Download Chapter 9 Answers

14. Let e 5 “Evanston is the capital of Illinois.”
Let c 5 “Chicago is the capital of Illinois.”
Let s 5 “Springfield is the capital of Illinois.”
Then in symbols, the given statements are:
Substitute x 5 1 into either equation to find y:
y 5 3x 2 1
y 5 3(1) 2 1
y52
,e → ,c
Therefore, the coordinates of the intersection
point are (1, 2).
16. Statements
Reasons
1. CGF and DGE bisect 1. Given.
each other at G.
2. G is the midpoint of
2. Definition of
bisector.
CGF and of DGE.
3. CG > GF and
3. Definition of
midpoint.
DG > GE
4. /CGD > /EGF
4. Vertical angles are
congruent.
5. nCGD > nEGF
5. SAS.
6. CD > EF
6. Corresponding
sides of congruent
triangles are
congruent.
s∨c
,e
Since ,e is true, ,c is true by the Law of
Detachment.
Since ,c is true, c is false.
Since c is false and s ∨ c is true, s is true by the
Law of Disjunctive Inference.
Therefore, Springfield is the capital of Illinois.
Part IV
15. Substitute the first equation for y in the second
equation, and solve for x:
x 1 2(3x 2 1) 5 5
x 1 6x 2 2 5 5
7x 5 7
x51
Chapter 9. Parallel Lines
9-1 Proving Lines Parallel (page 334)
8. Vertical angles (/7 and /5) are congruent. If
two coplanar lines are cut by a transversal so that
the interior angles on the same side of the
transversal are supplementary, then the lines are
parallel.
Applying Skills
Writing About Mathematics
1. Corresponding angles
2. Yes. This is the contrapositive of Theorem 9.1a: If
two coplanar lines cut by a transversal are not
parallel, then the alternate interior angles formed
are not congruent.
Developing Skills
3. If two coplanar lines are cut by a transversal so
that the alternate interior angles formed are
congruent, then the two lines are parallel.
4. If two coplanar lines are cut by a transversal so
that the corresponding angles are congruent,
then the two lines are parallel.
5. If two coplanar lines are cut by a transversal so
that the interior angles on the same side of the
transversal are supplementary, then the lines are
parallel.
6. Vertical angles (/2 and /4) are congruent. If two
coplanar lines are cut by a transversal so that the
interior angles on the same side of the transversal
are supplementary, then the lines are parallel.
7. Vertical angles (/2 and /4) are congruent. If
two coplanar lines are cut by a transversal so that
the corresponding angles are congruent, then the
lines are parallel.
g
g
g
9. Given: EF intersects AB and CD; /1 > /5.
g
g
Prove: AB 7CD
Statements
g
1. AB is not parallel
Reasons
1. Assumption.
g
to CD.
2. /1 > /3
3. /3 is not congruent
to /5.
4. /1 is not congruent
to /5.
5. /1 > /5
g
g
6. AB 7 CD
291
2. Vertical angles are
congruent.
3. If two coplanar
lines cut by a
transversal are not
parallel, then the
alternate interior
angles formed are
not congruent.
4. Transitive property.
5. Given.
6. Contradiction.
g
g
g
g
g
10. Given: AB ' EF and CD ' EF
g
Prove: AB 7 CD
Statements
g
g
1. AB ' EF and
g
g
g
g
4. AB y CD
Statements
1. Given.
g
5. nCEA > nDEB
Reasons
g
1. EF intersects AB
1. Given.
g
and CD;
/1 > /2.
2. /1 > /3
2. Definition of
perpendicular lines.
3. Right angles are
congruent.
3. /2 > /3
4. If two coplanar
lines are cut by a
transversal so that
the alternate interior
angles are congruent,
then the two lines are
parallel.
g
c. Statements
1. /ECA > /EDB
2. CA y DB
Reasons
1. Part b.
2. If two coplanar lines
are cut by a transversal so that the
alternate interior
angles formed are
congruent, then the
two lines are
parallel.
4. If two coplanar lines
are cut by a transversal
so that the
corresponding angles
are congruent, then the
two lines are parallel.
Writing About Mathematics
1. a. Yes. The inverse of Theorem 9.1a states: If two
coplanar lines are cut by a transversal so that
the alternate interior angles formed are not
congruent, then the lines are not parallel. The
contrapositive of this statement, which has the
same truth value, is: If two coplanar lines cut
by a transversal are parallel, then the alternate
interior angles formed are congruent. This is
the same as Theorem 9.1b.
b. Yes. The inverse of Theorem 9.6 states: If a
transversal is not perpendicular to one of two
parallel lines, then it is not perpendicular to
the other. Alternate interior angles of parallel
lines are congruent, so if the transversal does
not form a right angle with one of the parallel
lines, it will not form a right angle with the
other.
2. The measures of the angles formed by the
parallel lines and the transversal are all 90
degrees. Alternate interior angles of parallel lines
are congruent, so their measures are equal. If the
angles are also supplementary, then they must
both be right angles.
2. Definition of
bisector.
3. Definition of
midpoint.
4. Vertical angles are
congruent.
5. SAS.
Reasons
1. Part a.
2. Corresponding parts
of congruent angles
are congruent.
2. Vertical angles are
congruent.
3. Transitive property.
9-2 Properties of Parallel Lines
(pages 340–341)
Reasons
1. Given.
b. Statements
1. nCEA > nDEB
2. /ECA > /EDB
g
4. AB y CD
11. m/A 1 m/B 5 3x 1 (180 2 3x) 5 180. Since
interior angles on the same side of the
transversal AB are supplementary, AD y BC.
