-4

46. Using Congruent Triangles: CPCTC
GEOMETRY LESSON 4-4
36. 95; 85
37. The slope of line m is the same as the slope of line n.
38. not possible
39. not possible
4-4

47. Using Congruent Triangles: CPCTC
GEOMETRY LESSON 4-4
1. What does “CPCTC” stand for?
Use the diagram for Exercises 2 and 3.
2. Tell how you would show ABM ACM.
3. Tell what other parts are congruent by CPCTC.
Use the diagram for Exercises 4 and 5.
4. Tell how you would show RUQ TUS.
5. Tell what other parts are congruent by CPCTC.
Corresponding parts of congruent triangles are congruent.
You are given two pairs of s and AM AM by the Reflexive Prop., so ABM ACM by ASA.
AB AC, BM CM, B C
You are given a pair of s and a pair of sides and RUQ TUS because vertical angles are , so RUQ TUS by AAS.
RQ TS, UQ US, R T
4-4

48. Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
(For help, go to the Lesson 3-3.)
1. Name the angle opposite AB.
2. Name the angle opposite BC.
3. Name the side opposite A.
4. Name the side opposite C.
5. Find the value of x.
4-5

49. Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
Solutions
1. The angle opposite AB is the angle whose side is not AB: C
2. The angle opposite BC is the angle whose side is not BC: A
3. The side opposite A is the side that is not part of A: BC
4. The side opposite C is the side that is not part of C: BA
5. By the Triangle Exterior Angle Theorem, x = = 105°.
4-5

50. Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
Examine the diagram below. Suppose that you draw XB YZ. Can you use SAS to prove XYB XZB? Explain.
It is given that XY XZ.
By the definition of perpendicular, XBY = XBZ.
By the Reflexive Property of Congruence, XB XB.
However, because the congruent angles are not included between the congruent corresponding sides, the SAS Postulate does not apply.
You cannot prove the triangles congruent using SAS.
4-5

51. Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
Explain why ABC is isosceles.
ABC and XAB are alternate interior angles formed by XA, BC, and the transversal AB. Because XA || BC, ABC XAB.
The diagram shows that XAB ACB. By the Transitive Property of Congruence, ABC ACB.
You can use the Converse of the Isosceles Triangle Theorem to conclude that AB AC.
By the definition of an isosceles triangle, ABC is isosceles.
4-5

52. Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
Suppose that mL = y. Find the values of x and y.
MO LN The bisector of the vertex angle of an isosceles triangle is the perpendicular bisector of the base.
x = 90 Definition of perpendicular
mN = mL Isosceles Triangle Theorem
mL = y Given
mN = y Transitive Property of Equality
mN + mNMO + mMON = 180 Triangle Angle-Sum Theorem
y + y + 90 = 180 Substitute.
2y + 90 = 180 Simplify.
2y = 90 Subtract 90 from each side.
y = 45 Divide each side by 2.
Therefore, x = 90 and y = 45.
4-5

53. Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
Suppose the raised garden bed is a regular hexagon. Suppose that a segment is drawn between the endpoints of the angle marked x. Find the angle measures of the triangle that is formed.
Because the garden is a regular hexagon, the sides have equal length, so the triangle is isosceles.
By the Isosceles Triangle Theorem, the unknown angles are congruent.
Example 4 found that the measure of the angle marked x is 120. The sum of the angle measures of a triangle is 180.
If you label each unknown angle y, y + y = 180.
y = 180
2y = 60
y = 30
So the angle measures in the triangle are 120, 30 and 30.
4-5

54. Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
Pages 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

55. 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
; 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
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

56. 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 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

57. 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

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

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

60. Isosceles and Equilateral Triangles
GEOMETRY LESSON 4-5
Use the diagram for Exercises 1–3.
1. If mBAC = 38, find mC.
2. If mBAM = mCAM = 23, find mBMA.
3. If mB = 3x and mBAC = 2x – 20, find x.
4. Find the values of x and y. 71 90 25
5. ABCDEF is a regular hexagon. Find mBAC.
x = 60
y = 9 30
4-5
4-5

