# Chapter 6 ⋅

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Chapter 6 ⋅
```Chapter 6
Chapter 6 Maintaining Mathematical Proficiency (p. 303)
1. Slope perpendicular to y =
​ 13 ​x
—
− 5 is −3.
y = mx + b
y = −3x + b
1 = −3
1 = −9 + b
10 = b
An equation of the line is y = −3x + 10.
⋅
3+b
y = mx + b
y=x+b
−3 = 1
−3 = 4 + b
−7 = b
An equation of the line is y = x − 7.
⋅4 + b
1.a.Check students’ work.
b. Check students’ work.
c. Check students’ work (for sample in text, CA ≈ 1.97,
d. Every point on the perpendicular bisector of a segment is
—
—
CB ≈ 1.97); For all locations of C, CA
​ ​ and CB
​ ​ have the
same measure.
equidistant from the endpoints of the segment.
2.a.Check students’ work.
3. Slope perpendicular to y = −4x + 13 is —
​ 14 ​.
5. A midsegment of a triangle is a segment that connects the
midpoints of two sides of the triangle.
6.1 Explorations (p. 305)
2. Slope perpendicular to y = −x − 5 is 1.
4. An altitude of a triangle is the perpendicular segment from
a vertex to the opposite side or to the line that contains the
opposite side.
y = mx + b
1
y = —
​   ​ x + b
4
1
−2 = —
​   ​  (−1) + b
4
1
−2 = −​ —  ​ + b
4
1
4(−2) = 4​ −—
​   ​   ​+ 4b
4
−8 = −1 + 4b
⋅
(  )
−7 = 4b
−7
​   ​ = b
—
4
An equation of the line is y = —
​ 14 ​x − —
​ 74 ​.
b. Check students’ work.
c. Check students’ work (for sample in text, DE ≈ 1.24,
DF ≈ 1.24); For all locations of D on the angle bisector, ​
—​ and FD
—
ED
​ ​ have the same measure.
d. Every point on an angle bisector is equidistant from both
sides of the angle.
3. Any point on the perpendicular bisector of a segment is
equidistant from the endpoints of the segment. Any point on
the angle bisector is equidistant from the sides of the angle.
​ is 5 units, which is the same
4. The distance point D is from ​AB
​.  Point D is in the angle bisector,
as the distance D is from ​AC
so it is equidistant from either side of the angle. Therefore, ​
—​  ≅ DF
—
DE
​ ​.
6.1 Monitoring Progress (pp. 307–309)
4. w ≥ −3 and w ≤ 8, or −3 ≤ w ≤ 8
1. WZ = YZ
WZ = 13.75
5. m > 0 and m < 11, or 0 < m < 11
By the Perpendicular Bisector Theorem, WZ = 13.75.
6. s ≤ 5 or s > 2
2.
7. d < 12 or d ≥ −7
8. yes; As with Exercises 6 and 7, if the graphs of the two
inequalities overlap going in opposite directions and the variable
only has to make one or the other true, then every number on
the number line makes the compound inequality true.
Chapter 6 Mathematical Thinking (p. 304)
1. A perpendicular bisector is perpendicular to a side of the
triangle at its midpoint.
WZ = YZ
4n − 13 = n + 17
3n = 30
n = 10
YZ = n + 17
YZ = 10 + 17
YZ = 27
By the Perpendicular Bisector Theorem, YZ = 27.
3. WX = —
​ 12 ​WY
1
2. An angle bisector divides an angle of the triangle into two
WX = —
​ 2 ​(14.8)
WX = 7.4
3. A median of a triangle is a segment from a vertex to the
midpoint of the opposite side.
4. DA = 6.9 by the Angle Bisector Theorem.
Geometry
Worked-Out Solutions
189
Chapter 6
5.
3z + 7 = 2z + 11
z + 7 = 11
z=4
CD = 2z + 11 = 2 4 + 11 = 8 + 11 = 19
⋅
​  is the angle bisector of ∠ ABC and
6. Because AD = CD, BD​
m∠ ABC = 2m∠ CBD. Therefore, m∠ ABC = 2(39) = 78°.
7. no; In order to use the Converse of the Angle Bisector
—
Theorem (Thm. 6.4), PS
​ ​ would have to be perpendicular
—


to ​QP​ , and RS
​ ​ would have to be perpendicular to QR​
​  .
−1 − (−5) −1 + 5 4
​
= ​ —
​
= ​ —   ​ = 1
​
8. Slope: m = —
3 − (−1)
3+1
4
The slope of the perpendicular line is m = −1.
( −12+ 3 −5 +2 (−1) ) ( 22 −62 )
​,
​
​
​   ​  ​ = (1, −3)
midpoint = ​ —
​
​ = ​ —
​   ​ , —
—
y = mx + b
⋅
⋅ (1) + b
y = −1 x + b
−3 = −1
−3 = −1 + b
−2 = b
So, an equation of the perpendicular bisector is y = −x − 2.
6.1 Exercises (pp. 310–312)
Vocabulary and Core Concept Check
—
—
—
​  ⊥ VW
​ ​ ≅ WD
​ ​ and UX​
​ ​, point U
6. UW = 55; Because VD
—
is on the perpendicular bisector of VW
​ ​. So, by the
Perpendicular Bisector Theorem (Thm. 6.1), VU = WU.
VU = UW
9x + 1 = 7x + 13
2x = 12
x=6
UW = 7
⋅ 6 + 13 = 55
7. yes; Because point N is equidistant from L and M, point N is
—
on the perpendicular bisector of LM
​ ​ by the Converse of the
Perpendicular Bisector Theorem (Thm. 6.2). Because only
—
​  must be
one line can be perpendicular to LM
​ ​ at point K, NK​
—
​  .
the perpendicular bisector of LM
​ ​, and P is on NK​
8. no; You would need to know that either LN = MN or
LP = MP.
​  ⊥ ML​
​ .
9. no; You would need to know that PN​
10. yes; Because point P is equidistant from L and M, point P is
—
on the perpendicular bisector of LM
​ ​ by the Converse of the
—
—
Perpendicular Bisector Theorem (Thm. 6.2). Also, LN
​ ​ ≅ MN
​ ​,
—
 is a bisector of LM
so ​PN​
​ ​. Because P can only be on one of
—
 is the perpendicular bisector of LM
the bisectors, ​PN​
​ ​.
 ​, BD​
11. Because D is equidistant from BC​
​  and ​BA
​  bisects
∠ ABC by the Converse of the Angle Bisector Theorem
(Thm. 6.4). So, m∠ ABD = m∠ CBD = 20°.
1. Point C is in the interior of ∠DEF. If ∠DEC and ∠CEF are
congruent, then EC​
​  is the bisector of ∠DEF.
—
—
​ is an angle bisector of ∠ PQR, PS
​  , and SR
​  .
12. ​QS
​ ​ ⊥ QP​
​ ​ ⊥ QR​
So, by the Angle Bisector Theorem (Thm. 6.3), PS = RS = 12.
2. The question that is different is: Is point B collinear with
X and Z? B is not collinear with X and Z. Because the two
segments containing points X and Z are congruent, B is the
same distance from both X and Z, point B is equidistant from
—
X and Z, and point B is in the perpendicular bisector of XZ
​ ​.
m∠ KJL = m∠ MJK
7x = 3x + 16
4x = 16
Monitoring Progress and Modeling with Mathematics
x = 4
m∠ KJL = 7x = 7
​,  point H is on
​  ⊥ ​GJ
3. GH = 4.6; Because GK = KJ and HK​
—
the perpendicular bisector of GJ
​ ​. So, by the Perpendicular
Bisector Theorem (Thm. 6.1), GH = HJ = 4.6.
4. QR = 1.3; Because point T is equidistant from Q and S,
—
point T is on the perpendicular bisector of QS
​ ​ by the
Converse of the Perpendicular Bisector Theorem (Thm. 6.2).
So, by definition of segment bisector, QR = RS = 1.3.
​ and point D is equidistant from
​  ⊥ ​AC
5. AB = 15; Because DB​
A and C, point D is on the perpendicular bisector of AC
​—​ by the
Converse of the Perpendicular Bisector Theorem (Thm. 6.2).
By definition of segment bisector, AB = BC.
AB = BC
5x = 4x + 3
x=3
AB = 5 3 = 15
⋅
190
Geometry
Worked-Out Solutions
 bisects ∠ KJM.
13.​JL ​
Angle Bisector Theorem (Thm. 6.3)
Definition of angle bisector
⋅ 4 = 28°
14. EG​
​ bisects ∠ FEH.
Angle Bisector Theorem (Thm. 6.3)
FG = GHConverse of the Angle Bisector
Theorem (Thm. 6.4)
x + 11 = 3x + 1
−2x = −10
x=5
FG = 5 + 11 = 16
 , EH​
​ and ​EG​
​  bisects
15. yes; Because H is equidistant from ​EF
∠ FEG by the Angle Bisector Theorem (Thm. 6.3).
 , but
​ and ​EG​
16. no; Congruent segments connect H to both ​EF
 ,
​ and ​EG​
unless those segments are also perpendicular to ​EF
 .
​ and ​EG​
you cannot conclude that H is equidistant from ​EF
Chapter 6
—
—
17. no; Because neither BD
​ ​ nor DC
​ ​ are marked as perpendicular




to ​AB ​ or ​AC ​ respectively, you cannot conclude that DB = DC.
—
​ and
18. yes; D is on the angle bisector of ∠ BAC, DB
​ ​ ⊥ ​AB
—
​.  So, DB = DC by the Angle Bisector Theorem
​ ​  ⊥ ​AC
CD
(Thm. 6.3).
4
8
1 − (−7)
—
​
= —
​     ​ = −​ —  ​
22. Slope of YZ
​ ​:  m = —
​
7
−4 − 10 −14
The slope of the perpendicular line is m = —
​ 74 ​.
—
( 10 +2(−4) −72+ 1) ( 62 −62 )
​,
​
​
​   ​  ​ = (3, −3)
midpoint of YZ
​ ​  = ​ —
​
​ = ​ —
​   ​ , —
—
y = mx + b
7
y = —
​   ​ x + b
4
7
−3 = —
​   ​  3 + b
4
21
−3 = —
​   ​ + b
4
−12 = 21 + 4b
y=1 x+b
−33 = 4b
2 = 1 (4) + b
2=4+b
−2 = b
33
​ = b
−​ —
4
—
An equation of the perpendicular bisector of YZ
​ ​  is
7
33
y = —
​ 4 ​x − —
​  4  ​.
−1 − 5 −6
—
19. Slope of MN
​
= ​ —
​ = −1
​ ​:  m = —
​
7−1
6
The slope of the perpendicular line is m = 1.
1 + 7 5 + (−1)
8 4
​
​   ​   ​ = (4, 2)
midpoint = ​ —
​   ​, ​ —
​ = ​ —
​    ​, —
2
2
2 2
y = mx + b
(
) (  )
⋅
⋅
—
An equation of the perpendicular bisector of MN
​ ​  is y = x − 2.
12 3
12 − 0
—
​ = —
​   ​ = —
​   ​
20. Slope of QR
​ ​:  m = —
​

6 − (−2)
8
2
2
The slope of the perpendicular line is m = −​ —
3 ​.
—
( −22+ 6 0 +2 12 ) ( 24 122 )
​,
​
​
​   ​ = (2, 6)
midpoint of QR
​ ​  = ​ —
​
​ = ​ —
​   ​ , ​ —
—
22 = 3b
—
​   ​ = b
y = mx + b
2
y = −​ —  ​ x + b
3
2
6 = −​ —  ​  (2) + b
3
4
6 = −​ —  ​ + b
3
18 = −4 + 3b
⋅
—
​  , you
24. Because BP
​ ​ is not necessarily perpendicular to CB​
do not have sufficient evidence to say that BP = AP. The
reasoning should be: By the Angle Bisector Theorem
 and CA​
(Thm. 6.3), point P is equidistant from ​CB​
​ .
​  and ∠ APB is that PG​
​  is the
26.a.The relationship between PG​
angle bisector of ∠ APB.
22
b = —
​   ​
3
—
An equation of the perpendicular bisector of QR
​ ​  is
2
22
—
y = −​  3 ​x + —
​  3  ​.
4
1
8−4
—
​    ​ = —
​   ​
21. Slope of UV
​ ​:  m = —
​
​ = —
9 − (−3) 12 3
3
The slope of the perpendicular line is m = −​ —
1 ​ = −3.
−3 + 9 4 + 8
6 12
—
​,
​   ​  ​ = ​ —
​   ​ = (3, 6)
midpoint of UV
​ ​  = ​ —
​
​   ​ , ​ —
—
y = mx + b
y = −3x + b
6 = −3
6 = −9 + b
15 = b
—
—
​ will
​ ​ is not necessarily congruent to EC
​ ​, ​AB
23. Because DC
not necessarily pass through point C. The reasoning
​ ⊥ DE
​ is the
should be: Because AD = AE, and ​AB
​—​, ​AB
—
perpendicular bisector of DE
​ ​.
25. The Perpendicular Bisector Theorem (Thm. 6.1) will allow
—​  ≅ CD
—
​ ​.
22
3
(
⋅
2
2
) ( 2 2 )
⋅3 + b
—
An equation of the perpendicular bisector of UV
​ ​  is
y = −3x + 15.
b.m∠ APB gets larger. Covering the goal becomes more
difficult if the goalie remains at the same distance from
the puck on the perpendicular bisector. As the angle
increases, the goalie is farther away from each side of
the angle.