12. We are given that DC ' BC, so /BCD is a right
angle and m/BCD 5 90. Therefore,
m/BCD 1 m/ADC 5 180. Since interior same
angles on the same side of the transversal CD are
supplementary, AD y BC.
13. a. Statements
1. AB and CD bisect
each other at E.
2. E is the midpoint
of AB and of CD.
3. AE > EB,
CE > ED
4. /CEA > /DEB
g
g
Prove: AB y CD
Reasons
g
CD ' EF
2. /1 and /2 are right
angles.
3. /1 > /2
g
14. Given: EF intersects AB and CD; /1 > /2.
g
Developing Skills
3. 80
4. 150
6. 75
7. 115
9. 42
10. 80
12. 60
13. 44 and 136
292
5.
8.
11.
14.
120
50
65
21 and 21
20. Statements
15. a. m/1 5 70, m/2 5 110, m/3 5 70, m/4 5 50,
m/5 5 130, m/6 5 50, m/7 5 50, m/8 5 60,
m/9 5 70
b. 120
c. Yes
d. 120
e. 110
f. 130
g. 180
16. m/A 5 m/C 5 75; m/ABC 5 m/D 5 105
Applying Skills
17. Statements
Reasons
g
g
1. AB y CD and
Reasons
h
1. CE bisects exterior 1. Given.
/BCD
2. /BCE > /DCE
2. Definition of angle
bisector.
h
3. CE y AB
4. /DCE > /A
1. Given.
g
transversal EF
2. /3 > /5
3. /4 is the
supplement of /3.
4. /4 is the
supplement of /5.
18. Statements
g
g
5. /BCE > /B
6. /A > /B
21. a. Statements
1. /CAB > /DCA
g
2. AB y DCE
Reasons
g
g
1. AB y CD, EF ' AB
2. /AEF > /EFD
3. /AEF is a right
angle.
4. /EFD is a right
angle.
g
2. If two parallel lines
are cut by a
transversal, then the
alternate interior
angles formed are
congruent.
3. If two angles form
a linear pair, then they
are supplementary.
4. If two angles are
congruent, then their
supplements are
congruent.
g
5. EF ' CD
1. Given.
2. If two parallel lines are
cut by a transversal,
then the alternate
interior angles formed
are congruent.
3. Perpendicular lines
intersect to form right
angles.
4. Definition of
congruent angles.
b. Statements
g
1. AB y DCE
2. /CAB > /DCA,
/DCA > /ECB
3. /CAB > /ECB
4. /ECB > /ABC
5. Perpendicular lines
intersect to form right
angles.
19. When two parallel lines are cut by a transversal,
alternate interior angles are congruent. These
angles are also congruent to their respective
vertical angles. Since these vertical angles are
alternate exterior angles, alternate exterior
angles are congruent.
5.
6.
7.
8.
293
3. Given.
4. If two parallel lines
are cut by a transversal, then corresponding angles are
congruent.
5. If two parallel lines are
cut by a transversal,
then the alternate
interior angles formed
are congruent.
6. Substitution postulate.
Reasons
1. Given.
2. If two coplanar
lines are cut by a
transversal so that
the alternate interior angles formed
are congruent, then
the two lines are
parallel.
Reasons
1. Part a.
2. Given.
3. Transitive property.
4. If two parallel lines
are cut by a transversal, then the
alternate interior
angles formed are
congruent.
/CAB > /ABC
5. Transitive property.
m/ABC 1
6. Angles forming a
m/CBG 5 180
linear pair are
supplementary.
m/CAB 1
7. Substitution
m/CBG 5 180
postulate.
/CAB is the sup- 8. Definition of supplement of /CBG.
plementary angles.
22. If two parallel lines are cut by a transversal, then
the two interior angles on the same side of the
transversal are supplementary. Therefore, since
PQ y RS and /P is a right angle, its supplement,
/S, is also a right angle. Similarly, since QR y SP
and /P and /S are right angles, their respective
supplements, /Q and /R, are also right angles.
16. a. Slope of KL 5 213 , slope of MN 5 213
Slope of LM 5 22, slope of NK 5 3
Only one pair of sides has equal slopes.
Therefore, KLMN has only one pair of
parallel sides.
b. The slopes of the adjacent sides, 231 and 3, are
negative reciprocals, so the respective sides
are perpendicular and form right angles.
Therefore, /K and /N are right angles.
Hands-On Activity 1
a. If two coplanar lines are each perpendicular to
the same line, then they are parallel.
b. (1) y 5 41x 1 19
4
(2) y 5 12x 2 148
(3) y 5 291x 1 4
Hands-On Activity 2
1. Answers will vary.
a. A(a, b); B(c, d); C(e, f)
23. If two parallel lines are cut by a transversal, then
the two interior angles on the same side of the
transversal are supplementary. Therefore, since
KL y MN and /K is an acute angle measuring
less than 90°, its supplement, /N, must measure
more than 90° and be obtuse. Similarly, since
LM y NK and /K is acute, its supplement, /L,
must be obtuse. Since /N is an obtuse angle
measuring more than 90°, its supplement, /M,
must measure less than 90° and be acute.
9-3 Parallel Lines in the Coordinate Plane
(pages 345–347)
e b1f
b. Midpoint of AC 5 D A a 1
c , 2 B
e d1f
Midpoint of BC 5 E A c 1
2 , 2 B
Writing About Mathematics
1. They are the same line.
2. Vertical lines are parallel, but they do not have
the same slope since vertical lines do not have
slope.
