Congruent Figures

Solve each equation.

1.x + 6 = 25 2. x + 7 + 13 = 33

3.5x = 540 4. x + 10 = 2x

5.For the triangle at the right, use the Triangle

Angle-Sum Theorem to find the value of y.

(For help, go to page 24.)

GEOMETRY LESSON 4-1GEOMETRY LESSON 4-1

Congruent FiguresCongruent Figures

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 + 40 + 90 = 180; combine like terms: y + 130 = 180;

subtract 130 from both sides: y = 50

Congruent FiguresCongruent FiguresGEOMETRY LESSON 4-1GEOMETRY LESSON 4-1

4-1

List the corresponding sides in the same order.

List the corresponding vertices in the same order.


ABC QTJ. List the congruent corresponding

parts.

Angles: A Q B T C J

Sides: AB QT BC TJ AC QJ

4-1

XYZ KLM, mY = 67, and mM = 48. Find mX.

Use the Triangle Angle-Sum Theorem and the definition of congruent polygons to find mX.

mX = 65 Subtract 115 from each side.


mX + mY + mZ = 180 Triangle Angle-Sum Theorem

mZ = mM Corresponding angles of congruenttriangles that are congruent

mZ = 48 Substitute 48 for mM.

mX + 67 + 48 = 180 Substitute.

mX + 115 = 180 Simplify.

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.

4-1

Corresponding angles are not necessarily congruent, therefore you

cannot conclude that ABC CDE.

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 Givenb. CN DN Givenc. GN GN Reflexive Property of Congruenced. C D Givene. 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


Pages 182-185 Exercises

1. CAB DAB; C D; ABC ABD;AC AD; AB ABCB DB

2. GEF JHI; GFE JIH; EGF HJI;GE JH; EF HIFG IJ

3. BK

4. CM

5. ML

6. B

7. C

8. J

9. KJB

10. CLM

11. JBK

12. MCL

13. E, K, G, N

14. PO SI; OL ID;LY DE; PY SE

15. P S; O I; L D; Y E

16. 33 in.

17. 54 in.

18. 105

19. 77

4-1


20.36 in.

21.34 in.

22.75

23.103

24.Yes; RTK UTK, R U (Given) RKT UKTIf 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 .

s s

s

s

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

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.

38. JYB XCH

39. BCE ADE

40. TPK TRK

41. 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


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

s

s

s

s

s

s

48. a. 15

b.

4-1


49. 1.5

50. 4.25

51. 40

52. 72

53. Answers may vary. Sample:

54.

55. 100

56. RS = PQ

57. 1

58. 12

59. AB GH

4-1

In Exercises 1 and 2, quadrilateral WASH quadrilateral NOTE.

1. List the congruent corresponding parts.

2. mO = mT = 90 and mH = 36. Find mN.

3. Write a statement of triangle congruence.

4. Write a statement of triangle congruence.

5. Explain your reasoning in Exercise 4 above.

144


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.

Sample: DFH ZPR

Sample: ABD CDB

WA NO, AS OT, SH TE, WH NE;W N, A O, S T, H E

4-1

4-2


What can you conclude from each diagram?

1. 2. 3.

(For help, go to Lesson 2-5.)

Triangle Congruence by SSS and SASTriangle Congruence by SSS and SAS

Triangle Congruence by SSS and SASTriangle Congruence by SSS and SASGEOMETRY LESSON 4-2GEOMETRY 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

Write a paragraph proof.

Given: M is the midpoint of XY, AX AY

Prove: AMX AMY


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

AD BC. What other information do you need to prove ADC BCD?


Solution 1: If you know AC BD, you can prove ADC BCD by SSS.

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 2: If you know ADC BCD, you can prove ADC BCD by SAS.

4-2

Copy the diagram. Mark what is given on the diagram.


Given: RSG RSH, SG SH From the information given, can you prove RSG RSH? Explain.