61. Congruence in Right Triangles
GEOMETRY LESSON 4-6
(For help, go to the Lessons 4-2 and 4-3.)
Tell whether the abbreviation identifies a congruence statement.
1. SSS 2. SAS 3. SSA
4. ASA 5. AAS 6. AAA
Can you conclude that the two triangles are congruent? Explain.
4-6

62. Congruence in Right Triangles
GEOMETRY LESSON 4-6
Solutions
1. SSS names the Side-Side-Side Theorem.
2. SAS names the Side-Angle-Side Theorem.
3. SSA does not name a theorem or postulate.
4. ASA names the Angle-Side-Angle Postulate.
5. AAS names the Angle-Angle-Side Theorem.
6. AAA does not name a congruence theorem or postulate.
4-6

63. Congruence in Right Triangles
GEOMETRY LESSON 4-6
Solutions (continued)
7. The side that they share is congruent to itself by the Reflexive Property of Congruence. The two right angles are congruent to each other, and two other corresponding sides are marked congruent. The two triangles are congruent by SAS.
8. Two pairs of congruent sides are marked congruent. The included angles are congruent because vertical angles are congruent. Thus, the two triangles are congruent by SAS.
4-6

64. Congruence in Right Triangles
GEOMETRY LESSON 4-6
One student wrote “ CPA MPA by SAS” for the diagram below. Is the student correct? Explain.
The diagram shows the following congruent parts.
CA MA
CPA MPA
PA PA
There are two pairs of congruent sides and one pair of congruent angles, but the congruent angles are not included between the corresponding congruent sides.
The triangles are not congruent by the SAS Postulate, but they are congruent by the HL Theorem.
4-6

65. Congruence in Right Triangles
GEOMETRY LESSON 4-6
XYZ is isosceles. From vertex X, a perpendicular is drawn to YZ, intersecting YZ at point M. Explain why XMY XMZ.
4-6

66. Congruence in Right Triangles
GEOMETRY LESSON 4-6
Write a two–column proof.
Given: ABC and DCB are right angles, AC DB
Prove: ABC DCB
Statements Reasons
1. ABC and DCB are 1. Given right angles.
2. ABC and DCB are 2. Definition of a right triangle right triangles.
3. AC DB 3. Given
4. BC CB 4. Reflexive Property of Congruence
5. ABC DCB 5. If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and a leg of another right triangle, then the triangles are congruent. (HL Theorem).
4-6

67. Congruence in Right Triangles
GEOMETRY LESSON 4-6
Pages Exercises
1. ABC DEF by HL. Both are rt. , AC DF, and CB FE.
2. SPR QRP by HL. Both are rt. , SP QR (given) and PR PR by the Reflexive Prop. of .
3. LMP OMN by HL. Both are rt. because vert. are ; LP NO, and LM OM. s
4. AEB DCB by HL. Both are rt AB BD and EB CB
by the def. of midpt.
5. T and Q are rt. .
6. RX RT or XV TV
7. TY ER or RT YE
8. Right are needed, either A and G or
AQC and GJC. s
9. BC FA
10. RT NQ
11. a. Given
b. Def. of rt.
c. Reflexive Prop. of
d. Given
e. HL
4-6

68. Congruence in Right Triangles
GEOMETRY LESSON 4-6
12. a. suppl. are rt.
b. Def. of rt.
c. Given
d. Reflexive Prop. of
e. HL
13. PS PT so S T by the Isosc. Thm.
PRS PRT.
PRS PRT by AAS.
14. Yes; RS TU and RT TV. s
15. Yes; PM PM and
PMW is a rt. since JP || MW.
16. a. Given
b. Def. of
c. MLJ and KJL are rt. .
d. Given
e. LJ LJ
f. HL s
17. a. Given
b. IGH
c. Def. of rt.
d. I is the midpt. of HV.
e. Def. of midpt.
f. IGH ITV
18. HL; each rt. has a
hyp. and side.
19. x = 3; y = 2
20. x = –1; y = 3
4-6