Developing Skills
3. Perpendicular
4. Parallel
5. Parallel
6. Neither parallel nor perpendicular
7. Parallel
8. Perpendicular
9. y 5 23x 1 4
10. y 5 31x 1 4
11. y 5 12x 1 3
Applying Skills
2b
c. Slope of AB 5 dc 2
a
212
3
(2, 5)
1
b. 221
e. 3
c.
d.
e.
15. a.
21; since BC ' AB and DA ' AB, BC y DA.
1; since DA ' AB and DA ' DC, AB y DC.
y5x12
f. y 5 2x
g. (21, 1)
Slope of PQ 5 14 , slope of RS 5 14
b1f
2
2
The slope of the midsegment is equal to the
slope of the third side of the triangle.
Therefore, they are parallel.
2. Results will vary.
9-4 The Sum of the Measures of the
Angles of a Triangle (pages 351–352)
12. y 5 23
13. a.
d.
g.
14. a.
d1f
2 2 2
2b
Slope of DE 5 c 1
5 dc 2
a
e
2a1e
Writing About Mathematics
1. Yes. The sum of the angles of a triangle is 180
degrees. Therefore, if one angle has a measure
greater than 90 degrees, then the sum of the
other two angles must be less than 90 degrees,
and they must both be acute.
2. No. By definition, exactly two angles can be
supplementary. Since a triangle has three angles,
and no angle of a triangle can measure 0 degrees,
the sum of the measure of any two angles must
be less than 180 degrees.
Developing Skills
3. Yes
4. No
5. No
6. Yes
7. 40
8. 50
9. 54
10. 50
11. 80
12. 45
13. 52
14. 35
c. y 5 221x 1 6
f. y 5 3x 2 1
b. 21
Slope of QR 5 232 , slope of SP 5 232
Therefore, PQ y RS and QR y SP.
b. The slopes of the adjacent sides, 14 and 232 , are
not negative reciprocals. Therefore, no sides
are perpendicular and PQRS does not have a
right angle.
294
15. 20
16. 140
17. 90
18. 54
19. 120
20. m/ACD 5 60; m/ACB 5 120
21. m/ACD 5 90; m/ACB 5 90
22. m/ACD 5 60; m/B 5 20
23. m/B 5 95; m/ACD 5 135; m/ACB 5 45
Applying Skills
24. 46°, 67°, 67° 25. 70°, 55°, 55° 26. 20°, 80°, 80°
27. m/N 5 58; measure of exterior angle 5 122
28. a. m/A 5 72; m/B 5 67; m/C 5 41
b. BC
29. m/A 5 m/C 5 30; m/B 5 120; measure of
exterior angle 5 60
30. Let nABC be a triangle with /BCA the exterior
angle at C. Then m/A 1 m/B 1 m/C 5 180
since the sum of the measures of the angles of a
triangle is 180. Also, /BCA and /C form a linear
pair, so m/BCA 1 m/C 5 180. By the
substitution postulate, m/A 1 m/B 1 m/C 5
m/BCA 1 m/C. By the subtraction postulate,
m/A 1 m/B 5 m/BCA, or the measure of an
exterior angle of a triangle is equal to the sum of
the measures of the nonadjacent interior angles.
9-5 Proving Triangles Congruent by Angle,
Angle, Side (pages 356–357)
Writing About Mathematics
1. No, it simply means that SSA is insufficient to
prove the triangles congruent. The triangles may
or may not be congruent.
2. Yes. AC and BC are perpendicular, so /C is a
right angle and measures 90°. Since the sum of
the measures of the angles of a triangle is 180°,
the measures of /CAB and /CBA must be 90°,
which makes them complementary.
Developing Skills
3. Yes, by AAS.
4. Yes, by AAS.
5. No. AAA is insufficient to prove congruence.
6. Yes, by AAS.
7. No. SSA is insufficient to prove congruence.
8. Yes, by SAS.
Applying Skills
9. Let nABC > nDEF, so /A > /D and
AC > DF. Draw CG, the altitude from /A in
nABC, and FH, the altitude from /F in nDEF,
to form right triangles nABG and nDFH. Two
right triangles are congruent if the hypotenuse
and an acute angle of one right triangle are
congruent to the hypotenuse and an acute angle
of the other right triangle, so nACG > nDFH.
Corresponding parts of congruent triangles are
congruent, so the altitudes, CG and FH, are
congruent.
31. a. Graph
b. Graph
c. BC is a vertical line, so BC 5 |21 2 2| 5 3. CD
is a horizontal line, so CD 5 |21 2 2| 5 3.
Since BC 5 CD, BC > CD and nBDC is
isosceles. Horizontal and vertical lines are
perpendicular, so /BCD is a right angle.
d. m/BDC 5 45; each acute angle of an
isosceles right triangle measures 45°.
10. Let nABC > nDEF, so /A > /D, AC > DF,
and AB > DE. Draw CG, the median from /A
in nABC, and FH, the median from /F in
nDEF, to form triangles nABG and nDFH.
Since G is the midpoint of AB and H is the
midpoint of DE, AG 5 21AB and DH 5 12DE.
Halves of congruent segments are congruent, so
AG > DH, and nACG > nDFH by SAS.
Corresponding parts of congruent triangles are
congruent, so the altitudes, CG and FH, are
congruent.
g
e. AB is a horizontal line and, therefore,
perpendicular to BC . By the partition
postulate, m/DBA 5 m/DBC 1 m/ABC.