—
27. Draw XY
​ ​, using a radius that is greater than —
​ 12 ​the distance
—
of XY
​ ​. Draw two arcs of equal radii, using X and Y as
centers, so that the arcs intersect. Draw a line through both
intersections of the arcs.
Set a compass at 3 centimeters Z
by its own scale or with a ruler.
3 cm
Retaining this setting, place the
3.9 cm
3.9 cm
compass point on the midpoint
—
of XY
​ ​ and mark the point on
the perpendicular bisector as
X 2.5 cm
2.5 cm Y
point Z. The distance between
—
point Z and XY
​ ​ is 3 centimeters
because of the compass setting
and in this example, XZ and YZ are both equal to
3.9 centimeters. This construction demonstrates the
Perpendicular Bisector Theorem (Thm. 6.1).
Geometry
Worked-Out Solutions
191
Chapter 6
28. Because every point on a compass arc is the same distance
from one endpoint, and every point on the other compass
arc with the same setting is the same distance from the other
endpoint, the line connecting the points where these arcs
intersect contains the points that are equidistant from both
endpoints. You know from the Converse of the Perpendicular
Bisector Theorem (Thm. 6.2) that the set of points that are
equidistant from both endpoints make up the perpendicular
bisector of the given segment.
3x = 54
x = 18
​ bisects ∠ BAC.
1. Given
2. Definition of angle
bisector
—
—
​,  DC
​  3. Given
3. DB
​ ​ ⊥ ​AB
​ ​ ⊥ ​AC
4. ∠ ABD and ∠ACD are 4. Definition of
right angles.
perpendicular lines
5. ∠ ABD ≅ ∠ACD
−1 − 7
5. Right Angles Congruence
Theorem (Thm. 2.3)
6. —
0
5−5
—
​ = —
​     ​ = 0
​ ​:  m = —
​

30. B; Slope of MN
−8
6. Reflexive Property of
Congruence (Thm. 2.1)
The slope of the perpendicular line is undefined.
​,
​   ​  ​ = ​ —
​   ​   ​ = (3, 5)
midpoint of MN
​ ​  = ​ —
​
​   ​ , —
—
The equation of a line that has a slope that is undefined
—
REASONS
29. B; (3x − 9)° = 45°
STATEMENTS
( 7 + 2(−1) 5 +2 5 ) ( 62 102 )
7. AAS Congruence
Theorem (Thm. 5.11)
8. —
​DB ​ ≅ —
​DC ​
8. Corresponding parts of
congruent triangles are
congruent.
through the point (3, 5) is x = 3. So, the point that lies on the
perpendicular bisector is (3, 9).
31. no; In isosceles triangles, for example, the ray that has an
endpoint of the vertex and passes through the base (the
opposite side of the vertex) is not only an angle bisector of
the vertex, but also a perpendicular bisector of the base.
32. Given CA = CB
—
Prove Point C lies on the perpendicular bisector of AB
​ ​.
9. DB = DC
9. Definition of congruent
segments
B
b.Given BD = CD,
—
​,
​ ​ ⊥ ​AB
DB
​
​—​ ⊥ ​AC
DC
 bisects ∠ BAC.
Prove ​
D
A
C
C
STATEMENTS
A
B
P
​  such that point P is on ​
Given isosceles △ACB, construct CP​
—​ and CP​
—
​  ⊥ AB
AB
​ ​. So, ∠ CPB and ∠ CPA are right angles by
the definition of perpendicular lines, and △CPB and △CPA
—
—
—
—
are right triangles. Also, because AC
​ ​  ≅ BC
​ ​ and CP
​ ​ ≅ CP
​ ​
by the Reflexive Property of Congruence (Thm. 2.1),
△CPB ≅ △CPA by the HL Congruence Theorem (Thm. 5.9).
—
—
So, AP
​ ​ ≅ BP ​
​  because corresponding parts of congruent
triangles are congruent, which means that point P is the
—
midpoint of AB
​—​, and CP​
​  is the perpendicular bisector of AB
​ ​.
—
 bisects ∠ BAC, DB
​, DC
​
33.a.Given ​
​ ​  ⊥ ​AB
​—​  ⊥ ​AC
Prove DB = DC
1. Given
2. ∠ ABD and ∠ ACD
are right angles.
2. Definition of
perpendicular lines
3. △ ABD and △ACD
are right triangles.
3. Definition of a right
triangle
4. —
​BD ​ ≅ —
​CD ​
4. Definition of congruent
segments
5. —
5. Reflexive Property of
Congruence (Thm. 2.1)
6. △ ABD ≅ △ACD
6. HL Congruence Theorem
(Thm. 5.9)
7. Corresponding parts of
congruent triangles are
congruent.
B
D
A
​ bisects ∠ BAC.
C
192
Geometry
Worked-Out Solutions
REASONS
​,
​DB ​ ⊥ ​AB
1. BD = CD, —
—
​
​DC ​ ⊥ ​AC
8. Definition of angle
bisector
Chapter 6
— —
c.First, WV
​ ​ ≅ WV
​ ​ by the Reflexive Property of Congruence
34.a.Roosevelt School; Because the corner of Main and
3rd Street is exactly 2 blocks of the same length from each
hospital, and the two streets are perpendicular, 3rd Street
is the perpendicular bisector of the segment that connects
the two hospitals. Because Roosevelt school is on 3rd
Street, it is the same distance from both hospitals by the
Perpendicular Bisector Theorem (Thm. 6.1).
39. The triangle is isosceles because it has two congruent sides.
b. no; Because the corner of Maple and 2nd Street is
approximately the midpoint of the segment that connects
Wilson School to Roosevelt School, and 2nd Street is
perpendicular to Maple, 2nd Street is the perpendicular
bisector of the segment connecting Wilson and Roosevelt
Schools. By the contrapositive of the Converse of the
Perpendicular Bisector Theorem (Thm. 6.2), the Museum
is not equidistant from the two schools because it is not on
2nd Street.
35.a.y = x
b. y = −x
c. y = ​ x ​
36. no; In spherical geometry, all intersecting lines meet in two
points which are equidistant from each other because they
are the two endpoints of a diameter of the circle.
—​ ≅ CD
—
—​ ≅ CE
—
37. Given ​
​ ​  and ​AE
​ ​
—
—
AB ​ ≅ CB
​ ​
Prove ​
—
—
—
—
(Thm. 2.1). Then, because XW
​ ​ ≅ ZW
​ ​ and XV
​ ​ ≅ ZV
​ ​,
△WVX ≅ △WVZ by the SSS Congruence Theorem
(Thm. 5.8). So, ∠ VXW ≅ ∠ VZW because corresponding
parts of congruent triangles are congruent.
Maintaining Mathematical Proficiency
40. The triangle is scalene because no sides are congruent.
41. The triangle is equilateral because all sides are congruent.
42. The triangle is an acute triangle because all angles measure
less than 90°.
43. The triangle is a right triangle because one angle measures 90°.
44. The triangle is obtuse because one angle measure is greater
than 90°.
6.2 Explorations (p. 313)
B
4
3
A
2
D
−1
— —
— —
Perpendicular Bisector Theorem (Thm. 6.2), both points
—
​  is
D and E are on the perpendicular bisector of AC
​ ​. So, DE​
—
—
—
the perpendicular bisector of AC
​ ​. So, if AB
​ ​ ≅ CB
​ ​, then by the
Converse of the Perpendicular Bisector Theorem (Thm. 6.2),
​  . So, points D, E, and B are collinear.
point B is also on DE​
Conversely, if points D, E, and B are collinear, then by the
Perpendicular Bisector Theorem (Thm. 6.2), point B is also
—
—
—
on the perpendicular bisector of AC
​ ​. So, AB
​ ​ ≅ CB
​ ​.
3
4
5
6
7
a.The perpendicular bisectors of the sides of △ABC all
c.The circle passes through all three vertices of △ABC.
intersect at one point.
A
E
4
B
3
D
2
1
X
Y
0
V
−2
P
−1
0
1
2
3
4
5
6
7
−1
W
C
−2
−3
Z
2
−3
—
38. Given Plane P is a perpendicular bisector of XZ
​ ​ at point Y.
—
—
—
—
XW ​  ≅ ZW
​ ​  b. ​
XV ​  ≅ ZV
​ ​  c. ∠ VZW ≅ ∠ VZW
Prove a. ​
C
1
−2
​ ​ ≅ CD
​ ​ and AE
​ ​ ≅ CE
​ ​, by the Converse of the
0
−1
C
A
0
−2
D
1
B
E
—
—
bisector of XZ
​—​ at point Y, YW
​ ​ is a perpendicular bisector of ​
—
XZ ​ by definition of a plane perpendicular to a line. So, by
—​ ≅ ZW
—
the Perpendicular Bisector Theorem (Thm. 6.1), ​XW
​ ​.
a.Because YW
​ ​ is on plane P, and plane P is a perpendicular
—
b.Because YV​
​  is on plane P, and plane P is a perpendicular
—
bisector of XZ
​—​ at point Y, YV
​ ​ is a perpendicular bisector of ​
—
XZ ​ by definition of a plane perpendicular to a line. So, by
—​ ≅ ZV
—
the Perpendicular Bisector Theorem (Thm. 6.1), ​XV
​ ​.
a. The angle bisectors all intersect at one point.
c.distance ≈ 2.06; The circle passes through exactly one
point of each side of △ABC.
3. The perpendicular bisectors of the sides of a triangle meet
at a point that is the same distance from each vertex of the
triangle. The angle bisectors of a triangle meet at a point that
is the same distance from each side of the triangle.
Geometry
Worked-Out Solutions
193
Chapter 6
6.2 Monitoring Progress (pp. 315–318)
4.
1. The pretzel distributor is located at point F, which is the
circumcenter of △ABE.
3x + 8 = 7x + 2
QM = 3x + 8 = 3
B
F
A
E
S(−6, 5)
R(−2, 5)
6
4
y=2
−6
x
( −2 +2 (−6) 5 +2 5 ) ( −82 102 )
−2 + (−2) 5 + (−1)
—
​,
​
​
midpoint of RT
​ ​  = (​  ​
)​ = (​  ​ −42 ​, ​ 24 ​  )​ = (−2, 2)
2
2
—
​,
​ —
​  ​ = ​ —
​   ​ = (−4, 5)
midpoint of RS
​ ​  = ​ —
​
​   ​,  ​ —
The equation of the perpendicular bisector of RS
​ ​  through
— —
—  —
—
its midpoint (−4, 5) is x = −4, and the equation of the
perpendicular bisector of RT
​—​ through its midpoint (−2, 2) is
y = 2. The point of intersection of the two perpendicular
bisectors is (−4, 2). So, the coordinates of the circumcenter
of △RST is (−4, 2).
3. Graph △WXY.
3

2
9

2
3

2
21

2
By the Incenter Theorem, QM = QN = QP and
−2
2
(0, −1)
E
6.2 Exercises (pp. 319–322)
Vocabulary and Core Concept Check
1. When three or more lines, rays, or segments intersect in
the same point, they are called concurrent lines, rays, or
segments.
Monitoring Progress and Modeling with Mathematics
4 x
3. Because G is the circumcenter of △ABC, AG = BG = CG.
Therefore, because AG = 9, BG = 9.
y = −1
−4
4. Because G is the circumcenter of △ABC, AG = BG = CG.
Therefore, because GC = 11, GA = 11.
x=0
−6
L
2. The triangle that does not belong is the fourth triangle
because it shows the incenter of the triangle. The other three
show the circumcenter.
y X(1, 4)
W(−1, 4)
−4
⋅ — ​  ​ + 8 = — ​  ​ + 8 = 4.5 + 8 = 12.5
QN = 7x + 2 = 7 ⋅ ​( —
​  ​  )​ + 2 = —
​  ​ + 2 = 10.5 + 2 = 12.5
T(−2, −1)
x = −4
x = —
​ 46 ​ = —
​ 23 ​
5. Draw two angle bisectors and label the intersection of the
bisectors, the incenter, as L. Draw a perpendicular segment
from the incenter L to any one side of the triangle. Label that
point E. Draw a circle with center L and radius LE. It should
touch all sides of the triangle. The location of the lamppost is
at L.
y
(−4, 2)
−4x = −6
QP = 12.5 units.
2. Graph △RST.
QM = QN
Y(1, −6)
( −12+ 1 4 +2 4) ( 20 28)
—​  = ​  ​ 1 + ​,
​
midpoint of ​XY
( 2 1 ​ 4 +2(−6)
)​ = ​( ​ 22 ​ , ​ −22 ​  )​ = (1, −1)
—
​,
​   ​   ​ = ​ —
​   ​   ​ = (0, 4)
midpoint of ​WX ​  = ​ —
​
​   ​ , —
—
5. Because P is the incenter of △XYZ, PA = PB = PC.
Therefore, because PC = 9, PB = 9.
—
The equation of the perpendicular bisector of WX
​ ​  through its
6. Because P is the incenter of △XYZ, KP = HP = FP.
Therefore, because KP = 15, HP = 15.
— —
—  —
midpoint (0, 4) is x = 0, and the equation of the perpendicular
bisector of XY
​—​ through its midpoint (1, −1) is y = −1. The
point of intersection of the two perpendicular bisectors is
(0, −1). So, the coordinates of the circumcenter of △WXY
are (0, −1).
