Download Chapter 4 Answers

Chapter 4. Congruence of Line Segments,Angles, and Triangles
4-1 Postulates of Lines, Line Segments,
and Angles (pages 139–140)
7. a. Given: AB > CD and EF > CD
Prove: AB > EF
Writing About Mathematics
1. Points E and F must name the same point.
2. No. There is only one positive real number
that is the length of a given line segment. A
different number is the length of a different
line segment.
b. Statements
1. AB > CD
2. EF > CD
3. AB > EF
8. a. Given: 2EF 5 DB and GH 5 21DB
Prove: EF 5 GH
Developing Skills
3. a. Given: AB 5 AD and DC 5 AD
Prove: AB 5 DC
b. Statements
1. AB 5 AD
2. DC 5 AD
3. AB 5 DC
b. Statements
1. 2EF 5 DB
2. EF 5 12DB
Reasons
1. Given.
2. Given.
3. Transitive property
of equality.
3. GH 5 21DB
4. EF 5 GH
b. Statements
1. CD 5 2CE
2. CE 5 CF
3. CD 5 2CF
4. CB 5 2CF
5. CD 5 CB
Reasons
1. Given.
2. Given.
3. Transitive property
of equality.
5. a. Given: m/1 1 m/2 5 90, m/A 5 m/2
Prove: m/1 1 m/A 5 90
b. Statements
1. m/1 1 m/2 5 90
2. m/A 5 m/2
3. m/1 1 m/A 5 90
4. m/2 5 m/A
5. m/1 5 m/2
Reasons
1. Given.
2. Given.
3. Substitution postulate.
4. Given.
5. Transitive property of
equality.
10. a. Given: RT 5 RS, RD 5 21RT, RE 5 12RS
Prove: RD 5 RE
Reasons
1. Given.
2. Given.
3. Substitution
postulate.
6. a. Given: m/A 5 m/B, m/1 5 m/B,
m/2 5 m/A
Prove: m/1 5 m/2
b. Statements
1. m/1 5 m/B
2. m/A 5 m/B
3. m/1 5 m/A
Reasons
1. Given.
2. Halves of equal
quantities are equal.
3. Given.
4. Transitive property of
equality.
9. a. Given: CE 5 CF, CD 5 2CE, CB 5 2CF
Prove: CD 5 CB
4. a. Given: AD > CD and BD > CD
Prove: AD > BD
b. Statements
1. AD > CD
2. BD > CD
3. AD > BD
Reasons
1. Given.
2. Given.
3. Transitive property of
congruence.
b. Statements
1. RD 5 21RT
2. RT 5 RS
3. RD 5 21RS
Reasons
1. Given.
2. Given.
3. Substitution postulate.
4. RE 5 12RS
5. RD 5 RE
4. Given.
5. Substitution postulate.
11. a. Given: AD 5 BE and BC 5 CD
Prove: AC 5 CE
Reasons
1. Given.
2. Given.
3. Transitive property
of equality.
4. Given.
5. Transitive property
of equality.
b. Statements
1. AD 5 BE
2. BC 5 CD
3. AD 2 CD
5 BE – BC
4. AD 5 AC 1 CD,
BE 5 BC 1 CE
5. AC 5 CE
246
Reasons
1. Given.
2. Given.
3. Subtraction postulate.
4. Partition postulate.
5. Substitution postulate.
c. The shortest distance between two points is
the line segment between them. Hence, its
length is shorter than the length of the arc.
12. a. Given: /CDB > /CBD and
/ADB > /ABD
Prove: /CDA > /CBA
b. Statements
1. /CDB > /CBD
2. /ADB > /ABD
3. /CDB 1 /ADB
> /CBD 1 /ABD
4. /CDA
> /CDB 1 /BDA,
/CBA
> /CBD 1 /DBA
5. /CDA > /CBA
Reasons
1. Given.
2. Given.
3. Addition postulate.
20. Yes. Through two distinct points, one and only
one line can be drawn.
4. Partition postulate.
4-2 Using Postulates and Definitions in
Proofs (pages 142–144)
Writing About Mathematics
1. The symbol ABC refers to a line segment with
endpoints A and C with point B lying between
them. The symbol ACB refers to a line segment
with endpoints A and B with point C lying
between them. These symbols could not refer to
segments of the same line.
2. Yes. Since m/ABD 5 90, then:
m/ABD 1 m/DBC 5 180 or m/DBC 5 90
m/DBC 1 m/CBE 5 180 or m/CBE 5 90
m/ABE 1 m/CBE 5 180 or m/ABE 5 90
5. Substitution
postulate.
13. m/QSR 5 37
14. m/PST 5 50
15. m/PST 5 45
Applying Skills
16. a. Answers will vary.
b. Given: A triangle is equilateral.