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


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

s


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

10. U, V

11. WU

12. X

13. XZ, YZ

14. LG MN

15. T V or RS WU

16. DC CB

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.

22. ANG RWT; SAS

23. KLJ MON; SSS

24. Not possible; need H P or DY TK.

25. JEF SVF or JEF SFV; SSS

26. BRT BRS; SSS

27. 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


32.

33. a. Vertical

are .

b. Given

c. Def. of

midpt.

d. Given

e. Def. of

midpt.

f. SAS

s


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 .

36. ISP PSO; ISP OSP by SAS.

37. IP PO; ISP OSP by SSS.

s

s

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

s

s

41. 1. 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


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

51. FG

52. 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

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

49. E

50. AB


56. If x2 = 9, then x = 3. The statement is true but the converse is false.

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

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

(For help, go to the Lesson 4-2.)


Triangle Congruence by ASA and AASTriangle Congruence by ASA and AAS

4-3

Triangle Congruence by ASA and AASTriangle Congruence by ASA and AASGEOMETRY LESSON 4-3GEOMETRY 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 mC = mF = x and mB = mE = y, then mA = 180 – x – y = mD, so A = D.

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.


Therefore, FNI CAT GDO by ASA.

If F C, then F C G

The diagram shows N A D and FN CA GD.

4-3


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

Write a Plan for Proof that uses AAS.

Given: B D, AB || CD

Prove: ABC CDA


By the Reflexive Property, AC AC so ABC CDA by AAS.

Then ABC CDA if a pair of corresponding sides are congruent.

Because AB || CD, BAC DCA by the Alternate Interior Angles Theorem.

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

4. ABC CDA 4. AAS Theorem

2. BAC DCA 2. If lines are ||, then alternate interior angles are .

4-3

3. AC CA 3. Reflexive Property of Congruence



1. PQR VXW2. 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

12. FDE GHI; DFE HGI

13. a. UWV

b. UW

c. right

d. Reflexive

14. B D

15. MU UN

16. PQ QS

17. WZV WZY

4-3


4-3

18. a. Vert. are .

b. Given

c. TQ QR

d. AAS

19. PMO NMO; ASA

20. UTS RST; AAS

21. ZVY WVY; AAS

22. TUX DEO; AAS

23. The are not because no sides are .

s

s

24. 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.

s

s

s

s

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

s

s


29. a. Def. of

b. All right are

.

c. QTP

STR

d. Def. of midpt.

e. AAS

30.

s

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

s

34. Yes; by ASA since BDH FDH by def. of bis. and DH DH by the Reflexive Prop. of .


36. a. Check students’ work.

b. Answers may vary; most likely ASA.

4-3


37. AEB CED, BEC DEA, ABC CDA, BAD DCB

38. AEB CED, BECDEA, ABCCDA, ABDDCA, BADDCB, ABD DCB, CBA DAB, BCD ADC

39. They are bisectors; ASA.

40.

41.

42. D

43. F

1320

44. [2] a. RPQ SPQ, RQP

SQP (Def. of bisector)

b. ASA


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 2 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

46. ONL MLN; SAS

47. not possible

s

s s

s s

s

s

48. AC and CB

49. If corr. are , then the lines are ||.

50. 56 photos

51. 36 photos

52. 60% more paper

s

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.

3. 4. 5.

RF

S and U


GHI PQRAAS

not possible ABX ACXAAS

4-3

4-4


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.


Using Congruent Triangles: CPCTCUsing Congruent Triangles: CPCTC

Using Congruent Triangles: CPCTCUsing Congruent Triangles: CPCTCGEOMETRY LESSON 4-4GEOMETRY 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

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

(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 CRSCL CR

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



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. SSSb. 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

s

4-4

9. SPT OPT by SAS because TP TP by Reflexive Prop. of ; S O by CPCTC.

10. PNK MNL by SAS because KNP LNM by vert. are ; KP LM by CPCTC.

11. CYT RYP by AAS; CT RP by CPCTC.

s


12. 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

15. 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.