194
Geometry
Worked-Out Solutions
Chapter 6
7. Graph △ABC.
y
9. Graph △HJK.
x=5
12
y
C(8, 10)
H(−10, 7)
(5, 8)
4
4
A(2, 6) B(8, 6)
8
12
−12
x
( 2 +2 8 6 +2 6 ) ( 102 122 )
8 + 8 6 + 10
16 16
—
​
midpoint of BC
​ ​  = (​  ​   ​, ​
​ = ​  ​   ​,  ​   ​   ​ = (8, 8)
2
2 ) (2 2)
—
​  ​ = ​ —
​   ​   ​ = (5, 6)
midpoint of AB
​ ​  = ​ —
​   ​, ​ —
​   ​,  —
The equation of the perpendicular bisector of AB
​ ​  through
— —
—  —
—
its midpoint (5, 6) is x = 5, and the equation of the
perpendicular bisector of BC
​—​ through its midpoint (8, 8) is
y = 8. The point of intersection of the two perpendicular
bisectors is (5, 8). So, the coordinates of the circumcenter of
△ABC are (5, 8).
8. Graph △DEF.
y
4
−4
x
E(−1, −1)
y = −5
F(−7, −9)
−12
(−4, −5)
(  2
)
2
−8 −2
= (​  ​   ​,  ​   ​  )​ = (−4, −1)
2 2
−7 + (−7) −1 + (−9)
—
​,
​
​
midpoint of DF
​  ​ = (​  ​
)​
2
2
−14 −10
= (​  ​   ​, ​   ​
​ = (−7, −5)
2
2 )
−7 + (−1) −1 + (−1)
—
​,
​
​
midpoint of DE
​  ​ = ​ —
​
​
—
—  —
— —
—  —
—
The equation of the perpendicular bisector of DE
​  ​ through
its midpoint (−4, −1) is x = −4, and the equation of
the perpendicular bisector of DF
​— ​through its midpoint
(−7, −5) is y = −5. The point of intersection of the two
perpendicular bisectors is (−4, −5). So, the coordinates of
the circumcenter of △DEF are (−4, −5).
x
−4
4
7−3
−10 − (−6) −4
—
The slope of the line perpendicular to JH
​ ​  is m = 1.
​,
​ —
​  ​
midpoint of JH
​ ​  = ​ ——
​
​     ​ = −1
Slope of JH: m = ——
​
​ = —
—
7+3
( −10 + (−6)
2
2 )
−16 10
= ​( ​   ​, ​   ​  )​ = (−8, 5)
2 2
—  —
y = mx + b
5=1
5 = −8 + b
13 = b
D(−7, −1)
−8
4
J(−6, 3)
x = −4
K(−2, 3)
y=8
4
8
⋅ (−8) + b
—
The equation of the line perpendicular to JH
​ ​  is y = x + 13.
—
1
4
7−3
2
−10 − (−2) −8
—
The slope of the line perpendicular to HK
​  ​ is m = 2.
​,
​ —
​  ​
midpoint of HK
​  ​ = ​ ——
​
​     ​ = −​ —  ​
Slope of HK
​ ​:  m = ——
​
​ = —
—
—  —
y = mx + b
⋅ (−6) + b
5=2
5 = −12 + b
17 = b
7+3
( −10 + (−2)
2
2 )
−12 10
= (​  ​   ​, ​   ​  )​ = (−6, 5)
2 2
—
The equation of the line perpendicular to HK
​  ​ is y = 2x + 17.
—
3+3
( −6 + (−2)
2
2 )
−8 6
= (​  ​   ​,  ​    ​  )​ = (−4, 3)
2 2
​,
​ —
​  ​
midpoint of JK
​ ​  = ​ —
​
The equation of the perpendicular bisector of JK
​ ​ through its
—  —
—
midpoint (−4, 3) is x = −4.
The intersection of y = x + 13 and x = −4: The intersection of y = 2x + 17 and x = −4: So, the coordinates of the circumcenter of △HJK are (−4, 9).
y = −4 + 13 = 9, the point of intersection is (−4, 9).
y = 2(−4) + 17 = 9, the point of intersection is (−4, 9).
Geometry
Worked-Out Solutions
195
Chapter 6
10. Graph △LMN.
y
2
−2
11.
4
6
8 x
N(8, −6)
M(5, −3)
−4
−6
L(3, −6)
—
−3 − (−6) −3 + 6 3
5−3
2
2
2
—
The slope of the line perpendicular to LM
​ ​  is m = −​ —
3 ​.
​
= —
​
​
= —
​   ​
Slope of LM
​ ​:  m = —
​
(  2 2 )
8 9
= (​  ​    ​, −​   ​  )​ = (4, −4.5)
2 2
3 + 5 −6 + (−3)
—
​
midpoint of LM
​ ​  = ​ —
​   ​, ​ —
​
—
—
y = mx + b
2
9
−​ —  ​ = −​ —  ​  4 + b
2
3
9
8
−​ —  ​ = −​ —  ​ + b
2
3
9
8
6​ −—
​   ​   ​ = 6​ −—
​   ​   ​ + 6 b
2
3
−27 = −16 + 6b
⋅
(  ) (  )
−11 = 6b
11
​ = b
−​ —
6
2
—
The equation of the line perpendicular to LM
​ ​  is y = −​ —
​ 11
​.
3 ​x − —
6
−6 − (−3) −6 + 3 −3
—
​
= —
​
​
= —
​   ​ = −1
Slope of MN
​ ​:  m = —
​
8−5
3
3
—
The slope of the line perpendicular to MN
​ ​  is m = 1.
5 + 8 −3 + (−6)
13 9
—
​
midpoint of MN
​ ​  = ​ —
​   ​, ​ —
​ = ​ —
​   ​,  −​ —  ​   ​
(
2
y = mx + b
9 13
​ + b
−​ —  ​ = ​ —
2
2
9
13
2​ −—
​   ​   ​ = 2​ —
​   ​   ​ + 2b
2
2
−9 = 13 + 2b
) (  2 2 )
2
(  ) (  )
−11 = b
—
The equation of the line perpendicular to MN
​ ​  is y = x − 11.
−6 − (−6) −6 + 6 0
—
​
= —
​
​
= —
​   ​ = 0
Slope of LN
​ ​:  m = —
​
8−3
5
5
3 + 8 −6 + (−6)
—
​
midpoint of LN
​ ​  = ​ —
​   ​, ​ —

​
2
2
11 −12
11
​   ​  ​ = ​ —
= ​ —
​   ​,  —
​   ​,  −6  ​
2 2
2
—
The equation of the line perpendicular to LN
​ ​  is x = —
​ 11
​.
2
2
11
11
The intersection of y = −​ —

x
​
−
​

​
and
x
=
​

:
​
—
—
3
6
2
(
(
)
) (
)
—
y = −​ —
​  2  ​  ​ − —
​  6  ​ = −​ —
​  6  ​ = −​ —
3 ​​ —
6  ​ − —
6  ​ = −​  2  ​, the point of
intersection is ​ —
​ 11
​, −​ —
2  ​  ​. So, the coordinates of the
2
(  )
11
(
22
11
)
33
11
(
11
11
)
circumcenter of △LMN are ​ —
​  2  ​, −​ —
2  ​  ​.
196
Geometry
Worked-Out Solutions
11
3x = 9
x = —
​ 93 ​ = 3
⋅ 3 − 2 = 18 − 2 = 16
⋅ 3 + 7 = 16
ND = 6x − 2 = 6
NE = 3x + 7 = 3
By the Incenter Theorem, ND = NE = NF and NF = 16 units.
12. NG = NH
x + 3 = 2x − 3
−x = −6
x=6
NG = x + 3 = 6 + 3 = 9
NH = 2x − 3 = 2
By the Incenter Theorem, NG = NH = NJ and NJ = 9 units.
⋅ 6 − 3 = 12 − 3 = 9
NK = NL
2x − 2 = −x + 10
3x = 12
x=4
NK = 2x − 2 = 2
NL = −x + 10 = −4 + 10 = 6
By the Incenter Theorem, NK = NL = NM and NM = 6 units.
⋅4 − 2 = 8 − 2 = 6
14. NQ = NR
2x = 3x − 2
−1x = −2
x=2
⋅2 = 4
⋅2 − 2 = 6 − 2 = 4
NQ = 2x = 2
NR = 3x − 2 = 3
By the Incenter Theorem, NQ = NR = NS and NS = 4 units.
15.
−22 = 2b
2 11
6x − 2 = 3x + 7
13.
⋅
ND = NE
PX = PY
3x + 2 = 4x − 8
−1x = −10
x = 10
⋅ 10 + 2 = 30 + 2 = 32
⋅ 10 − 8 = 40 − 8 = 32
PX = 3x + 2 = 3
PY = 4x − 8 = 4
By the Incenter Theorem, PX = PY = PZ and PZ = 32 units.
16.
PX = PZ
4x + 3 = 6x − 11
−2x = −14
x=7
PX = 4x + 3 = 4
⋅ 7 + 3 = 28 + 3 = 31
PZ = 6x − 11 = 6 ⋅ 7 − 11 = 42 − 11 = 31
By the Circumcenter Theorem, PX = PY = PZ and PY = 31 units.
Chapter 6
17. Sample answer: 22. Construct an angle bisector of ∠BCA and ∠ABC. Label
the intersection of the two angle bisectors as D. Draw a
—
perpendicular line from D to AB
​ ​, label that point E. Using D
—
as the center and DE
​ ​ as the radius, construct a circle. Point
D is the incenter of △ABC.
4
A
3
2
D
1
B
0
−1
C
0
1
2
3
5
4
5
6
4
−1
A
3
−2
E
2
−1
A
3
D
1
0
0
B
0
1
2
3
5
4
6
−1
2
1
B
2
3
5C
4
6
−1
0
4
−1
D
1
23. Construct an angle bisector of ∠ABC and ∠CAB. Label
the intersection of the two angle bisectors as D. Draw a
—
perpendicular line from D to AB
​ ​, label that point E. Using D
—
as the center and DE
​ ​ as the radius, construct a circle. Point
D is the incenter of △ABC.
4
6
5
3
A
4
2
3
1
B
0
1
C
2
3
5
4
1
6
D
0
−1
0
1
2
C3
B
4
5
6
7
8
9
−1
−2
2
D
0
−1
A
24. Construct an angle bisector of ∠ABC and ∠BCA. Label
the intersection of the two angle bisectors as D. Draw a
—
perpendicular line from D to BC
​ ​, label that point E. Using D
—
as the center and DE
​ ​ as the radius, construct a circle. Point
D is the incenter of △ABC.
5
4
A
3
2
D
5
1
B
4
C
0
−1
0
1
2
3
4
5
6
A
3
−1
2
21. Construct an angle bisector of ∠ABC and ∠BCA. Label
the intersection of the two angle bisectors as D. Draw a
—
perpendicular line from D to AC
​ ​, label that point E. Using D
—
as the center and DE
​ ​ as the radius, construct a circle. Point
D is the incenter of △ABC.
A
3
E
2
D
1
C
0
−1
B
0
1
2
E
0
−1
B
0
1
2
C
3
4
5
6
−1
25. Because point G is the intersection of the angle bisectors, it
—
—
is the incenter. But, because GD
​ ​ and GF
​ ​ are not necessarily
perpendicular to a side of the triangle, there is not sufficient
—
evidence to conclude that GD
​—​ and GF
​ ​ are congruent.
Point G is equidistant from the sides of the triangle.
5
4
D
1
3
4
5
−1
6
26.Because point T is the intersection of the perpendicular
bisectors, it is the circumcenter and is equidistant from the
vertices of the triangle, not necessarily the sides.
TU = TW = TY
Geometry
Worked-Out Solutions
197
Chapter 6
27. You could copy the positions of the three houses, and
connect the points to draw a triangle. Then draw the three
perpendicular bisectors of the triangle. The point where the
perpendicular bisectors meet, the circumcenter, should be the
location of the meeting place.
33. Graph △ABC.
y
12
A(2, 5)
4
28. To find the location of the fountain, use the Incenter
Theorem. The incenter of the triangle is equidistance from
the sides of the triangle. That point will be the same distance
from each edge of the koi pond. So, place the fountain at the
incenter of the pond.
C(12, 3)
−4
D
C
B
29. The circumcenter of a scalene triangle is sometimes inside
the triangle. If the scalene triangle is obtuse or right, then
the circumcenter is outside or on the triangle, respectively.
However, if the scalene triangle is acute, then the
circumcenter is inside the triangle.
30. If the perpendicular bisector of one side of a triangle
intersects the opposite vertex, the triangle is always
isosceles. If the perpendicular bisector of one side of a
triangle intersects the opposite vertex, then it divides the
triangle into two congruent triangles. So, two sides of the
original triangle are congruent because corresponding parts
of congruent triangles are congruent.
31. The perpendicular bisectors of a triangle intersect at a point
that is sometimes equidistant from the midpoints of the
sides of the triangle. This only happens when the triangle is
equilateral.
32. The angle bisectors of a triangle intersect at a point that is
always equidistant from the sides of the triangle. This is the
Incenter Theorem (Thm. 6.6).
4
8
12
x
−4
—
6−5 1
6−2 4
—
The slope of the line perpendicular to AB
​ ​  is m = −4.
​  ​ = ​ —
​   ​ = (4, 5.5)
midpoint of AB
​ ​  = ​ —
​   ​, ​ —
​   ​ , ​ —
5.5 = −4 4 + b
5.5 = −16 + b
21.5 = b
A
B(6, 6)
8
​ = —
​   ​
Slope of AB
​ ​:  m = —
​

—
5+6
8 11
( 2 + 6
2
2 ) (2 2 )
y = mx + b
⋅
—
The equation of the line perpendicular to AB
​ ​  is
y = −4x + 21.5.
—
1
−2
3−5
5
12 − 2
10
—
The slope of the line perpendicular to AC
​ ​  is m = 5.