Since m/DBC 5 45 and m/ABC 5 90,
m/DBA 5 135.
32. In hexagon ABCDEF, draw AD . This divides
the hexagon into two quadrilaterals, ABCD
and ADEF. The sum of the measures of the
angles in each quadrilateral is 360°. Therefore,
the sum of the measures of the angles in
ABCDEF 5 360 1 360 5 720°.
h
11. Let nABC > nDEF, so /A > /D, AC > DF,
and /C > /F. Draw CG, the bisector of /A in
nABC, and FH, the bisector of /F in nDEF, to
form triangles nABG and nDFH. By the
definition of angle bisector, m/ACG 5 21m/C
h
33. BD bisects /ABC, so /ABD > /CBD. DB
bisects /ADC, so /ADB > /CDB. BD > BD
by the reflexive property of congruence, so
nABD > nCBD by SAS. Corresponding parts
of congruent triangles are congruent, so
/A > /C.
and m/DFH = 12m/F. Halves of congruent
angles are congruent, so /ACG > /DFH, and
295
nACG > nDFH by AAS. Corresponding parts
of congruent triangles are congruent, so the
altitudes, CG and FH, are congruent.
/A > /D. Since right angles are congruent,
/B > /E. Therefore, nABC > nDEF by AAS.
h
12. Statements
16. Let P be a point on the bisector BP of /ABC.
By the definition of angle bisector, /ABP >
/CBP. By the reflexive property of congruence,
Reasons
h
1. /A > /C and BD is 1. Given.
the bisector of /ABC.
2. /ABD > /CBD
2. Definition of angle
bisector.
3. BD > BD
3. Reflexive property.
4. nABD > nCBD
4. AAS.
5. /ADB > /CDB
5. Corresponding parts
of congruent triangles are congruent.
h
6. DB bisects /ADC.
13. Statements
1. AB y CD, AB > CD,
and AB ' BEC.
2. CD ' BEC
h
h
BP > BP. Draw PA ' BA and PC ' BC . Then
/BAP and /BCP are right angles and
congruent. Therefore, nABP > nCBP by AAS.
Corresponding parts of congruent triangles are
congruent, so PA > PC and PA 5 PC. By defih
h
nition, P is equidistant from BA and from BC .
17. Through a point E on side BC, draw DE parallel
to AC. If two parallel lines are cut by a
transversal, then the corresponding angles are
congruent. Therefore, /A > /BED and
/C > /BDE. By the reflexive property of
congruence, /B > /B. If AAA were sufficient
to establish congruence, then nEBD would be
congruent to nABC. But AB . DB and
CB . DB, so nEBD is not congruent to nABC.
6. Definition of angle
bisector.
Reasons
1. Given.
2. If a transversal is
perpendicular to one
of two parallel lines,
then it is perpendicular to the other.
3. /B and /C are right 3. Definition of
angles.
perpendicular lines.
4. /B > /C
4. Right angles are
congruent.
5. /AEB > /DEC
5. Vertical angles are
congruent.
6. nABE > nDCE
6. AAS.
7. BE > CE, AE > DE 7. Corresponding parts
of congruent angles
are congruent.
8. AED and BEC bisect 8. Definition of
each other.
bisector.
9-6 The Converse of the Isosceles Triangle
Theorem (pages 360–362)
Writing About Mathematics
1. Yes. The contrapositive of Theorem 7.3, which is
also true, states: If the measures of the angles
opposite two sides of a triangle are equal, then
the lengths of the sides opposite these angles are
equal.
2. Yes. The sum of the measures of the angles of a
triangle is 180°, so if two angles measure 45° and
90°, then the third must measure 45°. Since two
angles of the triangle have equal measures, they
are congruent, so the sides opposite these angles
must be congruent. By definition, the right
triangle is isosceles.
Developing Skills
3. a. 70
b. Isosceles
4. a. 30
b. Isosceles
5. a. 65
b. Isosceles
6. a. 60
b. Not isosceles
7. x 5 4
8. PR 5 RQ 5 21
9. m/R 5 72; m/N 5 36
10. AB 5 AC 5 36; BC 5 46
11. x 1 10 1 2x 1 2x 2 30 5 180, so x 5 40. The
three angles then measure 50°, 80°, and 50°. Since
two of the angles of the triangle measure 50°, the
triangle has two congruent angles and, therefore,
14. a. Under the translation T9,0, A(26, 0) → D(3, 0),
B(21, 0) → E(8, 0), and C(25, 2) → F(4, 2).
Distance is preserved under translation, so
nABC > nDEF by SSS.
b. Answers will vary. Under rx=1, A(26, 0) → (8, 0),
B(21, 0) → (3, 0), and C(25, 2) → (7, 2).
Under rx=5.5, (8, 0) → D(3, 0), (3, 0) → E(8, 0),
and (7, 2) → F(4, 2).
Distance is preserved under reflection, so
nABC > nDEF by SSS.
15. Let nABC and nDEF be right triangles with /B
and /E the right angles, AC > DF, and
296
two congruent sides. By definition, the triangle is
isosceles.
12. x 1 35 1 2x 2 10 1 3x 2 15 5 180, so x 5 25. The
three angles all measure 60°. Therefore, all of the
angles are congruent and the triangle is
equiangular. If a triangle is equiangular, then it is
equilateral.
13. 3x 1 18 1 4x 1 9 1 10x 5 180, so x 5 9. The
measures of the three angles are 45°, 45°, and 90°.