17. KLP KLQ because all rt are ; KL KL by Reflexive Prop. of ; and PKL QKL by def. of bisect; the are by ASA.

s

s

s

s

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

4-4


20. 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

s

s

s

4-4


26. 1. 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


34. ASA

35. AAS

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

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.

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

You are given two pairs of s and AM AM by the Reflexive Prop., so ABM ACM by ASA.

4-4



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.

Isosceles and Equilateral TrianglesIsosceles and Equilateral Triangles

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 = 75 + 30 = 105°.

Isosceles and Equilateral TrianglesIsosceles and Equilateral TrianglesGEOMETRY LESSON 4-5GEOMETRY LESSON 4-5

4-5

Examine the diagram below. Suppose that you draw XB YZ.

Can you use SAS to prove XYB XZB? Explain.

By the definition of perpendicular, XBY = XBZ.

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.

It is given that XY XZ.


By the Reflexive Property of Congruence, XB XB.

4-5

Explain why ABC is isosceles.

By the definition of an isosceles triangle, 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.

4-5

Suppose that mL = y. Find the values of x and y.

mN =mLIsosceles Triangle Theorem

mL = y Given

mN + mNMO + mMON=180Triangle Angle-Sum Theorem

mN = yTransitive Property of Equality

y + y + 90 =180Substitute. 2y + 90 =180

Simplify. 2y = 90 Subtract 90 from each side.y = 45 Divide each side by 2.

Therefore, x = 90 and y = 45.

MO LNThe bisector of the vertex angle of an isosceles triangle is the perpendicular bisector of the base.x = 90Definition of perpendicular


4-5

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, 120 + y + y = 180.120 + 2y = 180

2y = 60y = 30

So the angle measures in the triangle are 120, 30 and 30.


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.

4-5



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. Thm.

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


10. x = 4 ; y = 60

11. x = 36; y = 36

12. x = 92; y = 7

13. 64

14. 2

15. 42

16. 35

17. 150; 15

18. 24, 48, 72, 96, 120

12

12

19. a.

30, 30, 120

b. 5; 30, 60, 90, 120, 150

c. Check students’ work.

20. 70

21. 50

22. 140

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


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. 4-3. 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 = 2735. 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


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


43. a. 5

b.

44. 0 < measure of base < 45

45. 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 = 60 so m PAB =

60. Since PAB and QAB are

compl., m QAB = 30. QAB

is isosc. so m

AQB = 120.


4-5


50. RC = GV; RC GV by CPCTC since RTC GHV by ASA.

51. AAS

52. SSS

53. 24 sides

4-5

Use the diagram for Exercises 1–3.

1.If mBAC = 38, find mC.2.If mBAM = mCAM = 23, find mBMA. 3.If mB = 3x and mBAC = 2x – 20, find x.

4. Find the values of x and y. 5.ABCDEF is a regular hexagon. Find mBAC.

71

90

25

x = 60y = 9

30

4-5


4-5

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.

7. 8.

Congruence in Right TrianglesCongruence in Right Triangles

Solutions1.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.

Congruence in Right TrianglesCongruence in Right TrianglesGEOMETRY LESSON 4-6GEOMETRY LESSON 4-6

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

One student wrote “ CPA MPA by SAS” for the diagram

below. Is the student correct? Explain.

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.


The diagram shows the following congruent parts.

CA MA

CPA MPA

PA PA

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

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 triangleright 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



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 s

s s

s s

s

4. AEB DCB by HL. Both are rt. .AB BD and EB CBby 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 s

s

s

9. BC FA

10. RT NQ

11. a. Given

b. Def. of rt.

c. Reflexive Prop.

of

d. Given

e. HL

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 andRT TV.

s 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


21. whether the 7-yd side is the hyp. or a leg

22. 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

s

24.