Prove: The measures of the sides are equal.
c. If triangle is equilateral then all of its sides are
congruent. Congruent sides have equal
measures.
Developing Skills
3. a. Given: AB > CB, FD bisects AB, and FE
bisects CB.
Prove: AD > CE
17. a. Answers will vary.
b. Given: Points E and F are distinct and two
lines intersect at E.
Prove: The lines do not intersect at F.
c. Two lines can intersect at only one point.
If they intersect at both E and F, then E and
F must name the same point. But we are
given that E and F are distinct. This is a contradiction, so the lines do not intersect at F.
b. Statements
1. AB > CB
2. AB 5 CB
3. FD bisects AB,
FE bisects CB.
4. D is the midpoint
of AB, E is the
midpoint of CB.
5. AD 5 12AB,
CE 5 21CB
6. AD 5 CE
18. a. Answers will vary.
b. Given: A line through a vertex of a triangle is
perpendicular to the opposite side.
Prove: The line separates the triangle into two
right triangles.
c. A line through a vertex of a triangle that is
perpendicular to the opposite side separates it
into two triangles. Perpendicular lines
intersect to form right angles, so each of the
triangles has a right angle. A triangle with a
right angle is a right triangle.
7. AD > CE
19. a. Answers will vary.
b. Given: Two points on a circle are endpoints of
a line segment.
Prove: The length of the line segment is
shorter than the length of the portion
of the circle with the same endpoints.
247
Reasons
1. Given.
2. Definition of
congruent segments.
3. Given.
4. Definition of
bisector.
5. Definition of
midpoint.
6. Halves of equal
quantities are
equal.
7. Definition of
congruent segments.
h
h
Statements
C is the midpoint
of BD.
2. AB > BC,
BC > CD
3. AB 5 BC,
BC 5 CD
4. AB 5 BC 5 CD
4. a. Given: CA bisects /DCB, AC bisects /DAB,
and /DCB > /DAB.
Prove: /CAB > /DCA
b. Statements
1. /DCB > /DAB
2. m/DCB
5 m/DAB
Reasons
1. Given.
2. Definition of
congruent angles.
h
3. CA bisects /DCB, 3. Given.
Reasons
2. Definition of
midpoint.
3. Definition of
congruent segments.
4. Transitive property
of equality.
h
AC bisects /DAB
4. m/CAB
5 21m/DAB,
m/DCA
5 12m/DCB
5. m/CAB
5 m/DCA
6. /CAB > /DCA
8. a. Given: P and T are distinct points, P is the
midpoint of RS.
Prove: T is not the midpoint of RS.
b. Postulate 4.9 states that a line segment has
only one midpoint. P is the midpoint of RS,
and T does not name P. Therefore, T is not the
midpoint.
4. Definition of angle
bisector.
5. Halves of equal
quantities are
equal.
6. Definition of
congruent angles.
9. a. Given: DF > BE
Prove: DE > BF
b. Statements
1. DF > BE
2. EF > FE
3. DF 2 EF
> BE 2 FE
4. DF > DE1EF,
BE > BF1FE
5. DE > BF
5. a. Given: AD > BE
Prove: AE > BD
b. Statements
1. AD > BE
2. AE > AD 1 DE,
BD > BE 1 ED
3. AE > BE 1 DE
4. AE > BD
Reasons
1. Given.
2. Partition postulate.
3. Substitution
postulate.
4. Substitution
postulate.
4. AC 5 BD
b. Statements
1. AD > BC
2. AD 5 BC
Reasons
1. Given.
3. E is the midpoint
of AD.
F is the midpoint
of BC.
4. AE 5 21AD,
2. Partition postulate.
3. Substitution
postulate.
4. Substitution
postulate.
FC 5 21BC
5. AE 5 FC
7. a. Given: ABCD is a segment, B is the midpoint
of AC, and C is the midpoint of BD.
Prove: AB 5 BC 5 CD
b. Statements
1. B is the midpoint
of AC.
5. Substitution
postulate.
10. a. Given: AD > BC, E is the midpoint of AD,
and F is the midpoint of BC.
Prove: AE > FC
6. a. Given: ABCD is a segment and AB 5 CD.
Prove: AC 5 BD
b. Statements
1. AB 5 CD
2. AC 5 AB 1 BC,
BD 5 BC 1 CD
3. AC 5 BC 1 CD
Reasons
1. Given.
2. Symmetric property.
3. Subtraction
postulate.
4. Partition postulate.
6. AE > FC
Reasons
1. Given.
248
Reasons
1. Given.
2. Definition of
congruent segments.
3. Given.
4. Definition of
midpoint.