Since the triangle has a right angle, it is a right
triangle. Since two of the angles of the triangle
measure 40°, the triangle has two congruent
angles and, therefore, two congruent sides. By
definition, the triangle is isosceles.
14. 120
15. 360
Applying Skills
16. Statements
Reasons
1. /ABP > /PCD
1. Given.
2. /ABP and /PBC are 2. If two angles form a
supplements. /PCD
linear pair, then
and /PCB are
they are supplesupplements.
mentary.
3. /PBC > /PCB
3. If two angles are
congruent, then
their supplements
are congruent.
4. PB > PC
4. If two angles of a
triangle are congruent, then the sides
opposite these angles are congruent.
5. nBPC is isosceles.
5. Definition of
isosceles triangle.
18. Statements
h
1. BE bisects /DBC.
2. m/DBC 5 2m/DBE
h
3. BE y AC
4. /A > /DBE
5. m/A 5 m/DBE
6. m/DBC
5 m/A 1 m/C
7. 2m/A
5 m/A 1 m/C
8. m/A 5 m/C
9. /A > /C
10. AB > CB
19. Statements
1. /PBC > /PCB,
/APB > /DPC
2. PB > PC, AB > CD
Reasons
1. Given.
2. Definition of angle
bisector.
3. Given.
4. If two parallel lines
are cut by a transversal, then the
corresponding angles are congruent.
5. Definition of
congruent angles.
6. Exterior angle
theorem.
7. Substitution
postulate.
8. Subtraction
postulate.
9. Definition of
congruent angles.
10. If two angles of a
triangle are
congruent, then the
sides opposite
these angles are
congruent.
Reasons
1. Given.
2. If two angles of a
triangle are congruent, then the sides
opposite these angles are congruent.
3. /ABP and /PBC
3. If two angles form a
are supplements.
linear pair, then
/DCP and /PCB are
they are supplesupplements.
mentary.
4. /ABP > /DCP
4. If two angles are
congruent, then
their supplements
are congruent.
5. nABP > nDCP
5. AAS.
6. AP > DP
6. Corresponding parts
of congruent triangles are congruent.
17. Statements
1. /PAB > /PBA
2. PA > PB
Reasons
1. Given.
2. If two angles of a
triangle are congruent, then the sides
opposite these angles are congruent.
3. PA 5 PB
3. Definition of congruent segments.
4. P is on the perpen4. A point is on the
dicular bisector of AB.
perpendicular
bisector of a line
segment if and only
if it is equidistant
from the endpoints
of the line segment.
20. In nABC, /A > /B. Draw CD, the altitude
from /C. By definition, CD ' AB, so /ADC
and /BDC are right angles and congruent.
297
h
CD > CD by the reflexive property of
congruence. nACD > nBCD by AAS.
Therefore, CA > CB because corresponding
parts of congruent triangles are congruent.
e. Yes. Extend CP through P to a point M.
Since angles forming a linear pair are supplementary and m/APC 5 m/BPC 5 100, we
have that m/APM 5 m/BPM 5 80. Thus,
h
/APM > /BPM and CP bisects /APB.
6. m/TRS 5 70; m/RST 5 60; m/SPT 5 125
7. a–c. Graph
d. No. Since /A is not congruent to /B,
/PAB and /PBA (halves of these angles)
are also not congruent. If the measures of
two angles of a triangle are unequal, the
lengths of the sides opposite these angles
are unequal. Therefore, in nAPB, since
m/PAB Þ m/PBA, AP Þ BP. Similarly,
AP Þ CP and BP Þ CP.
e. Equilateral
9-7 Proving Right Triangles Congruent by
Hypotenuse, Leg (pages 365–367)
Writing About Mathematics
1. In two right triangles, the hypotenuse and one
leg, both legs, or one leg and its adjacent acute
angle must be congruent for the triangles to be
proved congruent.
2. PD , PQ. The distance from a point to a line is
the length of the perpendicular from the point to
the line. Therefore, /PDQ is a right angle and
PD can be considered a leg of right triangle.
Then, PQ is the hypotenuse of the triangle, and
the hypotenuse is longer than either leg.
Developing Skills
3. a. m/BCA 5 80
b. m/PAN 5 20; m/PBN 5 30; m/APB 5 130
c. m/PCL 5 40; m/PBL 5 30; m/BPC 5 110
d. m/PAM 5 20; m/PCM 5 40;
m/APC 5 120
e. No. /APB and /BPL form a linear pair so
are supplementary. Since m/APB 5 130,
m/BPL 5 50. Similarly, /APC and /CPL
form a linear pair so are supplementary.
Since m/APC 5 120, m/CPL 5 60.
m/BPL Þ m/CPL, so they are not congruent.
Applying Skills
8. Statements
h
1. PD ' BA at D,
h
PF ' BC at F.
2. /PDB and /PFB
are right angles.
3. PD 5 PF
4. PD > PF
5. BP > BP
6. nABP > nPFB
7. /ABP > /CBP
h
4. a.
b.
c.
d.
Therefore, AL does not bisect /CPB.
m/A 5 45; m/B 5 45
m/PAB 5 22.5; m/PBA 5 22.5;
m/APB 5 135
m/PBC 5 22.5; m/PCB 5 45;
m/BPC 5 112.5
m/PAC 5 22.5; m/PCA 5 45;
m/APC 5 112.5
2. Definition of
perpendicular lines.
3. Given.
4. Definition of congruent segments.
5. Reflexive property.
6. HL.
7. Corresponding parts
of congruent triangles
are congruent.