​, ​ —
​  ​ = ​ —
​   ​   ​ = (7, 4)
midpoint of AC
​ ​  = ​ —
​
​   ​,  —
​ = —
​   ​ = −​ —   ​
Slope of AC
​ ​:  m = —
​

—
y = mx + b
4=5 7+b
4 = 35 + b
5+3
14 8
( 2 + 12
2
2 ) ( 2 2)
⋅
−31 = b
—
The equation of the line perpendicular to AC
​ ​  is y = 5x − 31.
​ = —
​   ​ = −​ —   ​
Slope of BC
​  ​: m = —
​

—
1
−3
3 − 6
2
12 − 6
6
—
The slope of the line perpendicular to BC
​ ​  is m = 2.
​, ​ —
​  ​ = ​ —
midpoint of BC
​ ​  = ​ —
​
​   ​,  ​ —  ​   ​ = (9, 4.5)
—
6+3
18 9
( 6 + 12
2
2 ) ( 2 2)
y = mx + b
4.5 = 2 9 + b
4.5 = 18 + b
−13.5 = b
⋅
—
The equation of the line perpendicular to BC
​ ​  is
Find the intersection of y = −4x + 21.5 and y = 5x − 31.
−4x + 21.5 = 5x − 31
−9x + 21.5 = −31
y = 2x − 13.5.
198
Geometry
Worked-Out Solutions
−9x = −52.5
35
x = —
​   ​
6
11
35
​
y = 5 —
​   ​ − 31 = −​ —
6
6
⋅
(
11
35
)
The point of intersection is ​ —
​  6  ​, −​ —
6  ​  ​. So, the coordinates of
the circumcenter of △ABC are
(
​ —
​  35
​,
6
11
)
−​ —
6  ​  ​.
Chapter 6
34. Graph △DEF.
F(−2, −2)
−8
−4
y
x
−9 + (−5) −5 + (−9)
−14 −14
—
​,
​
​
​  ​
midpoint of DE
​  ​ = ​ —
​
​ = ​ —
​   ​, ​ —
—
(
2
= (−7, −7)
2
) (  2
−7 = 1
−7 = −7 + b
The equation of the line perpendicular to DE
​ ​  is y = x.
—
−2 − (−5) −2 + 5 3
—
​ = —
​
​ = —
​   ​
Slope of DF
​  ​: m = —
​

−2 − (−9) −2 + 9 7
7
—
The slope of the line perpendicular to DF
​ ​  is m = −​ —  ​.
(
2
) (
3
)
11 7
−9 + (−2) −5 + (−2)
—
​,
​ —
​
midpoint of DF
​  ​ = ​ —
​
​= ​−—
​   ​,  −​ —  ​   ​
⋅ (
⋅ (  )
2
2
2
)
⋅ (  )
​, —
​
​ = —
​   ​
Slope of EF
​ ​:  m = —
​

—
−2 − (−9) −2 + 9
−2 − (−5) −2 + 5
7
10x = −49
35
x = −​ —  ​ = −4.9
10
y = −4.9
The point of intersection is (−4.9, −4.9). So, the coordinates
of the circumcenter of △DEF are (−4.9, −4.9).
36. 242 + (14x)2 = 252
576 + 196x2 = 625
x = 6
196x2 = 49
49
x2 = —
​ 196
​
7
x = —
​ 14
​ = —
​ 12 ​
The value of x that will
The value of x that will make N the incenter is 6.make N the incenter is —
​ 12 ​.
37. The circumcenter of any right triangle is located at the
midpoint of the hypotenuse of the triangle.
y
A(0, 2b)
7
3
MAB(0, b)
B(0, 0)
49
Find the intersection of y = x and y = −​ —
​  3  ​.
3 ​x − —
7
49
​   ​
x = −​ —  ​ x − —
3
3
3x = −7x − 49
−7 = b
3
—
The equation of the line perpendicular to EF
​ ​  is y = −​ —
7 ​x −7.
35. 352 + (2x)2 = 372
1225 + 4x2 = 1369
4x2 = 144
x2 = 36
−98 = 6b
98
​ = b
−​ —
6
49
​ = b
−​ —
3
7
—
T
he equation of the line perpendicular to DF
​  ​ is y = −​ —
​ 49
​.
3 ​x − —
3
)
−14 = 2b
0=b
)
(  ) (  )
)
⋅ (−7) + b
y = mx + b
7
7
11
−​ —  ​ = −​ —  ​  ​ −—
​   ​   ​ + b
2
3
2
7 77
​   ​ + b
−​ —  ​ = —
2
6
7
77
6 ​ −—
​   ​   ​ = 6 ​ —
​   ​   ​ + 6b
2
6
−21 = 77 + 6b
⋅ (
2
11
2
y = mx + b
3
11
7
​ = −​ —  ​  ​ −—
−​ —
​   ​   ​ + b
2
7
2
11 21
​ = —
​   ​ + b
−​ —
2
14
11 3
—
−​   ​ = ​ —   ​ + b
2
2
11
3
2​ −—
​   ​   ​ = 2​ —
​   ​   ​ + 2b
2
2
−11 = 3 + 2b
—
3
( −5 +2 (−2) −9 +2 (−2) ) (  72
—
−9 − (−5) −9 + 5 −4
−5 − (−9) −5 + 9
4
—
The slope of the line perpendicular to DE
​  ​ is m = 1.
​   ​
​,
​ —
​
midpoint of EF
​ ​  = ​ —
​
​= ​ −​ —  ​ , −​ —
​ = —
​
​ = —
​   ​ = −1
Slope of DE
​ ​:  m = —
​

y = mx + b
−8
E(−5, −9)
The slope of the line perpendicular to EF
​ ​  is m = −​ —
7 ​.
−4
D(−9, −5)
—
MAC (a, b)
MBC (a, 0)
C(2a, 0) x
Let A(0, 2b), B(0, 0), and C(2a, 0) represent the vertices of
a right triangle where ∠ B is the right angle. The midpoint
—
—
of AB ​
​  is M​AB
— ​( 0, b). The midpoint of BC ​
​  is M​BC
— ​( a, 0). The
—
—
midpoint of AC
​ ​  is M​AC
— ​( a, b). Because AB
​ ​ is vertical, its
perpendicular bisector is horizontal. So, the equation of the
horizontal line passing through M​— ​( 0, b) is y = b. Because​
—​ is horizontal, its perpendicularABbisector is vertical. So,
BC
the equation of the vertical line passing through M​BC
— ​( a, 0)
is x = a. The circumcenter of △ABC is the intersection of
perpendicular bisectors, y = b and x = a, which is (a, b).
—
This point is also the midpoint of AC ​
​ .
Geometry
Worked-Out Solutions
199
Chapter 6
— bisects ∠ CAB, BD ​
—
38. Given △ ABC, ​
​  bisects ∠ CBA,
—
—
—
—
—
—
​  ⊥ ​AB ​, DF ​
DE ​
​ ⊥
BC ​
​ , DG ​
​  ⊥ CA​
​  .
Prove The angle bisectors intersect at D, which is
—
—
—
equidistant from AB ​
​ , BC ​
​ , and CA ​
​ .
C
G
A
42.a.The archaeologists need to locate the circumcenter of
the three stones because that will be the center of the
circle that contains all three stones. In order to locate the
circumcenter, the archaeologists need to find the point of
concurrency of the perpendicular bisectors of the sides of
the triangle formed by the three stones.
F
coordinates of the point at which the archaeologists
should look for the fire pit are (7, 7).
E
B
10
—
—
—
—
—
—
Because DE
​ ​ ⊥ AB
​ ​, DF
​ ​  ⊥ BC
​ ​, and DG
​ ​  ⊥ CA
​ ​, ∠ DFB, ∠ DEB,
∠ DEA, and ∠ DGA are congruent right angles. Also, by
definition of angle bisector, ∠ DBF ≅ ∠ DBE and
—
—
—
∠ DAE ≅ ∠ DAG. In addition, DB
​—​ ≅ DB
​ ​ and DA
​ ​ ≅ DA
​ ​  by
the Reflexive Property of Congruence (Thm. 2.1). So,
△DFB ≅ △DEB and △DEA ≅ △DGA by the AAS
Congruence Theorem (Thm. 5.11). Next, because
corresponding parts of congruent triangles are congruent,
—
—
—
​—​ ≅ DE
DF
​ ​ and DG
​ ​ ≅ DE
​ ​. By the Transitive Property of
—
—
Congruence (Thm. 2.1), DF
​—​ ≅ DE
​ ​ ≅ DG
​ ​. So, point D is
—
—
—
equidistant from AB
​ ​, BC
​ ​, and CA
​ ​. Because D is equidistant
—
from CA
​—​ and CB
​ ​, by the Converse of the Angle Bisector
Theorem (Thm. 6.4), point D is on the angle bisector of
∠ ACB. So, the angle bisectors intersect at point D.
39. The circumcenter is the point of intersection of the
perpendicular bisectors of the sides of a triangle, and it is
equidistant from the vertices of the triangle. In contrast, the
incenter is the point of intersection of the angle bisectors of a
triangle, and it is equidistant from the sides of the triangle.
41.a.To determine the location of the pool so that it touches
​  ​.
the edges, construct two angle bisectors ​ RD​
​ and QD​
Construct a perpendicular bisector through point D to QR ​
​—.
Label the intersection E. With D as the center, construct a
​—. If the incenter point were to move
in any direction, the circle contained within the triangle
would become smaller and not touch all three sides of the
triangle. So, the circle with the center at the incenter is the
largest circle that touches all three sides.
(
)
E
Q
R
D
P
y
A(2, 10)
8
B(13, 6)
D(7.09, 6.87)
6
4
2
C(6, 1)
2
4
6
8
10
12
x
43. B; by the Perpendicular Bisector Theorem
44. no; When you find the circumcenter of three of the points
and draw the circle that circumscribes those three points, it
does not pass through the fourth point. An example of one
circle is shown.
A
40. no; Because the incenter is the center of an inscribed circle,
it must be inside the triangle.
b.This is a circumcenter problem. The approximate
D
B
E
C
D
45. yes; In an equilateral triangle, each perpendicular bisector
passes through the opposite vertex and divides the triangle
into two congruent triangles. So, it is also an angle bisector.
46. The incenter is at the center of the hub of the windmill where
the blades, acting as angle bisectors, connect to the hub.
47.a.All triangles have exactly three angle bisectors and
three perpendicular bisectors. Only three ​ —
​  20
​  ​ segments
3
are needed to represent them on an equilateral triangle
because each perpendicular bisector also bisects the
opposite angle.
(  )
b.All six segments are needed to represent the three angle
bisectors and three perpendicular bisectors of a scalene
triangle because none of the perpendicular bisectors will
also bisect an angle.
b.Yes, and the radius would be decreased by 1 foot. You
would keep the center of the pool as the incenter of the
triangle, but you would make the radius of the pool at
least 1 foot shorter.
200
Geometry
Worked-Out Solutions
Chapter 6
48. Sample answer:  ​, ​ —
​
​   ​ = (13, 11)
midpoint of PN
​  ​ = ​ ​  —
​ = ​ —
​   ​,  ​ —
y = mx + b
1
11 = —
​   ​  13 + b
3
1
3 11 = 3 —
​   ​  13 + 3b
3
33 = 13 + 3b
53.13°
75 ft
45 ft
60 ft
90°
⋅
⋅ ⋅
⋅
36.87°
( 10 +2 16 20 2+ 2 ) ( 262 222 )
—
20 = 3b
49. To determine the radius of the circle, the angle bisectors
would be used.
20
3
1
—
The equation of the line perpendicular to PN
​  ​ is y = —
​ 3 ​x + —
​ 20
​.
3
Find the intersection of y = −​ —
​ 3 ​x + —
​  3  ​.
2 ​x + 15 and y = —
E(4, 128)
y
1
1
2
y = −​ —  ​ x + 15
D(0, 0)
6 C(8, 0) x
The radius of the circle is approximately 3 inches.
50. To determine the coordinates of the center of the circle and the
radius of the circle, use perpendicular bisectors.
Graph △LMN.
y
24
E(13, 11)
P(10, 20)
F(6, 12)
8
−8
8
T(2, 4)
16
24 x
N(16, 2)
—
20 − 4 16
10 − 2
8
1
—
The slope of the line perpendicular to TP
​ ​  is m = −​ —
2 ​.
​, ​ —
​
​   ​   ​ = (6, 12)
midpoint of TP
​ ​  = ​ —
​
​ = ​ —
​   ​,  —
y = mx + b
12 = −​ —  ​  6 + b
​ = —
​   ​ = 2

Slope of TP
​ ​:  m = —
​
—
1
2
12 = −3 + b
15 = b
( 10 2+ 2 20 2+ 4 ) ( 122 242 )
⋅
—
The equation of the line perpendicular to TP
​ ​  is y = −​ —
2 ​x + 15.
​ = —
​    ​ = −3
Slope of PN
​  ​: m = —
​

​ = b
—
4
8
1
1
20
​   ​ x + —
​   ​
−​ —  ​ x + 15 = —
2
3
3
1
1
20
6 ​ −—
​   ​   ​x + 6 15 = 6 —
​   ​ x + 6 —
​   ​
2
3
3
−3x + 90 = 2x + 40
10
​
—
18
20 − 2
10 − 16 −6
—
The slope of the line perpendicular to PN
​  ​ is —
​ 13 ​.
1
⋅ (
)
⋅
⋅
20
⋅
−5x + 90 = 40
−5x = −50
x = 10
1
2
1
—
= −​   ​  10 + 15 = −5 + 15 = 10
2
The coordinates of the center of the circle are (10, 10).