5. Halves of equal
quantities are equal.
6. Definition of
congruent segments.
3. RS 5 2RN,
LM 5 2LN
4. RN 5 LN
5. RS 5 LM
11. a. Given: AC > DB, and AC and DB bisect each
other at E.
Prove: AE > EB
b. Statements
1. AC > DB
2. AC 5 DB
3. AC and DB bisect
each other at E.
4. E is the midpoint
of AC and of DB.
5. AE 5 12AC,
EB 5 21DB
6. AE 5 EB
7. AE > EB
Reasons
1. Given.
2. Definition of
congruent segments.
3. Given.
d. RS 5 LM 5 48
e. The distance from L to RS is LN. The distance
from a point to a segment is the length of a
perpendicular from the point to the segment.
In this case, it is LN 5 12 (48) 5 24.
4. Definition of
bisector.
5. Definition of
midpoint.
4-3 Proving Theorems About Angles
(pages 152–154)
6. Halves of equal
quantities are equal.
7. Definition of
congruent segments.
Writing About Mathematics
1. Yes. /AEC and /BED are supplements of
congruent angles /BEC and /AED. If two
angles are congruent, then their supplements are
congruent.
2. No. The converse is “If two angles are
supplementary, then the angles form a linear
pair.” Supplementary angles do not need to be
adjacent.
Developing Skills
3. Statements
Reasons
1. m/ACD
1. Given.
1 m/DCB 5 90
2. /B > /DCA,
2. Given.
/A > /DCB
3. m/B 5 m/DCB,
3. Definition of
m/A > m/DCB
congruent angles.
4. m/B 1 m/A 5 90 4. Substitution postulate.
5. /A and /B are
5. Definition of
complements.
complementary
angles.
h
12. a. Given: DR bisects /CDA, /3 > /1, /4 > /2
Prove: /3 > /4
b. Statements
Reasons
h
1. DR bisects /CDA. 1. Given.
2. /1 > /2
2. Definition of angle
bisector.
3. /3 > /1,
3. Given.
/4 > /2
4. /3 > /4
4. Substitution
postulate.
Applying Skills
13. a. m/CDE 5 114
b. m/FDE 5 76
c. m/CDG 5 76
d. /CDE is obtuse.
14. a. Answers will vary.
b. AB 5 BC 5 10
c. The distance from A to BD is AB. The
distance from a point to a segment is the
length of a perpendicular from the point to the
segment. In this case, it is AB 5 10.
15. a. Answers will vary.
b. Given: RS and LM bisect each other at N,
RS ' LM, and RN 5 LN.
c. Statements
1. RS and LM bisect
each other at N.
2. N is the midpoint
of RS and of LM.
3. Definition of
midpoint.
4. Given.
5. Doubles of equal
quantities are equal.
4. Statements
1. /BFC and /BFG
form a linear pair.
/ADE and /ADC
form a linear pair.
2. /BFC and /BFG
are supplementary.
/ADE and /ADC
are also supplementary.
3. /ADC > /BFG
4. /ADE > /BFG
Reasons
1. Given.
2. Definition of
bisector.
249
Reasons
1. Definition of a
linear pair.
2. If two angles form a
linear pair, then they
are supplementary.
3. Given.
4. If two angles are
congruent, then their
supplements are
congruent.
5. Statements
1. /ADB is a right
angle.
2. CE ' DBE
3. /CEB is a right
angle.
4. /ADB > /CEB
6. Statements
1. /ABC and /BCD
are right angles.
2. m/ABC 5 90 and
m/BCD 5 90
3. m/EBA 1 m/EBC
5 m/ABC
m/ECB 1 m/ECD
5 m/BCD
4. m/EBA 1 m/EBC
5 90
m/ECB 1 m/ECD
5 90
5. /EBA and /EBC
are complements.
/ECB and /ECD
are complements.
6. /EBC > /ECB
7. /EBA > /ECD
7. Statements
1. /ABC and /DBF
are right angles.
2. m/ABC 5 90 and
m/DBF 5 90
3. m/ABC 1 m/DBC
5 m/ABC
m/DBC 1 m/CBF
5 m/DBF
4. m/ABC 1 m/DBC
5 90
m/DBC 1 m/CBF
5 90
5. /ABC and /DBC
are complements.
/DBC and m/CBF
are complements.
6. /DBC > /DBC
Reasons
1. Given.
Statements
7. /ABC > /CBF
2. Given.
3. Definition of
perpendicular lines.
4. If two angles are right
angles, then they are
congruent.
8. Statements
Reasons
g
g
Reasons
1. Given.
1. EF intersects DC
at G.
2. /CGH and /DGE
are vertical angles.
3. /CGH > /DGE
2. Definition of right
angles.