25.

26.

27.

4-6


28. 1. 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

s

29. 1. 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

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

s

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 = 26 ; 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 s

33. Since BE EA andBE 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


37. D

38. [2] a. TFW TGW

b. RFW and RGW are rt. .


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

s

s

s

46. If two lines are ||, then corr. are .

47. If two lines are ||, then same-side int. are suppl.

s

s

4-6

For Exercises 1 and 2, tell whether the HL Theorem can be used to provethe 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 4. AMD CNB

Notpossible


Yes; use the congruent hypotenuses and leg BC to prove ABC DCB

LM LO AM CNor

MD NB

4-6


(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

Using Corresponding Parts of Congruent TrianglesUsing Corresponding Parts of Congruent Triangles

4-7

Using Corresponding Parts of Congruent TrianglesUsing Corresponding Parts of Congruent TrianglesGEOMETRY LESSON 4-7GEOMETRY 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 15 + 16, 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 correspondingcongruent angles, the triangles are congruent by AAS.

c. Since AZK DRS and SDR JTN, by the TransitiveProperty of , ZKA NJT.

4-7

Name the parts of their sides that DFG and EHG share.

These parts are HG and FG, respectively.

Parts of sides DG and EG are shared by DFG and EHG.


Identify the overlapping triangles.

4-7

Write a Plan for Proof that does not use overlapping triangles.

Given: ZXW YWX, ZWX YXWProve: ZW YX

You can prove these triangles congruent using ASA as follows:


Label point M where ZX intersects WY, as shown in the diagram. ZW YX by CPCTC if ZWM YXM.

Look at MWX. MW MX by the Converse of the Isosceles Triangle Theorem.

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.

4-7


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

Given: CA CE, BA DE

Write a two-column proof to show that CBE CDA.

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 = CD6. CB = CD 6. Substitution


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

7. CB CD 7. Definition of congruence 8. CBE CDA 8. SAS 9. CBE CDA 9. CPCTC

4-7



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

11. LQP PML; HL

4-7


12. RST UTS; SSS

13. QDA UAD; SAS

14. 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.

17. QET QEU by SAS if QT QU. QT and QU are corr. parts of QTB and QUB which are by ASA.

18. 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


21. a.

b.

22. a.

b.

23. ACE BCD by ASA; AC BC, A B (Given) C C (Reflexive Prop. of ) ACE BCD (ASA)

24. 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)

s

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

26. ABC FCG; ASA

27. a. Given

b. Reflexive Prop.

of

c. Given

d. ETI

4-7


27. (continued)e. IRE

f. CPCTC

g. Given

h. All rt. are .

i. TDI

j. ROE

k. CPCTC

s

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

4-7


4-7

29–30. Proofs may vary. Samples are given.

29. It is given that 1 2 and 3 4. 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.

30. 1. 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 .)

8. ADC EDG (ASA)

31. a. AD BC; AB DC; AE EC; DE EB

s


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

32. 1. 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

s

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 .


s

38. [4] a. HL

b.

c. x = 30. In ADC m A + m ADC + m ACD = 180. Substituting, 90 + x + x + x = 180.

Solving, x = 30.

d. 120; it is suppl. to a 60°

e. 6 m; DC = 2(AD)

4-7


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)

12

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)

13

53

56

4-7

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 BCProve: XD XC

XY

KSR MRSSAS

GHI IJGASA

XD XC by CPCTC if DXA CXB.This congruence holds by AAS if BAD ABC. Show BAD ABC by SSS.

4-7

4-A

Congruent TrianglesCongruent TrianglesGEOMETRY CHAPTER 4GEOMETRY 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 .


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

Congruent TrianglesCongruent TrianglesGEOMETRY CHAPTER 4GEOMETRY 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.

15. CFE DEF; SSS

16. TQS TRA; SAS


ABC CDA

4-A

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