9. When nADC is an isosceles right triangle,
AD > DC and /A and /C are congruent
complementary angles. Since DB ⊥ AC,
/BDC and /C are also complementary. If two
angles are complements of the same angle,
then they are congruent. Thus, /BDC >
/A > /C, and so DB > BC by the converse
of the Isosceles Triangle Theorem. By HL,
we can conclude that nABD > nDBC. Thus,
nABD > nDBC when nADC is an isosceles
right triangle.
10. If AL and BM are parallel, then /PAB and
/PBA are supplementary. By the partition
postulate, m/A 5 m/CAP 1 m/PAB and
m/B 5 m/CBP 1 m/PBA. Since a whole is
greater than any of its parts, m/PAB , m/A
and m/PBA , m/B. But in nABC, the sum of
h
e. Yes. Extend CP through P to a point M.
Since angles forming a linear pair are supplementary and m/APC 5 m/BPC 5 112.5, we
have that m/APM 5 m/BPM 5 67.5. Thus,
h
5. a.
b.
c.
d.
Reasons
1. Given.
/APM > /BPM and CP bisects /APB.
m/A 5 20; m/B 5 20
m/PAB 5 10; m/PBA 5 10; m/APB 5 160
m/PBC 5 10; m/PCB 5 70; m/BPC 5 100
m/PAC 5 10; m/PCA 5 70; m/APC 5 100
298
b. Statements
1. nQPR > nPSR
2. QP > SP
the measures of /A, /B, and /C is 180°. Thus,
m/A 1 m/B , 180, and so:
m/PAB1m/PBA , m/A1m/B , 180
Since /PAB and /PBA are supplementary,
m/PAB 1 m/PBA 5 180, and we arrive at
180 , m/A1m/B , 180, a contradiction. Thus,
the assumption is false, and AL and BM are not
parallel.
11. Statements
1. AB ' BD, BD ' DC
2. /ABD and /CDB
are right angles.
3. AD > CB
4. BD > BD
5. nABD > nCDB
6. /A > /C,
/ADB > /CBD
7. AD y CB
12. a. Statements
1. In nQRS, the
bisector of /QRS
is perpendicular to
QS at P.
2. /QRP > /SRP
3. RP > RP
4. /QPR and /PSR
are right angles.
5. /QPR > /PSR
6. nQPR > nPSR
7. QR > SR
8. nQRS is isosceles.
3. P is the midpoint
of QS.
Reasons
1. Part a.
2. Corresponding parts
of congruent triangles are congruent.
3. Definition of
midpoint.
13. Given: Isosceles triangle nABC with vertex /B,
D the midpoint of AC, DE ' BC, and
DF ' AB.
Prove: DE > DF
Proof: By the isosceles triangle theorem, since
AB > BC in nABC, /A > /C. Since D is the
midpoint of AC, AD > CD. Since DE ' BC and
DF ' AB, /CED and /AFD are right angles
and congruent. Therefore, nCED > nAFD by
AAS. Then DE > DF because corresponding
parts of congruent triangles are congruent.
Reasons
1. Given.
2. Definition of
perpendicular lines.
3. Given.
4. Reflexive property.
5. HL.
6. Corresponding
parts of congruent
triangles are
congruent.
7. If two coplanar
lines are cut by a
transversal so that
the alternate interior angles formed
are congruent, the
two lines are
parallel.
14. a. In ABCD, /A and /C are right angles, so /A
> /C. Since AB 5 CD, AB > CD. By the
reflexive property of congruence, BD > BD,
so nABD > nCDB by HL. Since
corresponding parts of congruent triangles
are congruent, AD > BC and AD 5 BC.
b. From part a, nABD > nCDB. Corresponding
parts of congruent triangles are congruent, so
/ABD > /CDB.
Reasons
1. Given.
c. From parts a and b, nABD > nCDB. Since
corresponding parts of congruent triangles
are congruent, /ADB > /CBD and
m/ADB 5 m/CBD. The acute angles of a
right triangle are complementary, so
m/CDB 1 m/CBD 5 90. By the substitution
postulate, m/CDB 1 m/ADB 5 90.
Since m/CDB 1 m/ADB 5 m/ADC,
m/ADC 5 90, and /ADC is a right angle.
2. Definition of angle
bisector.
3. Reflexive property.
4. Definition of
perpendicular lines.
5. Right angles are
congruent.
6. ASA.
7. Corresponding
parts of congruent
triangles are
congruent.
8. Definition of
isosceles triangle.
15. In ABCD, /ABC and /BCD are right angles
and therefore congruent. Since AC 5 BD,
AC > BD . By the reflexive property of
congruence, BC > BC , and nABC > nDCB by
HL. Corresponding parts of congruent triangles
are congruent, so AB > CD and AB 5 CD.
299
›
›
16. Since PB ' ABC and PD ' ADE , /ABP and
/ADP are right angles and congruent. We are
b. Since AB > BC in nABC and AE > ED in
nABD, the triangles are isosceles.
given that PB 5 PD, so PB > PD. AP > AP by
the reflexive property of congruence. Therefore,
nABP > nADP by HL. Corresponding parts of
congruent triangles are congruent, so /BAP >
/DAP or /CAS > /EAS. By the definition of
›
angle bisector, APS bisects /CAE.
c. Since nABC > nAED, and corresponding
parts of congruent triangles are congruent,
AC > AD and nDAC is isosceles.
22. a. From /L in LNMRST, draw diagonals LN,
LR, and LS. Since LMNRST is regular, it is
both equilateral and equiangular. Therefore,
LM > MN > LT > TS, /M > /T, and
nLMN > nLTS by SAS.