Distance between the center (10, 10) and T(2, 4):
⋅
——
distance = √
​
(10 − 2)2 + (10 − 4)2 ​
The radius of the circle is 10 units.
—
—
—
= ​√82 + 62 ​
=√
​ 64 + 36 ​
=√
​ 100 ​ = 10
⋅ ⋅
⋅⋅
⋅⋅
⋅⋅
⋅⋅
⋅⋅
⋅
⋅ ⋅
1
​   ​  AB AC
51. Total area of △BAC = —
2
1
​   ​  x AC
2
1
Area of △BDC = —
​   ​  x BC
2
1
Area of △BDA = —
​   ​  x AB
2
1
1
1
1
​    ​ AB AC = —
​   ​  x AC + —
​   ​  x BC + —
​   ​  x AB
—
2
2
2
2
1
1
​    ​ AB AC = —
​   ​  x(AC + BC + AB)
—
2
2
1
1
2 —
​    ​ AB AC = 2 —
​   ​  x(AC + BC + AB)
2
2
AB AC = x(AC + BC + AB)
⋅ ⋅
⋅ ⋅
⋅ ⋅ ⋅
⋅
AB ⋅ AC
——
​
​ = x
(AC + BC + AB)
⋅⋅
— —
The expression for x in terms of the lengths of AB
​ ​,  ​BC ​, and
AC ​
​  is ——

​    ​.
—
⋅
AB AC
(AC + BC + AB)
Geometry
Worked-Out Solutions
201
Chapter 6
Maintaining Mathematical Proficiency
(
) ( 2 2 )
58. 2x + 3y = 18
−3 + 3 5 + 5
0 10
—
​,
​   ​  ​ = ​ —
​   ​ = (0, 5)
52. midpoint of AB
​ ​  = ​ —
​
​   ​ , ​ —
—
2
2
———
—
—
AB = √
​
​( 3 − (−3) )2​­ + (5 − 5)2 ​ = √
​ 62 + 0 2 ​
=√
​ 36 ​ = 6
) (  )
(
2 + 10 −1 + 7
12 6
—
​, ​ —
​
​    ​  ​ = (6, 3)
​ ​  = ​ —
​
​ = ​ —
​   ​,  —
53. midpoint of AB
2
2
2 2
———
AB = √
​
​( 10 − 2 )​­2 + (​  7 − (−1) )​2 ​
—
—
—
—
=√
​ 82 + 82 ​
= ​√64 + 64 ​
=√
​ 128 ​ = 8​√ 2 ​ ≈ 11.3
(
(
)
−5 + 4 1 + (−5)
—
​,
​
​
​ ​  = ​ —
​
​
54. midpoint of AB
—
2
2
1
−1 −4
​  ​ = ​ −—
= ​ —
​   ​,  ​ —
​   ​ , −2  ​
2
2 2
) (
3y = −2x + 18
y = −​ —
​ 18
​
3 ​x + —
3
y = −​ —
3 ​x + 6
The slope of the new line is —
​ 2 ​.
y = mx + b
y = —
​ 32 ​x + b
−6 = —
​ 2 ​ (−8) + b
−6 = −12 + b
AB = √
​ ​( 4 − (−5) )​ + (−5 − 1)  ​
—
—
3
6=b
The equation of the line passing through P(−8, −6) and
perpendicular to y = 2x + 1 is y = —
​ 32 ​x + 6.
—
) (
12
)
AB = √
​ ​( 5 − (−7) )​ + (9 − 5)  ​
—
—
y = mx + b
y = −​ —
2 ​x + b
8 = −​ —
2 ​ 2 + b
8 = −1 + b
10
—
−12
−8
P(−8, −6)
y
8
2x + 3y = 18
4
8
12 x
−4
−8
P(2, 8)
y=
1
−2 x
+9
−12
1
1
6
⋅
4
y = 2x + 1
2
9=b
The equation of the line
passing through P(2, 8)
and perpendicular to
1
y = 2x + 1 is y = −​ —
2 ​x + 9.
2
4
6
8
10 x
57. The line y = −5 is horizontal. The equation of the line
passing through P(6, −3) and perpendicular to y = −5 is
x = 6. The slope of the perpendicular line is undefined so
the equation is vertical.
y
8
3
=√
​ 122 + 42 ​
=√
​ 144 +  16 ​
=√
​ 160 ​ ≈ 12.6
56. The slope of the new 1
line is −​ —
2 ​.
y
y = 2x + 6
——
2
2

⋅
−7 + 5 5 + 9
−2 14
—
​,
​   ​  ​ = ​ —
​   ​ = (−1, 7)
​ ​  = ​ —
​
​   ​,  ​ —
55. midpoint of AB
—
2
2
2 2
3
=√
​ 92 + (−6)2 ​
=√
​ 81 +  36 ​
=√
​ 117 ​ ≈ 10.8
(
2
)
———
2
2

2
2
4
x=6
−2
−4
−6
202
x
59. y + 3 = −4(x + 3)
y + 3 = −4x − 12
y = −4x − 15
The slope of the new line is —
​ 14 ​.
y = mx + b
y = —
​ 4 ​x + b
1 = —
​ 4 ​ (−4) + b
y
1
1
4
⋅
1
y = 4x + 2
1 = −1 + b
P(−4, 1)
2=b
−8
−6
−4
The equation of the line
passing through P(−4, 1)
and perpendicular to
y + 3 = −4(x + 3)
is y = —
​ 14 ​x + 2.
−2
2 x
−2
−4
y + 3 = −4(x + 3)
P(6, −3)
y = −5
Geometry
Worked-Out Solutions
Chapter 6
6.3 Explorations (p. 323)
2.a.Check students’ work.
1.a.Draw △ABC and plot the midpoints of each side and
7
6
6
B
5
5
F
A
4
E
B
4
D
3
G
3
G
2
D
2
0
C
0
1
2
3
4
5
6
7
8
b.The medians of a triangle are concurrent at a point inside
2
​ 23 ​.
—
BE ≈ 3, BG ≈ 2; The ratio is
The ratio of the length of the longer segment to the length segment is —
​ 2 ​of the longer segment ​ a = —
​ 2 ​b  ​. The median
equals a + b = —
​ 2 ​b + b = —
​ 2 ​b + —
​ 2 ​b = —
​ 2 ​b. Because the
segment, b, is —
​ 3 ​of the median.
of the whole median is 2 : 3 or —
​ 23 ​. If the shorter segment of
the median is a and the longer segment is b, the shorter
1
1
median is
​ 32 ​b,
—
1
2
(
1
3
)
by multiplying by the reciprocal, the longer
2
5
C
6
7
8
9
c.The altitudes that connect a vertex and a point on the
opposite side are all perpendicular to that side. If the
triangle is acute, the altitudes meet inside of the triangle.
If the triangle is a right triangle, the legs of the right
triangle are the altitudes, and therefore, meet at a point on
the triangle. If the triangle is obtuse, the altitudes meet at
a point on the outside of the triangle.
3. The medians meet at a point inside the triangle that divides
each median into two segments whose lengths have the ratio
1 : 2. The altitudes meet at a point inside, on, or outside the
triangle depending on whether the triangle is acute, right, or
obtuse.
—
4. The two segments of RU
​ ​ have lengths of 1 inch and 2 inches.
6.3 Monitoring Progress (pp. 325–327)
⋅
⋅
​ 13 ​ 2100 = 700 ft
1. PS = —
2
PC = —
​ 3 ​ 2100 = 1400 ft
⋅
1000 = —
​  ​ ⋅ BC
2 ⋅ 1000 = 2 ⋅ —
​  ​ ⋅ BC
2.
4
the shorter segment to the length of the longer segment is
1
1 : 2 or ​ —
​.
2
AD ≈ 6, AG ≈ 4; The ratio is —
​ 3 ​.
3
The altitudes meet at the same point.
c.Segment AD is divided into AG ≈ 4 and GD ≈ 2. So, the
​ 12 ​. Segment BE is divided into BG ≈ 2 and
GE ≈ 1, so the ratio is about —
​ 12 ​. The ratio of the length of
2
the triangle.
1
−1
9
−1
F
0
1
0
A
1
E
3.
PT = —
​ 12 ​ PA
1

2
800 = —
​ 12 ​ PA
1

2
⋅
⋅
2 ⋅ 800 = 2 ⋅ —
​  ​ ⋅ PA
BT = —
​ 12 ​ BC
2000 = BC
So, BC is 2000 feet.
TC =
⋅ BC
⋅ 2000
​ 12 ​
—
​ 12 ​
—
1

2
1600 = PA
So, PA is 1600 feet.
TA = PT + PA
TC =
TC = 1000
TA = 800 + 1600
TA = 2400
So, TC is 1000 feet.
So, TA is 2400 feet.
Geometry
Worked-Out Solutions
203
Chapter 6
4.
y
6.
G(4, 9)
A(0, 3)
8
6
4
y
4
2
F(2, 5)
M(5, 5)
−2
P(4, 5)
2
B(0, −2)
H(6, 1)
4
2
6
x
( 4 +2 6 9 +2 1 ) ( 102 102 )
—
​  ​ = ​ —
​   ​   ​ = (5, 5).
The midpoint of GH
​ ​ is ​ —
​   ​, ​ —
​   ​,  —
​ ​.
The centroid is —
​   ​ of FM
2 + (5 − 5)2 ​
FM = √
​ (5 − 2)

The centroid is —
​ 3 ​ 3 = 2, so the centroid P is 2 units to
—
2
3
——
—
—
=√
​ 32 ​ = √
​ 9 ​ = 3
2
⋅
the right of F, which are the coordinates (4, 5). So, the
coordinates of the centroid of △FGH are (4, 5).
5.
L(−1, 4)
X(−3, 3)
P(−1, 2)
−5
5
3
J(0, 1.5)
1
3 x
Z(−1, −2)
—
( 1 +2(−1) 5 +2(−2)) ( 02 32 )
​,
​
​
​   ​   ​ = (0, 1.5).
The midpoint of YZ
​ ​ is ​ —
​

​ = ​ —
​   ​ , —
—
1
−1.5
1.5 − 3
—
​ = —
​   ​ = −​ —  ​ .
The slope of XJ
​ ​ is ​ —

2
0 − (−3)
3
y = mx + b
1
1.5 = −​ —  ​  0 + b
2
1.5 = b
The centroid has the coordinates of the intersection of
​   ​ and x = −1.
y = −​ —  ​ x + —
⋅
C(6, −3)
—
1
−3 − (−2)
6
6−0
—
The slope of the line perpendicular to BC
​ ​ is 6 and the line
passes through vertex A(0, 3).
​
= −​ —  ​
slope of BC
​ ​  = —
​
y = mx + b
3=6
3=b
⋅0 + b
—
The equation of the line perpendicular to BC ​
​  is y = 6x + 3.
​
= ​ —
​ = −1
slope AC
​ ​  = —
​
—
−3 − 3 −6
6−0
6
—
The slope of the line perpendicular to AC
​ ​ is 1 and the line
passes through vertex B(0, −2).
y = mx + b
−2 = 1
−2 = b
⋅0 + b
—
The equation of the line perpendicular to AC ​
​  is y = x − 2.
Find the intersection of y = 6x + 3 and y = x − 2.
6x + 3 = x − 2
5x + 3 = −2
5x = −5
x = −1
y=6
So, the orthocenter is outside the triangle and the coordinates
⋅ (−1) + 3 = −6 + 3 = −3
are (−1, −3).
3
b = —
​    ​
2
1
3
— though J(0, 1.5) is y = −​ —
​ x + —
​   ​ .
The equation of ​XJ ​
2
2
−3 + 1 3 + 5
−2 8
—
​,
​   ​  ​ = ​ —
The midpoint of XY
​ ​ is ​ —
​
​   ​, ​ —  ​   ​ = (−1, 4).
—
2
2
2 2
6
4 − (−2)
—
​ = —
​   ​ = undefined.
The slope of ​ZL ​  is —
​

−1 − (−1) 0
—
The equation of ZL
​ ​ through J(−1, 4) is x = −1.
(
) (  )
1
3
2
2
1
3
​    ​
y = −​ —  ​ x + —
2
2
1
3
​    ​
y = −​ —  ​ (−1) + —
2
2
1 3 4
​   ​ = —
​   ​ = 2
y = —
​   ​ + —
2 2 2
So, the centroid has coordinates (−1, 2).
204
y Y(1, 5)
−3
6 x
2
Geometry
Worked-Out Solutions
7. △ JKL is a right triangle; therefore, the orthocenter is on the
triangle at the right angle (−3, 4).
K(−3, 4) y
L(5, 4)
3
−1
3
5 x
−3
J(−3, −4)
Chapter 6
8. Proving △ABD ≅ △CBD by the SSS Congruence Theorem
(Thm. 5.8) at the beginning of the proof would be the same.
But then you would state that ∠ ABD ≅ ∠ CBD because
corresponding parts of congruent triangles are congruent.
—
This means that BD
​ ​ is also an angle bisector by definition.
Vocabulary and Core Concept Check
1. The four types of triangular concurrencies are circumcenters,
incenters, centroids, and orthocenters. The circumcenters are
formed by the intersection of the perpendicular bisectors.
The incenters are formed by the intersection of the angle
bisectors. The centroids are formed by the intersection of the
medians. The orthocenters are formed by the intersection of
the altitudes.
2. The length of a segment from a vertex to the centroid is
two-thirds the length of the median from that vertex.