3. Partition postulate.
4. m/BHG
5 m/CGH
5. /BHG > /CGH
6. /BHG > /DGE
4. Substitution postulate.
9. Statements
1. /ABC and /DBC
form a linear pair.
/ACB and /ECB
form a linear pair.
2. /ABC and /DBC
are supplementary.
/ACB and /ECB
are supplementary.
3. /ABC > ACB
4. /DBC > /ECB
5. Definition of
complementary
angles.
6. Given.
7. If two angles are
congruent, then their
complements are
congruent.
Reasons
1. Given.
10. Statements
2. Definition of right
angles.
3. Partition postulate.
g
g
3. AEB ' CED
4. Substitution
postulate.
11. Statements
1. /ABC is a right
angle.
2. m/ABC 5 90
5. Definition of
complementary
angles.
1. Given.
2. Definition of
vertical angles.
3. Vertical angles are
congruent.
4. Given.
5. Definition of
congruent angles.
6. Transitive property of
congruency.
Reasons
1. Definition of a linear
pair.
2. If two angles form a
linear pair, then they
are supplementary.
3. Given.
4. If two angles are
congruent, their
supplements are
congruent.
Reasons
g
1. AEB and CED
intersect at E.
2. /AEC > /CEB
g
Reasons
7. If two angles are
congruent, then their
complements are
congruent.
1. Given.
2. Given.
3. If two angles
intersect to form
congruent adjacent
angles, then they are
perpendicular.
Reasons
1. Given.
2. Definition of right
angles.
3. m/DBA 1 m/CBD 3. Partition postulate.
5 m/ABC
(Cont.)
6. Reflexive property of
congruence.
250
Statements
4. m/DBA 1 m/CBD
5 90
5. /DBA and /CBD
are complementary.
6. /BAC and /DBA
are complementary.
7. /BAC > /CBD
Reasons
4. Substitution postulate.
2. No. nRST > nSTR implies that /R > /S,
/S > /T, /T > /R, RS > ST, ST > TR, and
TR > RS. However, this is not necessarily true.
5. Definition of
complementary angles.
6. Given.
Developing Skills
3. AB > CB, BD > BD, DA > DC, /A > /C,
/ABD > /DBC, /ADB > /CDB
4. AD > CB, DB > DB, AB > DC, /A > /C,
/ADB > /CBD, /ABD > /CDB
5. AB > EB, BD > BC, AD > EC, /A > /E,
/ABD > /EBC, /D > /C
6. Reflexive property
7. Symmetric property
8. Transitive property
9. Symmetric property
10. Symmetric property
7. If two angles are
complements of the
same angle, then they
are congruent.
12. m/AED 5 70, m/DEB 5 110, m/AEC 5 110
13. m/DEB 5 m/AEC 5 120, m/AED
5 m/CEB 5 60
14. m/BEC 5 m/DEA 5 75,
m/AEC 5 m/DEB 5 105
15. m/CEB 5 m/DEA 5 45,
m/BED 5 m/AEC 5 135
16. a. y 5 20, x 5 30
b. m/RPL 5 50, m/LPS 5 130,
m/MPS 5 50
17. If two angles are straight angles, then they both
measure 180 degrees. Hence, they both have the
same measure. By definition of congruent angles,
the two angles are congruent.
4-5 Proving Triangles Congruent Using
Side, Angle, Side (pages 160–161)
Writing About Mathematics
1. No. Suppose that they are. Let t1 represent the
top of the first pole, b1 represent the point where
the first pole meets the ground, and w1 represent
the point where the wire of the first pole meets
the ground. Define t2, b2, and w2 similarly for the
second pole. If the poles have the same height,
then t1b1 > t2b2. Since the poles are both
perpendicular to the ground, /t1b1w1 > /t2b2w2.
Since the point where the wires meet the ground
is 5 feet away from the foot of the poles,
b1w1 5 b2w2 5 5 feet, and so b1w1 > b2w2.
Therefore, nt1b1w1 > nt2b2w2 by SAS. In
particular, t1w1 > t2w2, but these represent the
lengths of the wires. The assumption is false, and
the heights of the poles are unequal.
2. No. Two non-congruent triangles can have one,
two, or three angles congruent. Also, two noncongruent triangles can have one side congruent,
or two sides congruent if the angle between them
is not congruent.
18. If two angles, /1 and /2, are both supplements
of the same angle, /3, then m/1 1 m/3 5 180
and m/2 1 m/3 5 180. By the substitution
postulate, m/1 1 m/3 5 m/2 1 m/3. By the
subtraction postulate, m/1 5 m/2 so /1 > /2.