9-8 Interior and Exterior Angles of
Polygons (pages 371–372)
b. nLMN > nLTS and corresponding parts
of congruent triangles are congruent, so
LN > LS. Because LMNRST is regular,
NR > RS. By the reflexive property of
congruence, LR > LR so nLNR > nLSR
by SSS.
Writing About Mathematics
1. Yes. A diagonal may be drawn from every vertex
of a polygon to every other vertex except for
itself and the two adjacent vertices. Therefore,
each vertex of an n-sided polygon is an endpoint
of (n 2 3) diagonals.
2. Yes. Each vertex is an endpoint of (n – 3)
diagonals, so n vertices are the endpoints of
n(n 2 3) diagonals. A diagonal has two
endpoints, so there are n2 (n 2 3) diagonals.
Developing Skills
3. a. 180
b. 900
c. 1,260
4. a. 360
b. 1,080
c. 540
d. 1,440
5. a. 360
b. 360
c. 360
d. 360
6. a. 90
b. 90
7. a. 72
b. 108
8. a. 60
b. 120
9. a. 45
b. 135
10. a. 40
b. 140
11. a. 30
b. 150
12. a. 18
b. 162
13. a. 10
b. 170
14. a. 847
b. 17137
15. a. 12
b. 8
c. 6
d. 3
16. a. 4
b. 6
c. 9
d. 18
17. a. 3
b. 4
c. 5
d. 7
e. 10
f. 17
g. 12
h. 22
Applying Skills
18. 8 sides
19. 9 sides
20. No. If two angles were greater than 180 degrees,
then the sum of the interior angles would already
exceed 360 degrees, which is the sum of the
measures of the angles of a quadrilateral.
21. a. From /A in ABCDE, draw diagonals AC and
AD to form nABC and nAED. Since
ABCDE is regular, it is both equilateral
and equiangular. Therefore,
AB > BC > AE > ED, /B > /E, and
nABC > nAED by SAS.
c. m/M 5 m/T 5 120;
m/MLN 5 m/MNL 5 m/TLS
5 m/TSL 5 m/NLR 5 m/SLR 5 30;
m/LSR 5 m/LNR 5 90;
m/LRS 5 m/LRN 5 60
23. a. Slope of AB 5 slope of CD 5 21
Slope of BC 5 slope of DA 5 1
The slopes of adjacent sides are negative
reciprocals, therefore AB ' BC, BC ' CD,
CD ' DA, and /A, /B, /C, and /D are
right angles.
b. The x-axis is horizontal and the y-axis is
vertical, so they intersect to form right
angles, which are congruent, so
/AOB > /BOC > /COD > /DOA.
Also, AO 5 CO 5 BO 5 DO 5 2, so
AO > CO > BO > DO and
nAOB > nBOC > nCOD > nDOA
by SAS.
c. Quadrilateral ABCD is equiangular because
it contains four right angles. It is equilateral
because each of its sides is the hypotenuse of
a congruent right triangle. Therefore, it is
regular.
Hands-On Activity
a. No
b. Yes
c. The angle bisectors of a polygon are concurrent
if and only if the polygon is regular.
Review Exercises (pages 375–376)
1. 15
5. a. 18
300
2. 30
b. 72
3. 24
4. 55
c. 72
d. 108
e. 72
6.
9.
12.
15.
18.
21.
30
7. 6
8. 36
42, 48, 90
10. 120
11. 80
40
13. 154
14. 60
28
16. 20, 80, 80
17. 90
30
19. 12
20. 1,260
In nABC, /C is a right angle so m/C 5 90.
Therefore, m/B , 90 since /B are
complementary so /B must be acute. If the
measures of two angles of a triangle are unequal,
the lengths of the sides opposite these angles are
unequal and the longer side lies opposite the
larger angle. Therefore, AB . AC.
Cumulative Review (pages 376–378)
Part I
1. 4
6. 3
Part II
2. 2
7. 1
3. 1
8. 1
4. 3
9. 1
5. 2
10. 2
11. AC; m/C 5 45 and m/B 5 90. If the measures
of two angles of a triangle are unequal, the
lengths of the sides opposite these angles are
unequal and the longer side lies opposite the
larger angle. Therefore, AC is the longest side.
12. Since a point on the perpendicular bisector of a
segment is equidistant from the endpoints of the
segment, AP 5 BP. Therefore, AB > BP and
nABP is isosceles.
Part III
22. Since AEB and CED bisect each other,
AE > EB and CE > ED. Vertical angles are
congruent, so /AEC > /BED. Therefore,
/AEC > /BED by SAS. Corresponding parts
of congruent triangles are congruent, so
/EAC > /EBD. If two coplanar lines are cut by
a transversal so that the alternate interior angles
formed are congruent, then the two lines are
parallel. Therefore, AC y BD.
13. Since ABCD is equilateral,
AB > BC > CD > DE. By the reflexive
property of congruence, AC > AC, so
nABC > nADC by SSS. Corresponding parts
of congruent triangles are congruent, so
/BAC > /DAC and /BCA > /DCA. By the
definition of angle bisector, AC bisects /DAB
and /DCB.
23. In nBPA, BP > CP, so by the isosceles triangle
theorem /PBC > /PCB. /PBA and /PBC
form a linear pair, so they are supplementary.
Also, /PCD and /PCB form a linear pair, so
they are supplementary. If two angles are
congruent, then their supplements are
congruent, so /PBA > /PCD. It is given that
/APB > /DPC, so nABP > nDCP by ASA.