Monitoring Progress and Modeling with Mathematics
3. PN =
​ 23 ​QN
—
​ 23 ​ 9
—
​ 18
​ = 6
—
3
⋅
PN =
PN =
PN = 6 units
⋅ QN
⋅ 9
​ 13 ​
—
​ 13 ​
—
4. PN =
PN =
PN =
​ 23 ​QN
—
​ 23 ​ 21
—
​ 42
​ = 14
—
3
⋅
PN = 14 units
⋅ QN
⋅ 21
QP =
QP =
QP = 3
QP = 7
QP = 3 units
QP = 7 units
⋅
​ 23 ​ 30
—
​ 60
​ = 20
—
3
QP =
​ 13 ​
—
​ 13 ​
—
5. PN = —
​ 23 ​QN
QP =
2
6. PN = —
​ 3 ​QN
PN =
⋅
​ 23 ​ 42
—
​ 84
​ = 28
—
3
PN =
PN =
PN = 20 units
QP =
QP =
​ 13 ​
—
​ 13 ​
—
QP = 10
QP = 14
QP = 10 units
QP = 14 units
⋅ QN
⋅ 30
6.3 Exercises (pp. 328–330)
PN =
PN = 28 units
QP =
QP =
⋅ QN
⋅ 42
​ 13 ​
—
​ 13 ​
—
⋅
5 = —
​  ​ ⋅ CE
3 ⋅ 5 = 3 ⋅ —
​  ​ ⋅ CE
⋅
⋅
3 ⋅ 11 = 3 ⋅ —
​  ​ ⋅ CE
7. DE = —
​ 13 ​ CE
1

3
1

3
⋅
⋅
CD =
CD =
CD = 10 units
2
66
CD = —
​  3  ​ = 22
CD = 22 units
⋅
⋅
3 ⋅ 15 = 3 ⋅ —
​  ​ ⋅ CE
1

3
⋅
⋅
CD =
CD =
CD = 18 units
15 = —
​ 13 ​ CE
​ 23 ​ CE
—
​ 23 ​ 27
—
​ 54
​ = 18
—
3
DE = —
​ 13 ​ CE
CE = 27 units
CD =
10.
1

3
27 = CE
⋅
⋅
CD = —
​ 3 ​ 33
1

3
CE = 33 units
2
⋅
9 = —
​  ​ ⋅ CE
3 ⋅ 9 = 3 ⋅ —
​  ​ ⋅ CE
33 = CE
CD = —
​ 3 ​ CE
9. DE = —
​ 13 ​ CE
11 = —
​ 13 ​ CE
​ 23 ​ CE
—
​ 23 ​ 15
—
​ 30
​ = 10
—
3
CD =
CE = 15 units
DE = —
​ 13 ​ CE
1

3
15 = CE
8.
45 = CE
CE = 45 units
2
⋅
⋅
CD = —
​ 3 ​ CE
2
CD = —
​ 3 ​ 45
90
CD = —
​  3  ​ = 30
CD = 30 units
—
11. Because G is a centroid, BF ​
​  is a median and F is the
—
midpoint of AC ​
​ . Therefore, FC = 12 units.
—
12. Because G is a centroid, BF ​
​  is a median and BG is —
​ 23 ​ BF.
⋅
⋅
—
​  ​ ⋅ 6 = —
​  ​ ⋅ —
​  ​ ⋅ BF
BG = —
​ 3 ​ BF
⋅
2
6 = —
​ 23 ​ BF
3

2
3

2
2

3
9 = BF
Therefore, BF = 9 units.
—
13. Because G is a centroid, AE ​
​  is a median and AG = —
​ 23 ​ AE.
2
⋅
⋅
⋅
AG = —
​ 3 ​ AE
AG = —
​ 3 ​ 15
AG = 10
Therefore, AG = 10 units.
2
—
14. Because G is a centroid, AE ​
​  is a median and GE = —
​ 13 ​ AE.
1
⋅
⋅
⋅
GE = —
​ 3 ​ AE
GE = —
​ 3 ​ 15
GE = 5
Therefore, GE = 5 units.
1
Geometry
Worked-Out Solutions
205
Chapter 6
15.
8
6
4
16.
y
(
C(5, 7)
( )
11
5, 3
F(6.5, 4)
G(−2, 7)
7
−3 , 5
)
I(−4, 5)
A(2, 3)
2
y
8
F(1, 5)
4
E(−2.5, 4)
H(−6, 3)
D(5, 2)
2
B(8, 1)
4
2
6
−6
8 x
—
( 5 +2 8 7 +2 1 ) ( 132 82 )
​  ​ = ​ —
​   ​   ​ = (6.5, 4).
The midpoint of CB
​ ​ is ​ —
​   ​, ​ —
​   ​,  —
​ = —
​     ​ = —
​   ​ .
The slope of AF
​ ​  is —
​

—
y = mx + b
4−3
6.5 − 2
⋅
⋅ ⋅
⋅
1
4.5
2
9
−4
−2
​,
​   ​  ​ = ​ —
​   ​ = (−4, 5).
​ —
​
​   ​,  ​ —
—
​   ​ = 0.
The slope of FI
​ ​  is —
​
​ = —
( −6 +2(−2) 3 +2 7 ) ( −82 102 )
—
0
5−5
1 − (−1) 2
—
The equation of FI ​
​  through I(−4, 5) is y = 5.
​, ​ —
​  ​ = ​ —
The midpoint of HF
​  ​ is ​ —
​
​   ​,  ​ —   ​  ​ = (−2.5, 4).
​
​     ​ = 6.
The slope of GE
​ ​  is ——
​
​ = —
​ = —
2
4 = —
​   ​  6.5 + b
9
2
9 4 = 9 —
​   ​  6.5 + 9 b
9
36 = 2 6.5 + 9b
36 = 13 + 9b
y = mx + b
23 = 9b
4=6
23
​   ​ = b
—
9
​   ​.
The equation of AF ​
​  through F(6.5, 4) is y = —
​   ​ x + —
⋅
⋅
—
23
2
9
9
2+8 3+1
10 4
—
​  ​ = ​ —
​   ​   ​ = (5, 2).
The midpoint of AB
​ ​ is ​ —
​   ​, ​ —
​   ​,  —
2
2
2 2
2 − 3 −1
—
​
= —
​   ​ = undefined.
The slope of CD
​ ​  is —
​
5−5
0
(
) (  )
—
The equation of CD
​ ​  through D(5, 2) is x = 5.
The centroid has the coordinates of the intersection of
y = —
​ 29 ​x + —
​ 23
​ and x = 5.
9
​   ​
y = —
​   ​ x + —
23
2
9
9
2
23
y = —
​   ​  5 + —
​   ​
9
9
10 23
​   ​
y = —
​   ​ + —
9
9
33 11
​   ​
y = —
​   ​ = —
9
3
( −6 2+ 1 3 +2 5 ) ( −52 82 )
—
—
7−4
−2 − (−2.5)
3
−2 + 2.5
3
0.5
⋅ (−2.5) + b
4 = −15 + b
19 = b
—
The equation of GE
​ ​  through E(−2.5, 4) is y = 6x + 19.
The centroid has the coordinates of the intersection of
y = 6x + 19 and y = 5.
5 = 6x + 19
−14 = 6x
7
14
​ = −​ —  ​
x = −​ —
6
3
7
The centroid has coordinates ​ −—
​   ​ , 5  ​.
3
(  )
17.
y
S(5, 5)
4
⋅
2
(5, 1)
E(8, 1)
U(−1, 1)
−4
8
−2
(  )
11
The centroid has coordinates ​ 5, —
​  3  ​  ​.
Geometry
Worked-Out Solutions
12 x
D(5, −1)
T(11, −3)
—
(
) (  )
16 2
5 + 11 5 + (−3)
2
2 2
2
0
1−1
—
​   ​ = 0.
The slope of UE
​ ​  is —
​
​ = —
8 − (−1) 9
—
The equation of UE ​
​  through E(8, 1) is y = 1.
​
​, ​ —
The midpoint of ST
​ ​ is ​ —
​
​ = ​ —
​   ​,  ​ —   ​  ​ = (8, 1).
—
The midpoint of UT
​  ​ is
​,
​ —
​
​   ​  ​ = (5, −1).
​ —
​
​ = ​ —
​   ​,  —
​
= —
​   ​ = undefined.
The slope of SD
​ ​  is —
​
206
2 x
—
The midpoint of HG
​  ​ is
( −1 +2 11 1 + 2(−3) ) ( 102 −22 )
—
−1 − 5 −3
5−5
0
—
The equation of SD
​ ​  through D(5, −1) is x = 5.
The centroid has the coordinates of the intersection of x = 5
and y = 1, which is (5, 1).
Chapter 6
18.
4
3
19.
y X(1, 4)
L(0, 5) y
4
D(4, 3)
y=1
Z(2, 3)
2
(103, 3)
2
Y(7, 2)
E(4.5, 2.5)
1
M(3, 1)
−2
4
N(8, 1)
6
8 x
−2
2
4
6
8 x
—
(
y = 2x − 5
) (  )
8 6
1+7 4+2
2
2 2
2
3−3 0
—
​ = —
​   ​ = 0.
The slope of ZD
​ ​  is —
​

4−2 2
—
The equation of ZD ​
​  through D(4, 3) is y = 3.
​  ​ = ​ —
​    ​  ​ = (4.5, 2.5).
The midpoint of ZY
​ ​  is ​ —
​   ​, ​ —
​   ​ , —
—
(
) (  )
⋅
⋅
⋅ (
)
62 = 14b
62 31
​   ​
b = —
​   ​ = —
14
7
​   ​.
The equation of XE
​ ​  through E(4.5, 2.5) is y = −​ —  ​ x + —
—
3
31
7
7
The centroid has the coordinates of the intersection of
3
31
​   ​ and y = 3.
y = −​ —  ​ x + —
7
7
3
31
​   ​
y = −​ —  ​ x + —
7
7
3
31
​   ​
3 = −​ —  ​ x + —
7
7
3
31
7 3 = 7 ​ −—
​   ​   ​x + 7 ​ —
​   ​   ​
7
7
21 = −3x + 31
⋅
⋅ (
−10x =−3x
(0, −5)
​  ​ = ​ —
​   ​   ​ = (4, 3).
The midpoint of XY
​  ​is ​ —
​   ​, ​ —
​   ​ , —
9 5
2 + 7 3 + 2
2
2 2
2
3
15
1.5
4 − 2.5
—
​ = —
​
The slope of XE
​ ​  is —
​

​ = −​ —  ​ = −​ —  ​ .
35
7
1 − 4.5 −3.5
y = mx + b
3 9
2.5 = −​ —  ​  —
​    ​ + b
7 2
27
5
​   ​ = −​ —  ​ + b
—
14
2
27
5
14 —
​   ​ = 14 ​ −—
​   ​   ​ + 14b
14
2
35 = −27 + 14b
−4
)
⋅ (
—
1−1 0
8−3 5
—
The slope of the line perpendicular to MN
​ ​ is undefined and
passes through (0, 5). Therefore, the equation of that line
is x = 0.
1
4
5−1
—
​ = —
​     ​ = −​ —  ​ .
The slope of the line containing LN ​
​  is —
​

2
0 − 8 −8
—
The slope of the line perpendicular to LN
​ ​ is 2.
y = 2x + b
1=2 3+b
1=6+b
−5 = b
​ = —
​   ​ = 0.
The slope of the line containing MN
​ ​  is —
​

y = mx + b
⋅
—
The equation of the line perpendicular to LN ​
​  containing point
(3, 1) is y = 2x − 5.
The orthocenter is the intersection of x = 0 and y = 2x − 5.
y = 2x − 5
⋅0 − 5
y=2
y = −5
The orthocenter of △LMN is located on the outside and the
coordinates are (0, −5).
20. △ XYZ is a right triangle; therefore, the orthocenter is on the
triangle at the intersection of the legs.
y
Z(−3, 6)
)
6
4
Y(5, 2)
X(−3, 2)
10
​   ​ = x
—
3
10
The centroid has coordinates ​ —
​   ​,  3  .​
3
(  )
−2
2
4
x
The coordinates of the orthocenter are (−3, 2).
Geometry
Worked-Out Solutions
207
Chapter 6
21.
C(−1, 3)
(−1, 2)
y
6
4
(0, 73)
4
V(0, 4)
B(1, 0)
A(−4, 0)
−6
22.
y
x = −1
T(−2, 1)
−2
U(2, 1)
2 x
−2
2
8
2
y = 3x + 3
2
4 x
7
y = 3x + 3
—
0
0−0
1 − (−4) 5
—
The slope of the line perpendicular to AB ​
​  is undefined and
passes through (−1, 3). Therefore, the equation of that line is
x = −1.
3
3
3−0
—
​ = —
​     ​ = −​ —  ​ .
The slope of the line containing CB
​ ​  is —
​

2
−1 − 1 −2
2
—
The slope of the line perpendicular to CB ​
​  is —
​   ​ .
3
y = mx + b
​   ​ = 0.
The slope of the line containing AB ​
​  is —
​
​ = —
​   ​ = 0.
The slope of the line containing TU ​
​  is —
​
​ = —
0 = —
​   ​  (−4) + b
2
3
⋅
8
0 = −​ —  ​ + b
3
8
​   ​  = b
—
3
—
The equation of the line perpendicular to CB
​ ​  containing
8
2
​   ​ .
point (−4, 0) is y = —
​   ​ x + —
3
3
The orthocenter is the intersection of x = −1 and
8
2
​   ​ .
y = —
​   ​ x + —
3
3
8
2
​   ​
y = —
​   ​ x + —
3
3
2
8
y = —
​   ​  (−1) + —
​   ​
3
3
2 8
​    ​
y = −​ —  ​ + —
3 3
6
y = —
​   ​ = 2
3
The orthocenter of △ABC is inside the triangle with
coordinates (−1, 2).