19. If two angles, /1 and /2, are congruent and
/3 and /4 are their respective supplements, then
m/1 1 m/3 5 180 and m/2 1 m/4 5 180.
By the substitution postulate, m/1 1 m/3 5
m/2 1 m/4. Since m/1 5 m/2, by the
subtraction postulate, m/3 5 m/4. Therefore,
/3 > /4.
Applying Skills
20. 120°
23. 75°
21. 40°
24. 65°
22. 130°
25. 33°
Developing Skills
3. Yes
4. Yes
6. Yes
7. No
9. /D > /C
10. AD > DB
11. /EDA > /CDB
4-4 Congruent Polygons and
Corresponding Parts (page 157)
Writing About Mathematics
1. No. One pair of congruent corresponding sides is
not sufficient to prove two triangles congruent.
251
5.
8.
No
Yes
4-6 Proving Triangles Congruent Using
Angle, Side, Angle (pages 163–164)
Applying Skills
12. a. Answers will vary;
Given: ABC and DBE bisect each other.
b. Prove: nABE > nCBD
c. Statements
1. ABC and DBE
bisect each other.
2. B is the midpoint
of ABC and of
DBE.
3. AB > BC,
DB > BE
4. /ABE > /DBC
5. nABE > nCBD
Writing About Mathematics
1. Yes. Since the two triangles are congruent,
corresponding angles are congruent. Since one of
the angles in the first triangle is a right angle, one
of the angles in the second triangle must also be a
right angle.
2. a. No. The two triangles are not necessarily
congruent, as the figure shows.
Reasons
1. Given.
2. Definition of
bisector.
C
3. Definition of
midpoint.
4. Vertical angles are
congruent.
5. SAS.
13. a. Answers will vary;
Given: ABCD is a quadrilateral; AB 5 CD,
BC 5 DA; /DAB, /ABC, /BCD, and
/CDA are right angles.
b. Prove: Diagonal AC separates the
quadrilateral into two congruent
triangles.
c. Statements
1. AB 5 CD and
BC 5 DA
2. AB > CD and
BC > DA
3. /ABC and /ADC
are right angles.
4. /ABC > /ADC
5. nABC > nCDA
F
D
Reasons
1. Given.
3. /PQR > /RQS
4. RQ > RQ
5. nPQR > nRQS
E
b. /A cannot be congruent to /D. CB cannot
be congruent to FE.
Developing Skills
3. Yes. Two angles and the side between them in
one triangle are congruent to two angles and the
side between them in the other triangle.
4. Yes. Two angles and the side between them in
one triangle are congruent to two angles and side
between them in the other triangle.
5. No. Only one pair of congruent angles are given.
6. /ACD > /CAB
7. /AED > /CEB
8. DB > DB
Applying Skills
9. Statements
Reasons
1. /E > /C,
1. Given.
/EDA > /CDB
2. D is the midpoint
2. Given.
of EC.
3. ED > DC
3. Definition of
midpoint.
4. nDAE > nDBC
4. ASA.
2. Definition of
congruent segments.
3. Given.
4. Right angles are
congruent.
5. SAS.
14. a. Answers will vary;
Given: /PQR and /RQS form a linear pair,
/PQR > /RQS, and PQ 5 QS.
b. Prove: nPQR > nRQS
c. Statements
1. PQ 5 QS
2. PQ > QS
B
A
Reason
1. Given.
2. Definition of
congruent segments.
3. Given.
4. Reflexive property.
5. SAS.
252
10. Statements
h
1. DB bisects /ADC
the two sides of one triangle are congruent, then
two sides of the other triangle must be also
congruent and that triangle is isosceles.
2. Yes. If all corresponding sides are congruent,
then by SSS, the two triangles are congruent.
Reasons
1. Given.
h
and BD bisects
/ABC.
2. /ADB > /BDC,
/CBD > /DBA
3. DB > DB
4. nABD > nCBD
11. Statements
1. AD ' BC
2. /CDA > /BDA
3. AD > AD
h
4. AD bisects /BAC.
5. /CAD > /BAD
6. nADC > nADB
12. Statements
1. /DBC > /GFD
2. /DBC and /ABD
form a linear pair;
/GFD and /DFE
form a linear pair.
3. /DBC and /ABD
are supplementary;
/GFD and /DFE
are supplementary.
4. /DFE > /ABD
2. Definition of angle
bisector.
3. Reflexive property of
congruence.
4. ASA.
Developing Skills
3. Yes
4. Yes
5. No
6. BC > DC
7. AD > BC
8. FB > EC
9. SAS
10. Not enough information
11. ASA or SAS
12. SSS
13. Not enough information
14. Not enough information
15. No. One triangle can be acute and the other can
be obtuse.