Corresponding parts of congruent triangles are
congruent, so PA > PD.
g
g
14. We are given that AD y CB. If two parallel lines
are cut by a transversal, then the alternate
interior angles formed are congruent. Therefore,
/ECB > /EDA and m/ECB 5 m/EDA.
By the exterior angle theorem,
m/DEB 5 m/EBC 1 m/ECB.
By the substitution postulate,
24. In nBPC, /PBC > /PCB. If two angles of a
triangle are congruent, then the sides opposite
these angles are congruent, so PB > PC. It is
given that /PBC > /PCB. /PBA and /PBC
form a linear pair, so they are supplementary.
Also, /PCD and /PCB form a linear pair, so
they are supplementary. If two angles are
congruent, then their supplements are congruent,
so /PBA > /PCD. Since AB > DC, nABP >
nDCP by SAS. Corresponding parts of
congruent triangles are congruent, so PA > PD.
m/DEB 5 m/EBC 1 m/EDA.
Part IV
15.
3x 1 4x 1 8x 5 180
15x 5 180
x 5 12
3x 5 36, 4x 5 48, 8x 5 96
The smallest exterior angle is the supplement
of the largest interior angle, so its measure is
180 2 96 5 84.
16. The midpoint of AB 5 A 2121 7, 2221 6 B 5 (3, 2)
6 2 (22)
The slope of AB 5 7 2 (21) 5 1
The perpendicular bisector has a slope of 21 and
goes through (3, 2):
y 5 2x 1 b
2 5 23 1 b
b55
The equation of the perpendicular bisector is
y 5 2x 1 5.
25. No. The sum of the measures of the interior angles of a pentagon is 540 degrees, and angles A, B,
C, and D have measures which already sum to 540.
Since m/E Þ 0, Herbie’s pentagon is not possible.
Exploration (page 376)
Results will vary.
301
Chapter 10. Quadrilaterals
10-2 The Parallelogram (pages 383–385)
16. Statement
Reason
1. Parallelograms EBFD 1. Given.
and ABCD
2. ED > BF
2. Opposite sides of a
parallelogram are
congruent.
3. /E > /F,
3. Opposite angles of
/BAD > /DCB
a parallelogram are
congruent.
4. /BAD and /EAD
4. If two angles form a
are supplements,
linear pair, then
/DCB and /FCB are
they are supplesupplements.
mentary.
5. /EAD > /FCB
5. If two angles are
congruent, then
their supplements
are congruent.
6. nEAD > nFCB
6. AAS.
Writing About Mathematics
1. No. If opposite sides are supplementary, the
quadrilateral is not necessarily a parallelogram.
For example, consider a quadrilateral with angle
measures 30°, 40°, 150°, and 140°. Opposite
angles are supplementary but not congruent, and
so the quadrilateral is not a parallelogram.
2. No. Consider parallelogram ABCD with the
diagonals intersecting at the point E. The four
triangles formed are nABE, nBCE, nCDE, and
nADE. Since AE > EC and BE > ED, in order
for nABE to be congruent to nBCE, AB would
need to be congruent to adjacent side BC.
However, this is not necessarily true.
Developing Skills
3. a. 70, 110, 110
b. 65, 115, 115
c. 90, 90, 90
d. 50, 130, 130
e. 25, 155, 155
f. 12, 168, 168
4. x 5 50; m/A 5 80; m/B 5 100
5. x 5 34; m/A 5 78; m/B 5 102
6. x 5 70; m/A 5 m/C 5 110; m/B 5 m/D 5 70
7. x 5 22; m/A 5 m/C 5 66; m/B 5 m/D 5 114
8. x 5 15; m/A 5 m/C 5 75; m/B 5 m/D 5 105
9. x 5 120; m/A 5 m/C 5 60; m/B 5 m/D 5 120
10. x 5 5; AB 5 CD 5 27
11. x 5 2; y 5 2; AB = CD 5 12; BC 5 DA 5 8
12. x 5 3; AC 5 12
13. y 5 2; DB 5 9
17. In a parallelogram, consecutive angles are
supplementary. Thus, the corners adjacent to the
corner forming a right angle are also right angles.
In a parallelogram, opposite angles are
congruent. Thus, the corner opposite the corner
forming a right angle is also a right angle.
Therefore, Petrina’s floor has four right angles.
18. Suppose that ABCD is a parallelogram. Then
opposite angles /A and /C are congruent and
m/A 5 m/C. However, we are given that m/A
and m/C are not equal, which is a contradiction.
Therefore, the assumption is false, and ABCD is
not a parallelogram.
Applying Skills
14. In parallelogram ABCD, diagonal AC divides
the parallelogram into two congruent triangles
with nABC > nCDA. Since corresponding parts
of congruent triangles are congruent, AB > CD
and AD > BC. Therefore, the opposite sides of a
parallelogram are congruent.
19. B
C
A
15. In a parallelogram ABCD, diagonal AC divides
the parallelogram into two congruent triangles
with nABC > nCDA. Since corresponding parts
of congruent triangles are congruent, /B > /D.
Similarly, diagonal BD divides the parallelogram
into two congruent triangles with nABD >
nCDB, and /A > /C. Therefore, the opposite
angles of a parallelogram are congruent.
D
Q
R
P
S
20. Since PQ > AB, QR > BC, and /B > /Q,
nABC > nPQR by SAS. Since a diagonal
divides a parallelogram into two congruent
triangles, nABC > nCDA and nPQR > nRSP.
By the transitive property of congruence,
302