⋅
—
0
1−1
2 − (−2) 4
—
The slope of the line perpendicular to TU ​
​  is undefined and
passes through (0, 4). Therefore, the equation of that line
is x = 0.
3
3
4−1
—
​
= —
​     ​ = −​ —  ​ .
The slope of the line containing VU ​
​  is —
​
2
0 − 2 −2
2
—
The slope of the line perpendicular to VU ​
​  is —
​   ​ .
3
y = mx + b
⋅
2
1 = —
​   ​  (−2) + b
3
4
1 = −​ —  ​ + b
3
4
3 1 = 3 ​ −—
​    ​  ​ + 3 b
3
3 = −4 + 3b
⋅ (
⋅
)
⋅
7 = 3b
7
b = —
​   ​
3
—
The equation of the line perpendicular to VU
​ ​  containing
7
2
​   ​ .
point (−2, 1) is y = —
​   ​ x + —
3
3
7
2
​   ​ .
The orthocenter is the intersection of x = 0 and y = —
​   ​ x + —
3
3
7
2
​   ​
y = —
​   ​ x + —
3
3
2
7
y = —
​   ​  0 + —
​   ​
3
3
7
y = —
​   ​
3
The orthocenter of △TUV is inside the triangle with
7
coordinates ​ 0, —
​   ​   ​.
3
⋅
(  )
23. Construct the medians of an isosceles right triangle by
finding the midpoint of each side and connecting the
midpoint and the vertex opposite that midpoint. Where
the medians intersect is the location of the centroid. The
orthocenter is on the triangle at the right angle.
centroid
orthocenter
208
Geometry
Worked-Out Solutions
Chapter 6
24. Construct the medians of an obtuse scalene triangle by
finding the midpoint of each side and connecting the
midpoint and the vertex opposite that midpoint. Where the
medians intersect is the location of the centroid. Construct
the altitudes by drawing a perpendicular segment from each
vertex to the opposite side or to the line that contains the
opposite side. The orthocenter is located outside the triangle
where the altitudes intersect.
—
—
27. The length of DE
​ ​ should be —
​ 13 ​of the length of AE
​ ​ because it
is the shorter segment from the centroid to the side.
1
DE = —
​ 3 ​ AE
DE = —
​ 3 ​(18)
DE = 6
1
—
—
28. The length of DE
​ ​ is —
​ 12 ​of the length of AD
​ ​ because DE = —
​ 13 ​ AE
2
​ 3 ​ AE.
DE = —
DE = —
​ 2 ​(24)
DE = 12
centroid
1
1
—
29. Given Isosceles △ABC, AD ​
​  is an angle bisector of ∠ ABC.
orthocenter
25. Construct the medians of a right scalene triangle by finding
the midpoint of each side and connecting the midpoint
and the vertex opposite that midpoint. Where the medians
intersect is the location of the centroid. The orthocenter is on
the triangle at the right angle.
—
Prove ​
BD ​ is a median.
B
A
centroid
orthocenter
26. Construct the medians of an acute isosceles triangle by finding
the midpoint of each side and connecting the midpoint and the
vertex opposite that midpoint. Where the medians intersect is
the location of the centroid. Construct the altitudes by drawing
a perpendicular segment from each vertex to the opposite side.
The orthocenter is where the altitudes intersect.
centroid
D
C
STATEMENTS
REASONS
1. △ ABC is an
isosceles triangle.
1. Given
2. —
​AB ​ ≅ —
​BC ​
2. Definition of isosceles
triangle
3. —
​AD ​ is an angle bisector
of ∠ ABC.
3. Given
4. △ABD ≅ ∠ CBD
4. Definition of angle bisector
5. —
​BD ​ ≅ —
​BD ​
5. Reflexive Property of
Congruence (Thm. 2.1)
6. SAS Congruence
Theorem (Thm. 5.5)
6. △ABD ≅ △CBD
7. —
​CD ​
orthocenter
8. D is the midpoint of —
​AC  ​.
​BD ​ is a median.
9. —
7. Corresponding parts of
congruent triangles are
congruent.
8. Definition of midpoint
9. Definition of median
Geometry
Worked-Out Solutions
209
Chapter 6
30. Given Isosceles △ABC, —
—
​  is an altitude to AC ​
BD ​
​ .
35. The centroid and orthocenter are sometimes the same point.
The centroid and the orthocenter are not the same point
unless the triangle is equilateral.
B
—
Prove ​
BD ​ is a perpendicular bisector.
36. The centroid is always formed by the intersection of the
three medians. This is the definition of a centroid.
A
STATEMENTS
1. △ ABC is an
isosceles triangle.
2. —
​AB ​ ≅ —
​BC ​
3. —
​BD ​ is an
altitude to —
​AC ​.
4. —
​BD ​ ⊥ —
​AC ​
5. ∠ ADB and ∠ CDB
are right angles.
6. —
​BD ​ ≅ —
​BD ​
7. △ABD ≅ △CBD
8. —
​CD ​
9. D is the midpoint
of —
​AC  ​.
—
​BD ​ is a
10. perpendicular
bisector.
D
C
REASONS
1. Given
2. Definition of isosceles
triangle
3. Given
4. Definition of altitude
5. Definition of perpendicular
6. Reflexive Property of
Congruence (Thm. 2.1)
7. HL Congruence Theorem
(Thm. 5.9)
8. Corresponding parts of
congruent triangles are
congruent.
9. Definition of midpoint
10. Definition of perpendicular
bisector
31. The centroid is never on the triangle. Because medians are
always inside a triangle, and the centroid is the point of
concurrency of the medians, it will always be inside the
triangle.
32. The orthocenter is sometimes outside the triangle. An
orthocenter can be inside, on, or outside the triangle
depending on whether the triangle is acute, right, or obtuse.
33. A median is sometimes the same line segment as a
perpendicular bisector. A median is the same line segment
as the perpendicular bisector if the triangle is equilateral
or if the segment is connecting the vertex angle to the base
of an isosceles triangle. Otherwise, the median and the
perpendicular bisectors are not the same segment.
34. An altitude is sometimes the same line segment as an angle
bisector. An altitude is the same line segment as the angle
bisector if the triangle is equilateral or if the segment is
connecting the vertex angle to the base of an isosceles
triangle. Otherwise, the altitude and the angle bisector are
not the same segment.
210
Geometry
Worked-Out Solutions
37. Both segments are perpendicular to a side of a triangle, and
their point of intersection can fall either inside, on, or outside
of the triangle. However, the altitude does not necessarily
bisect the side, but the perpendicular bisector does. Also, the
perpendicular bisector does not necessarily pass through the
opposite vertex, but the altitude does.
38. All are segments that pass through the vertex of a triangle.
A median connects a vertex with the midpoint of the opposite
side. An altitude is perpendicular to the opposite side. An
angle bisector bisects the angle through which it passes.
The medians of a triangle intersect at a single point, and the
same is true for the altitudes and angle bisectors of a triangle.
Medians and angle bisectors always lie inside the triangle, but
altitudes may be inside, on, or outside of the triangle.
39. Area = —
​ 12 ​bh
The area of the triangle in solid red is
—
​ 2 ​ —
​ 2 ​ 3 = —
​  4  ​ = 6.75 square inches.
The special segment of the triangle used was the altitude.
1
⋅⋅
9
27
—
—
—
40. K is the centroid and DH
​ ​, EJ
​ ​,  and FG
​ ​ are medians.
a.EJ = 3KJ
b.DK = 2KH
c.FG = —
​ 2 ​FK
d.KG = —
​ 3 ​FG
F
3
J
1
K
H
D
G
41.
BD = —
​ 23 ​BF
2
⋅
42.
E
GD = —
​ 13 ​GC
1
4x + 5 = —
​ 3 ​ 9x
2x − 8 = —
​ 3 ​(3x + 3)
4x + 5 = 6x
2x − 8 = x + 1
5 = 2x
​ 52 ​ =
—
x=9
x
44.
DF = —
​ 12 ​BD
1
5x = 2(3x − 2)
4x − 1 = —
​ 2 ​(6x + 4)
5x = 6x − 4
4x − 1 = 3x + 2
−x = −4
x=3
x=4
Chapter 6
45.
—
51. Given △ABC is an equilateral triangle. BD
​ ​ is a median.
y
median
8
3
y2 = 4 x + 5
—
Prove ​
BD ​ is angle bisector, 6
4
A
median
(0, 2)
−8
−6
−4
2
median
4
y1 = 3x − 4
−4
y3 = −2 x − 4
x
y2
y-intercept (0, 5).
STATEMENTS
(
3. Given
4. —
​CD  ​
5. —
​BD  ​ ≅ —
​BD ​
)
2. Definition of equilateral
triangle
3. —
​BD ​ is a median.
20
3
C
1. Given
2. —
​AB ​ ≅ —
​AC ​  ≅ —
​BC ​
​ —
​  43 ​,
D
REASONS
1. △ ABC is an
equilateral triangle.
(  0  )​and the
y-intercept (0, −4).
= —
​  ​x + 5: Graph the x-intercept (​  −—
​  ​, 0  )​and the
4. Definition of median
5. Reflexive Property of
Congruence (Thm. 2.1)
—
y3 = −​ —
2 ​x − 4: Graph the x-intercept ​ −​  3 ​, 0  ​and the
6. △ABD ≅ △CBD
6. SSS Congruence
Theorem (Thm. 5.8)
The points of intersection that form the triangle are (4, 8),
∠ABD ≅ ∠CBD
7. Corresponding parts of
congruent triangles are
congruent.
form a linear pair and
are supplementary.
8. Linear Pair Postulate
are right angles.
9. If two angles are congruent
and supplementary then
they are right angles.
3
y-intercept (0, −4).
3
y1 = 3x − 4: Graph the x-intercept
6
−2
3

4
B
perpendicular bisector,
and altitude.
8
(−4, 2), and (0, −4). The equation of the median from y1
to the opposite vertex is y = 2. The equation of the median
from y2 to the opposite vertex is x = 0. The coordinates of
the centroid are (0, 2).
46. right triangle; The orthocenter of a right triangle is the vertex
of the right angle.
47. PE = —
​ 13 ​AE, PE = —
​ 12 ​AP, PE = AE − AP
10. —
​BD ​ ⊥ —
​AC ​
—
48.a.KM
​ ​ is a median. It contains the centroid.
​CD ​
11. —
—
b.KN
​ ​ is an altitude. It contains the orthocenter.
c.The area of △JKM is —
​ 2 ​ 9
△ KLM is —
​ 2 ​ 9 h = —
​ 2 ​h, which indicates that the two
areas are equal. Yes, triangles formed by the median will
1
1
⋅⋅
9
10. Definition of perpendicular
⋅ ⋅ h = — ​  ​h and the area of
11. Corresponding parts of
congruent triangles are
congruent.
9

2
—
​AD ​ is a perpendicular 13. Definition of
13. bisector.
perpendicular bisector
always have the same area because they will have the
same base length and height.
49. yes; If the triangle is equilateral, then the perpendicular
bisectors, angle bisectors, medians, and altitudes will all be
the same three segments.
50. centroid; Because the triangles formed by the median of any
triangle will always be congruent, the mass of the triangle on
either side of the median is the same. So, the centroid is the
point that has an equal distribution of mass on all sides.
14. —
14. Definition of altitude
52. The orthocenter, circumcenter, and the centroid are all inside
the triangle. They are all three distinct points. The three
concurrent points are all collinear.
y
A
4
orthocenter
3
circumcenter
2
1
B
centroid 1
12. Definition of midpoint
12. D is the midpoint
​BC ​.
of —
C
2
3
4
5
x
Geometry
Worked-Out Solutions
211
Chapter 6
53. Sample answer: The circle passes through nine significant
points of the triangle. They are the midpoints of the sides, the
midpoints between each vertex and the orthocenter, and the
points of intersection between the sides and the altitudes.
y
A
9
8
7
6
K
I
J
5
D
4
E
B
L
1
1
2
3
C
4
5
6
x
—
—
54. Given LP
​ ​ and MQ
​ ​ are medians of scalene △LMN. Point R
—
—
​  such


is on ​LP ​ such that LP ​
​  ≅ PR ​
​ . Point S is on MQ​
—
that MQ ​
​— ≅ QS ​
​ .
— —
—
—
—
b. NS ​
​  and NR ​
​  are both parallel to LM ​
​ .
c. R, N, and S are collinear.
Prove a. NS
​ ​  ≅ NR ​
​
R
M
REASONS
1. Given
2. Definition of median
3. ∠ LPM ≅ ∠ RPN,
∠ MQL ≅ ∠ SQN
3. Vertical Angles
Congruence Theorem
(Thm. 2.6)
4. △LPM ≅ △RPN,
△MQL ≅ △SQN
4. SAS Congruence
Theorem (Thm. 5.5)
6. ​—
NS  ​≅ —
​NR  ​
H
G
1. ​—
LP  ​and —
​MQ  ​ are
medians of scalene
△LMN; —
​LP  ​≅ —
​PR  ​,
​—
MQ  ​≅ —
​QS  ​
NP  ​≅ —
​MP  ​, —
​ LQ  ​≅ —
​NQ  ​
2. ​—
5. ​—
NR  ​≅ —
​LM  ​, —
​NS  ​ ≅ —
​LM  ​ 5. Corresponding parts of
congruent triangles are
congruent.