Reasons
1. Given.
2. If two lines are
perpendicular, then
they intersect to form
congruent adjacent
angles.
3. Reflexive property of
congruency.
C
4. Given.
5. Definition of angle
bisector.
6. ASA.
B
A
Reasons
1. Given.
2. Definition of a linear
pair.
F
E
D
3. Linear pairs of angles
are supplementary.
Applying Skills
16. Statements
1. AC ' CED and
BD ' CED
2. /ACE and /BDE
are right angles.
4. The supplements of
congruent angles are
congruent.
5. AE bisects FB at D. 5. Given.
6. D is the midpoint
6. Definition of bisector.
of FB.
7. FD > DB
7. Definition of
midpoint.
8. /FDE > /ADB
8. Vertical angles are
congruent.
9. nDFE > nDBA
9. ASA (steps 4, 7, 8).
3. /ACE > /BDE
4. /AEC > /BED
5. AEB bisects CED.
6. E is the midpoint
of CED.
7. CE > ED
4-7 Proving Triangles Congruent Using
Side, Side, Side (pages 166–167)
8. nEAC > nEBD
Writing About Mathematics
1. Yes. If two triangles are congruent, then
corresponding sides are congruent. Therefore, if
253
Reasons
1. Given.
2. Perpendicular
segments meet to
form right angles.
3. Right angles are
congruent.
4. Vertical angles are
congruent.
5. Given.
6. Definition of bisector.
7. Definition of
midpoint.
8. ASA (steps 3, 7, 4).
g
17. Statements
Reasons
1. nABC is equilateral. 1. Given.
2. AC > BC
2. In an equilateral
triangle, all sides are
congruent.
3. D is the midpoint
3. Given.
of AB.
4. AD > DB
4. Definition of
midpoint.
5. DC > DC
5. Reflexive property of
congruence.
6. nACD > nBCD
6. SSS.
18. Statements
1. PS 5 QS
2. PS > QS
3. RS ' PQ
4. /PSR > /QSR
5. SR > SR
g
3. AD > AD
h
4. AD is the bisector
of /BAC.
5. /BAD > /CAD
6. nABD > nACD
g
that MN and PM intersect at M. If they are
g
distinct lines, then P does not lie on MN. If they
coincide, then it does.
9. Yes. At a given point on a given line, only one
perpendicular can be drawn to the line.
Reasons
1. Assumption.
2. Definition of
congruent segments.
3. Given.
4. If two lines are
perpendicular, then
they intersect to form
congruent adjacent
angles.
5. Reflexive property of
congruence.
6. SAS.
7. Given.
10. Statements
1. m/A 5 m/D 5 50
2. /A > /D
3. AB 5 DE 5 10 cm
4. AB > DE
5. m/B 5 m/E 5 45
6. /B > /E
7. nABC > nDEF
6. nPSR > nQSR
7. nPSR is not
congruent to nQSR.
8. PS Þ QS
8. Contradiction.
19. Statements
1. AC 5 AB 5 15
2. AC > AB
g
5. LM and KM coincide; at a given point on a given
line, one and only one perpendicular can be
drawn to the line.
6. LM 1 MN , LR 1 RN; the shortest distance
between two points is the length of the line
segment joining these two points.
7. Yes. An angle has one and only one bisector.
8. Not necessarily. Two lines cannot intersect in
more than one point. However, we are only given
11. Statements
1. GEH bisects DEF.
2. E is the midpoint
of DEF.
3. DE > EF
Reasons
1. Given.
2. Definition of
congruent segments.
3. Reflexive property of
congruence.
4. m/D 5 m/F
5. /D > /F
6. /DEH > /GEF
4. Given.
7. nGFE > nHDE
5. Definition of angle
bisector.
6. SAS.
12. Statements
1. /B > /E
2. AB > DE
3. BC > EF
4. nABC > nDEF
5. nABC is not
congruent to nDEF.
6. /B is not congruent
to /E.
Review Exercises (pages 169–170)
1. The measure of the angle is 125° and its
complement is 55°.
2. m/PMN 5 48, m/LMP 5 132
3. m/K 5 m/Q 5 63
4. B 5 F; two lines cannot intersect in more than
one point.
254
Reasons
1. Given.
2. Definition of
congruent angles.
3. Given.
4. Definition of
congruent segments.
5. Given.
6. Definition of
congruent angles.
7. ASA.
Reasons
1. Given.
2. Definition of bisector.
3. Definition of
midpoint.
4. Given.
5. Definition of
congruent angles.
6. Vertical angles are
congruent.
7. ASA.
Reasons
1. Assumption.
2. Given.
3. Given.
4. SAS.
5. Given.
6. Contradiction.
14. We have a system of two equations:
Exploration (page 170)
1.