69. Congruence in Right Triangles
GEOMETRY LESSON 4-6
21. whether the 7-yd side is the hyp. or a leg
ABX ADX; HL
ABC ADC; SSS or SAS
BXC DXC; HL
23. a. Answers may vary. Sample: You could show that suppl. AXB and AXD are .
b. ABC ADC by SSS so BAC
DAC by CPCTC.
ABX ADX by SAS so AXB AXD. AXB is suppl. and to AXD so they are both rt. . s
24.
25.
26.
27.
4-6

70. Congruence in Right Triangles
GEOMETRY LESSON 4-6
EB DB; A and C are rt. (Given)
2. BEA BDC are rt (Def. of rt. )
3. B is the midpt. of AC (Given)
4. AB BC (Def. of midpt.)
5. BEA BDC (HL) s
LO bisects MLN, OM LM, ON LN, (Given)
2. M and N are rt. (Def. of )
3. MLO NLO (Def. of bis.)
4. M N (All rt. are .)
5. LO LO (Reflexive Prop. of )
6. LMO LNO (AAS) s
30. Answers may vary. Sample: Measure 2 sides of the formed by the amp. and the platform’s corner. Since the will be by HL or SAS, the are the same. s
4-6

71. Congruence in Right Triangles
GEOMETRY LESSON 4-6
31. a.
b. slope of DG = –1; slope of GF = –1; slope of GE = 1
c. EGD and EGF are rt. .
d. DE = ; FE = 26 s
31. (continued)
e. EGD EGF by HL. Both are rt. ,
DE FE, and EG EG.
32. An HA Thm. is the same as AAS with AAS corr. to the rt. , an acute , and the hyp. s
33. Since BE EA and BE EC, AEB and
CEB are both rt AB BC because ABC is equilateral, and BE BE. AEB CEB by HL.
34. No; AB CB because AEB CEB, but doesn’t have to be to AB or to CB.
35. A
36. H s
4-6

72. Congruence in Right Triangles
GEOMETRY LESSON 4-6
37. D
38. [2] a. TFW TGW
b. RFW and RGW are rt. .
[1] one part correct
39. isosceles
40. equilateral s
41. BC || AD because each slope = –1. BT BA, BA AD because product of slopes is –1.
42. If || lines then alt. int. are .
43. If two lines are ||, then same-side int. are suppl.
44. Vert. are .
45. If two lines are ||, then corr. are . s
46. If two lines are ||, then corr. are .
47. If two lines are ||, then same-side int. are suppl. s
4-6

73. Congruence in Right Triangles
GEOMETRY LESSON 4-6
For Exercises 1 and 2, tell whether the HL Theorem can be used to prove
the triangles congruent. If so, explain. If not, write not possible.
1. 2.
For Exercises 3 and 4, what additional information do you
need to prove the triangles congruent by the HL Theorem?
3. LMX LOX AMD CNB
Yes; use the congruent hypotenuses and leg BC to prove ABC DCB
Not possible
LM LO
AM CN or
MD NB
4-6

74. Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
(For help, go to the Lessons 1-1 and 4-3.)
1. How many triangles will the next two figures in this pattern have?
2. Can you conclude that the triangles are congruent? Explain.
a.  AZK and DRS   
b.  SDR and JTN   
c.  ZKA and NJT
4-7

75. Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
Solutions
1. For every new right triangle, segments connect the midpoint of the hypotenuse with the midpoints of the legs of the right triangle, creating two new triangles for every previous new triangle. The first figure has 1 triangle. The second has 1 + 2, or 3 triangles. The third has 3 + 4, or 7 triangles. The fourth will have 7 + 8, or 15 triangles. The fifth will have , or 31 triangles.
2. a.  Two pairs of sides are congruent. The included angles are congruent. Thus, the two triangles are congruent by SAS.
b. Two pairs of angles are congruent. One pair of sides is also congruent, and, since it is opposite a pair of corresponding congruent angles, the triangles are congruent by AAS.
c. Since AZK DRS and SDR JTN, by the Transitive Property of , ZKA NJT.
4-7

76. Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
Name the parts of their sides that DFG and EHG share.
Identify the overlapping triangles.
Parts of sides DG and EG are shared by DFG and EHG.
These parts are HG and FG, respectively.
4-7

77. Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
Write a Plan for Proof that does not use overlapping triangles.
Given: ZXW YWX, ZWX YXW
Prove: ZW YX
Look again at ZWM and YXM.  ZMW YMX because vertical angles are congruent, MW MX, and by subtraction  ZWM YXM, so ZWM YXM by ASA.
Look at MWX. MW MX by the Converse of the Isosceles Triangle Theorem.
Label point M where ZX intersects WY, as shown in the diagram. ZW YX by CPCTC if ZWM YXM.
You can prove these triangles congruent using ASA as follows:
4-7

78. Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
Write a paragraph proof.
Given: XW YZ, XWZ and YZW are right angles.
Prove: XPW YPZ
Plan: XPW YPZ by AAS if WXZ ZYW. These angles are congruent by CPCTC if XWZ YZW. These triangles are congruent by SAS.
Proof: You are given XW YZ. Because XWZ and YZW are right angles, XWZ YZW. WZ ZW, by the Reflexive Property of Congruence.
Therefore, XWZ YZW by SAS. WXZ ZYW by CPCTC, and XPW YPZ because vertical angles are congruent.
Therefore, XPW YPZ by AAS.
4-7

79. Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
Given: CA CE, BA DE
Write a two-column proof to show that CBE CDA.
Plan: CBE CDA by CPCTC if CBE CDA. This congruence holds by SAS if CB CD.
Proof:
Statements Reasons
1. BCE DCA 1. Reflexive Property of Congruence
2. CA CE, BA DE 2. Given
3. CA = CE, BA = DE 3. Congruent sides have equal measure.
4. CA – BA = CE – DE 4. Subtraction Property of Equality
5. CA – BA = CB, 5. Segment Addition Postulate CE – DE = CD
6. CB = CD 6. Substitution
7. CB CD 7. Definition of congruence
8. CBE CDA 8. SAS
9. CBE CDA 9. CPCTC
4-7

80. 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
4-7

81. 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.
4-7

82. 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
4-7

83. 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
4-7

84. 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
4-7

85. 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
4-7

86. 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)
4-7

87. 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
4-7

88. Using Corresponding Parts of Congruent Triangles
GEOMETRY LESSON 4-7
1. Identify any common sides and angles in AXY and BYX.
For Exercises 2 and 3, name a pair of
congruent overlapping triangles. State the
theorem or postulate that proves them congruent.
2. 3.
4. Plan a proof.
Given: AC BD, AD BC
Prove: XD XC XY
KSR MRS
SAS
GHI IJG
ASA
XD XC by CPCTC if DXA CXB. This congruence holds by AAS if BAD ABC. Show BAD ABC by SSS.
4-7
4-7

89. 9. Answers may vary. Sample: The corr. sides of the two may not be .
Congruent Triangles
GEOMETRY CHAPTER 4
Page 236
1. PAY APL
2. ONE OSE
3. SAS
4. HL
5. not possible
6. SSS
7. ASA
8. AAS
9. Answers may vary. Sample: The corr. sides of the two may not be .
10. Answers may vary. Sample:
CE HD; CO HF; EO DF; C H;
E D; O F s
11. No; the lengths may be different.
12. 36
13. AT || GS, so ATG
SGT because they are alt. int It is given that AT GS, and GT GT by the Reflexive Prop. of , so GAT TSG by SAS. s
4-A

90. LN by the Reflexive Prop. of , so OLN MLN by ASA. 15. CFE DEF; SSS
Congruent Triangles
GEOMETRY CHAPTER 4
14. Since LN bisects
OLM and ONM,
OLN MLN and
ONL MNL. LN
LN by the Reflexive Prop. of , so OLN
MLN by ASA.
CFE DEF; SSS
TQS TRA; SAS
17. Answers may vary. Sample:
ABC CDA
4-A