(2x 2 y) 1 (x 1 4y) 5 180
2x 2 y 5 x 1 4y
Solving for x in the second equation, x 5 5y.
Substituting for x in the first equation:
2(5y) 2 y 1 5y 1 4y 5 180
18y 5 180
y 5 10
2. We are given that STUVWXYZ is a cube. In a
cube, all edges are congruent. Therefore,
ST > TX > UT. In a cube, all faces are squares.
All angles in a square are right angles.
Therefore, /STX, /UTX, and /STU are right
angles. Right angles are congruent, so
/STX > /UTX > /STU. Then
nSTX > nUTX > nSTU by SAS.
Therefore, y 5 10 and x 5 5(10) 5 50.
Part IV
15. Let x 5 the measure of one angle, then
x
2 1 12 5 the measure of the other angle.
x 1 x2 1 12 5 90
3
2x
Cumulative Review (pages 170–173)
Part I
1. 4
2. 2
5. 4
6. 2
9. 4
10. 2
Part II
11. Statements
1. PQ bisects RS at M.
2. M is the midpoint
of RS.
3. RM > MS
4. /R > /S
5. /RMQ > /SMP
6. nRMQ > nSMP
12. Statements
1. DE 5 DG,
EF 5 GF
2. DE > DG,
EF > GF
3. DF > DF
3. 1
7. 3
5 78
x 5 52
Therefore, the angles measure 52° and 38°.
16. a. nDML > nENM > nFLN
b. We are given that nDEF is equilateral and
equiangular. Thus, /D > /E > /F and
DE > EF > FD. We are also given that M, N,
and L are the midpoints of the sides of the
triangle, and so 12DE 5 DM 5 ME,
1
1
2 EF 5 EN 5 NF, and 2 FD 5 FL 5 LD.
Since halves of congruent segments
are congruent, DM > ME > EN >
NF > FL > LD. Therefore, nDML >
nENM > nFLN by SAS.
4. 4
8. 1
Reasons
1. Given.
2. Definition of bisector.
3. Definition of
midpoint.
4. Given.
5. Vertical angles are
congruent.
6. ASA.
c. In part b, we showed that nDML > nENM >
nFLN. Since corresponding parts of
congruent triangles are congruent,
LM > MN > LN. Therefore, nMNL is
equilateral.
Reasons
1. Given.
d. By the partition postulate,
2. Definition of
congruent segments.
3. Reflexive property of
congruence.
4. SSS.
m/DLF 5 m/DLM 1 m/MLN 1 m/FLN
m/FNE 5 m/FNL 1 m/LNM 1 m/ENM
m/DME 5 m/EMN 1 m/LMN 1 m/DML
Since all straight angles are congruent and
/DLF, /FNE, and /DME are straight
angles, m/DLF 5 m/FNE 5 m/DME.
Thus, by the substitution postulate,
4. nDEF > nDGF
Part III
13. Yes. Let w 5 “Our team wins,” c 5 “We
celebrate,” and p 5 “We practice.” Since both
c ∨ p and ,p are true, by the Law of Disjunctive
Inference, c is true. Since ,w → ,c implies
c → w, w is true by the Law of Detachment.
Therefore, our team won.
m/DLM 1 m/MLN 1 m/FLN
5 m/FNL 1 m/LNM 1 m/ENM
5 m/EMN 1 m/LMN 1 m/DML
255
In part b, we showed that
nDML > nENM > nFLN. Since
corresponding parts of congruent triangles are
congruent, m/DML 5 m/FLN 5 m/ENM and
m/DLM 5 m/FNL 5 m/EMN. Thus, by the
substitution postulate,
Therefore, by the subtraction postulate, we have
that m/MLN 5 m/LNM 5 m/LMN or
/MLN > /LNM > /LMN, and so nNLM is
equiangular.
m/DLM 1 m/MLN 1 m/FLN
5 m/DLM 1 m/LNM 1 m/FLN
5 m/DLM 1 m/LMN 1 m/FLN
Chapter 5. Congruence Based on Triangles
5-1 Line Segments Associated with
Triangles (pages 177–178)
3. RS is a median.
4. S is the midpoint
of PQ.
5. PS > QS
6. nPSR > nQSR
Writing About Mathematics
1. In nABC, let /B be a right angle. The altitude
drawn to vertex B is DB. The altitudes drawn
to vertices A and C are the sides of the triangle,
BA and BC, and therefore intersect with
vertex B.
2. The altitudes intersect outside of the triangle.
8.
Developing Skills
3. a. Answers will vary.
b. /ACE > /BCE
c. AF > BF
d. /ADC and /BDC are right angles.
4. a. Answers will vary.
b. In acute triangles, the altitudes intersect inside
of the triangle. In obtuse triangles, the
altitudes intersect outside of the triangle. In
right triangles, the altitudes intersect at the
vertex of the right angle.
5. a. Answers will vary.
b. In all triangles, the angle bisectors intersect
inside of the triangle.
6. a. Answers will vary.
b. In all triangles, the medians intersect inside of
the triangle.
7.
E
D
9.
S
Statements
1. PR > QR
2. /P > /Q
G
F
Statements
Reasons
1. EG is an angle
1. Given.
bisector of nDEF.
2. EG bisects /DEF. 2. Definition of angle
bisector of a triangle.
3. /DEG > /FEG
3. Definition of angle
bisector.
4. EG > EG
4. Reflexive property.
5. EG is an altitude.
5. Given.
6. EG ' DF
6. Definition of an
altitude of a triangle.
7. /DGE > /FGE
7. If two lines are
perpendicular, then
they intersect to form
congruent adjacent
angles.
8. nDEF > nFEG
8. ASA (steps 3, 4, 7).
R
P
3. Given.
4. Definition of
median.
5. Definition of midpoint.
6. SAS (steps 1, 2, 5).
C
Q
Reasons
1. Given.
2. Given.
A
D
B
(Cont.)
256
Statements
1. CD is an altitude
of nABC.
2. CD ' AB
3. /ADC > /BDC
4. CD > CD
5. CD is a median
of nABC.
6. D is the midpoint
of AB.
7. AD > BD
12. Area (nAMC) 5 21bh 5 12AM(DC)
Reasons
1. Assumption.
Area (nBMC) 5 21bh 5 12MB(DC)
Since M is the midpoint of AB, AM 5 MB. By the
substitution postulate, 12AM(DC) 5 12MB(DC) .
Therefore, the median divides nABC into two
triangles with equal areas.
2. Definition of an
altitude of a triangle.
3. If two lines are
perpendicular, then
they intersect to form
congruent adjacent
angles.
4. Reflexive property.
5. Given.
13. The farmer can determine the midpoint of
one side of the piece of land and then build a
fence from that point to the opposite vertex of
the plot.
5-2 Using Congruent Triangles to Prove
Line Segments Congruent and Angles
Congruent (pages 179–181)
6. Definition of a
median.
7. Definition of a
midpoint.
8. SAS (steps 4, 3, 7).
9. Given.
Writing About Mathematics
1. Yes. Since nABE > nDEF, /A > /D and
/B > /E. Congruent angles have equal
measures, so if m/A 1 m/B 5 90, then
m/D 1 m/E 5 90 and /D and /E are also
complementary.
2. Yes. The legs of an isosceles triangle are
congruent. Therefore, since the leg of one
isosceles triangle is congruent to the leg of
another isosceles triangle, then the other legs are
also congruent. As the vertex angles are
congruent, the triangles are congruent by SAS.
Developing Skills
3. a. nRPM > nSPM
b. SAS
c. /RPM > /SPM, /R > /S, RP > SP
4. a. nABD > nCDB
b. SSS
c. /ABD > /CDB, /BDA > /DBC,
/A > /C
5. a. nABE > nCDE
b. SAS
c. /A > /C, /B > /D, AB > CD
6. a. nABE > nCDE
b. ASA
c. /B > /D, AB > CD, BE > DE
7. a. nPQR > nRSP
b. SAS
c. /QRP > /SPR, /RPQ > /PRS, PR > RP
8. a. nABE > nCDE
b. SSS
c. /A > /C, /B > /D, /BEA > /DEC
8. nADC > nBDC
9. nADC is not
congruent to
nBDC.
10. CD is not an
10. Contradiction
altitude of nABC.
(steps 8, 9).
Applying Skills
10. Assume that NO is an angle bisector of nLMO.
Then NO bisects /LNM and /LNO > /MNO.
NO > NO by the reflexive property of
congruence. It is given that NO is an altitude of
nLNM, so by definition, NO ' LM. Therefore,
/LON and /MON are right angles and
congruent. nLON > nMON by SAS.
LN > MN since corresponding parts of
congruent triangles are congruent. However,
nLMN is scalene, so this is a contradiction.
Therefore, the assumption is false and NO is not
an angle bisector.
11. Let the telephone pole and the two wires form a
pair of triangles with the ground, nACD and
nBCD. Side CD, the pole, is congruent to itself.
If the pole is perpendicular to the ground, then
CD ' AB. This forms two congruent right
angles, /ADC and /BDC. Since D is the
midpoint of AB, AD > BD. Therefore, nACD >
nBCD by SAS. The wires, AC and BC, are
corresponding parts of congruent triangles, so are
congruent and have equal length.
257