F
3
2
a. STATEMENTS
P
N
Q
L
S
6. Transitive Property of
Congruence (Thm. 2.1)
b.It was shown in part (a) that △LPM ≅ △RPN and
c.Because NS
​ ​ and NR
​ ​ are both parallel to the same segment, ​
△MQL ≅ △SQN. So, ∠ LMP ≅ ∠ RNP and
∠ MLQ ≅ ∠ SNQ because corresponding parts of
—
—
congruent triangles are congruent. Then, NS
​ ​  LM
​ ​  and
—
—

​ ​  LM
NR
​ ​ by the Alternate Interior Angles Converse (Thm. 3.6).
— —
—
LM ​, they would have to be parallel to each other by the
Transitive Property of Parallel Lines (Thm. 3.9). However,
because they intersect at point N, they cannot be parallel.
So, they must be collinear.
Maintaining Mathematical Proficiency
−3 1
3−6
—
55. Slope of AB ​
​ = —
​    ​ = —
​   ​
​ :  ​ —

−1 − 5 −6 2
−6
−6
1
3−9
—
Slope of CD ​
​
​
​   ​
​ :  ​ ——
​ = —
​ = —
​ = —
−16 − (−4) −16 + 4 −12 2
—
—
—
—
The slopes of AB
​ ​ and CD
​ ​ are equal, so AB ​
​   CD ​
​ .
1
−2
−2
4−6
—
​
​   ​ = −​ —   ​
​ = —
​ = —
​ :  ​ —
56. Slope of AB ​
4
5 − (−3) 5 + 3
8
3
1
−7 − (−10) −7 + 10
—
Slope of CD ​
​ = —
​
​ = —
​    ​ = —
​   ​
​ :  ​ ——

−2 − (−14) −2 + 14 12 4
—
—
—
The slopes of AB
​ ​ and CD
​ ​ are not equal, so AB ​
​  is not parallel
to CD ​
​—.
5
2 − (−3) 2 + 3
—
​
= ​ —
​
= —
​     ​ = −5
​ :  ​ —
57. Slope of AB ​
5−6
−1
−1
2+4
6
2 − (−4)
—
Slope of CD ​
​ = —
​
​ = —
​     ​ = −6
​ : ​  —

−5 − (−4) −5 + 4 −1
—
—
—
The slopes of AB
​ ​ and CD
​ ​ are not equal, so AB ​
​  is not parallel
to CD ​
​—.
−4
−4
2−6
—
​
​    ​ = 2
​ :  ​ —
​ = —
​ = —
58. Slope of AB ​
−7 − (−5) −7 + 5 −2
−5 − 1 −6
—
Slope of CD ​
​
= ​ —   ​ = 2
​ : —
​
4−7
−3
212
Geometry
Worked-Out Solutions
—
—
—
—
The slopes of AB
​ ​ and CD
​ ​ are equal, so AB ​
​  is parallel to CD ​
​ .
Chapter 6
6.1– 6.3 What Did You Learn? (p. 331)
4. Graph △ABC.
1. Sample answer: yes; You realize that if you construct the
—​, you have created two isosceles triangles that
segment ​AC
share vertices A and C, and a third triangle that shares the
same vertices. Then you look back at the Perpendicular
Bisector Theorem (Thm. 6.1) and its converse to see that points
D and E would have to be on the perpendicular bisector of ​—​. Then, in the same way, in order for B to also be on the same
AC
—
—
perpendicular bisector, AB
​ ​ and CB
​ ​ would have to be congruent.
2. Sample answer: These can be constructed with a compass
and a straightedge, or they can be constructed with geometry
software; If you construct them with geometry software, you
create a triangle that fits the description first. Then use the
software to draw the perpendicular bisectors of each side.
Next, label the point where these three lines meet. Finally,
draw a circle with its center at this point of intersection
that passes through one vertex of the triangle. It will
automatically pass through the other two vertices.
3. Sample answer: Right triangles are the only kind of
triangles that have one of the points of intersection on the
vertex of the triangle; While all segment types can be inside
the triangle, only the perpendicular bisectors and altitudes
can be on or outside the triangle.
—
—
,  point V is on the
​  ⊥ ​SU ​
1. Because SW ​
​  ≅ UW ​
​  and VW​
perpendicular bisector of SU ​
​—.
SV = UV
−6x + 11 = −1
−6x = −12
x=2
UV = 8x − 1 = 8
Perpendicular Bisector Theorem (Thm. 6.1)
2x + 11 = 8x − 1
—
 is an angle bisector of ∠ PSR, PQ ​
 , and RQ ​
 .
2. ​SQ ​
​— ⊥ ​SP ​
​  ⊥ ​SR ​
PQ = RQ
6x = 3x + 9
3x = 9
PQ = 6x = 6
Angle Bisector Theorem (Thm. 6.3)
x=3
⋅ 3 = 18
​  and GK​
 bisects
3. Because J is equidistant from GH​
​  , ​GJ ​
∠ HGK by the Angle Bisector Theorem (Thm. 6.3).
m∠ HGJ = m∠ JGK
x−4=3
x=7
m∠ JGK = 4x + 3 = 4
31° + m∠ GJK + 90° = 180°
y
2
−6
2 x
−2
B(−4, −4)
(−2, −4)
C(0, −4)
( −4 +2 (−4) 2 + 2(−4) )
−8 −2
= (​  ​   ​,  ​   ​  )​ = (−4, −1)
2 2
−4 + 0 2 + (−4)
—
​,
​
​
midpoint of AC ​
​  = (​  ​
)​
2
2
−4 −2
= (​  ​   ​,  ​   ​  )​ = (−2, −1)
2 2
—
​,
​ —
​
midpoint of AB ​
​  = ​ —
​
​
The equation of the perpendicular bisector of AB ​
​  through its
—  —
— —
—  —
midpoint (−4, −1) is y = −1.
3
6
2 − (−4)
—
​
= —
​     ​ = −​ —  ​
slope of AC ​
​  = —
​
2
−4 − 0
−4
—
The slope of the perpendicular line to AC ​
​  is —
​ 23 ​.
y = mx + b
2
−1 = —
​   ​  (−2) + b
3
4
−1 = −​ —  ​ + b
3
4
3 (−1) = 3 ​ −—
​   ​   ​ + 3b
3
−3 = −4 + 3b
⋅ 2 − 1 = 16 − 1 = 15
D(−2, −1)
A(−4, 2)
(−4, −1)
6.1–6.3 Quiz (p. 332)
—
⋅
⋅
⋅ (
)
1 = 3b
1
​   ​ = b
—
3
—
The equation of the perpendicular bisector of AC ​
​  through its
2
1
midpoint (−2, −1) is y = ​ —
​x + —
​
​.
3
3
2
1
Find the point of intersection of y = −1 and y = —
​ 3 ​x + —
​ 3 ​.
1
2
​   ​
y = —
​   ​ x + —
3
3
2
1
−1 = —
​   ​ x + —
​   ​
3
3
−3 = 2x + 1
−4 = 2x
x = −2
The coordinates of the circumcenter of △ABC are (−2, −1).
5x − 4 = 4x + 3
⋅ 7 + 3 = 28 + 3 = 31°
m∠ GJK + 121° = 180°
Triangle Sum Theorem
(Thm. 5.1)
m∠ GJK = 59°
Geometry
Worked-Out Solutions
213
Chapter 6
9.
5. Graph △DEF.
16
6
8
8
4
−2
16 x
—
0
5−5
11 − 3 8
—
The slope of the line perpendicular to DF ​
​  is undefined, so the
—
equation of the line perpendicular to DF ​
​  is x = 7.
—
( 3 +2 7 5 +2 9 ) ( 102 142 )
​  ​, ​ —
​   ​   ​ = (5, 7)
midpoint of DE ​
​  = ​ —
​   ​, ​ —
​   ​,  —
​ = —
​   ​ = 1
slope of DE ​
​  = —
​

—
9−5 4
7−3 4
—
The slope of the line perpendicular to DE ​
​  is −1.
y = mx + b
⋅5 + b
7 = −1
7 = −5 + b
12 = b
—
The equation of the line perpendicular to DE ​
​  through the
midpoint (5, 7) is y = −x + 12. The intersection of x = 7
and y = −x + 12 is y = −7 + 12 = 5. The coordinates of
the circumcenter of △DFE are (7, 5).
NQ = NR
2x + 1 = 4x − 9
−2x + 1 = −9
−2x = −10
x = 5
NU = NV
7.
−3x + 6 = −5x
6 = −2x
−3 = x
NQ = NR = NS
= 2x + 1
= −3 (−3) + 6
= 2 5 + 1
=9+6
= 11
= 15
8.
4x − 10 = 3x − 1
x − 10 = −1
NZ = NY = NW = 4x − 10 = 4
x=9
⋅ 9 − 10 = 36 − 10 = 26
L(5, −2)
​
​ —
​   ​, ​ —
​ = ​ —
​   ​,  ​ —  ​   ​ = (5, 2).
​   ​ = 0.
The slope of JD ​
​  is —
​
​ = —
( 5 +2 5 6 + 2(−2) ) ( 102 42 )
—
0
2−2
5 − (−1) 6
—
The equation of JD ​
​  through D(5, 2) is y = 2.
—
The midpoint F of JL ​
​  is
5 + (−1) −2 + 2
4 0
​,
​
​
​   ​   ​ = (2, 0).
​ —
​
​ = ​ —
​   ​ , —
—
2
2
2 2
6−0 6
—
​ = —
​   ​ = 2.
The slope of KF ​
​  is —
​

5−2 3
y = mx + b
0=2 2+b
0=4+b
−4 = b
—
The equation of KF ​
​  through F(2, 0) is y = 2x − 4.
The centroid has the coordinates of the intersection of y = 2
and y = 2x − 4.
2 = 2x − 4
6 = 2x
3=x
The coordinates of the centroid are (3, 2).
(
) (  )
⋅
10.
y
−8
−6
−4
−2
N(−4, −2)
H(−4, −4)
D(−6, −4)
⋅
NZ = NY
x
= −2x + 6
6
F(2, 0)
—
The midpoint D of KL ​
​  is
NU = NV = NT
⋅
4
−2
​ = —
​   ​ = 0
slope of DF ​
​  = —
​

6.
D(5, 2)
J(−1, 2)
G(9, 7)
F(11, 5)
(7, 5)
4 D(3, 5)
H(3, 2)
4
E(7, 9)
K(5, 6)
y
H(5, 7)
y
M(−8, −6)
2 x
−2
P(0, −4)
F(−4, −5)
−8
—
The midpoint D of MN ​
​  is
​,
​ —
​
​   ​  ​ = (−6, −4).
​ —
​
​ = ​ —
​   ​, —
(
) (
)
−4 + (−8) −2 + (−6)
−12 −8
2
2
2
2
−4 − (−4) 0
—
​
= —
​   ​ = 0.
The slope of DP ​
​  is —
​
0 − (−6)
6
—
The equation of DP ​
​  through D(−6, −4) is y = −4.
—
The midpoint F of MP ​
​  is
​
​  ​ = (−4, −5).
​,
​
​ —
​
​ = ​ —
​   ​,  ​ —
—
(
) (
)
−8 −10
0 + (−8) −4 + (−6)
2
2
2
2
−2 − (−5) −2 + 5 3
—
​ = —
​
​ = —
​   ​ = undefined.
The slope of NF ​
​  is —
​

−4 − (−4) −4 + 4 0
—
The equation of NF ​
​  through F(−4, −5) is x = −4.
The centroid has the coordinates of the intersection of y = −4
and x = −4. So, the coordinates of the centroid are (−4, −4).
214
Geometry
Worked-Out Solutions
Chapter 6
11.
13.a.The point of concurrency at the center of the circle is the
—
—
—
incenter. BG ​
​ , CG ​
​ , and AG ​
​  are angle bisectors.
y
T(−2, 5)
6
V(2, 5)
b.Sample answer: △BGF ≅ △BGE by HL Congruence
Theorem (Thm. 5.9).
1
y = −2 x + 4
c.Because △BGF ≅ △BGE and corresponding parts of
congruent triangles are congruent, BE = BF = 3 centimeters.
So, AE = 10 − 3 = 7 centimeters. Then you can use the
Pythagorean Theorem (Thm. 9.1) for △AEG to find EG,
which is the radius of the wheel.
U(0, 1)
−4
−2
2
4 x
—
0
5−5
2 − (−2) 4
—
The slope of the line perpendicular to TV ​
​  is undefined and
passes through U(0, 1). Therefore, the equation of that line
is x = 0.
AG2 = GE2 + EA2
82 = GE2 + 72
64 = GE2 + 49
15 = GE2
5−1 4
—
​
= —
​   ​ = 2.
The slope of the line containing UV ​
​  is —
​
GE = √
​ 15 ​ ≈ 3.9
The slope of the line perpendicular to UV ​
​  is −​ —  ​ .
y = mx + b
14.a.The point of concurrency used was the centroid.
​   ​ = 0.
The slope of the line containing TV ​
​  is —
​
​ = —
2−0
—
1
y = −​ —  ​ x + b
2
1
5 = −​ —  ​ (−2) + b
2
5=1+b
4=b
2
1
2
6.4 Explorations (p. 333)
—
The equation of the line perpendicular to UV ​
​  containing
point T(−2, 5) is y = −​ —  ​ x + 4.
The orthocenter is the intersection of x = 0 and y = −​ —  ​ x + 4.
1
2
1
2
1
y = −​ —  ​ x + 4
2
1
y = −​ —  ​  0 + 4
2
y=4
The coordinates of the orthocenter are located on the inside
b.The circumcenter should have been used because the
Circumcenter Theorem (Thm. 6.5) can be used to find a
point equidistant from the three points (the three cities).
—
⋅
1.a.Check students’ work.
— —
—
—
—
which indicates that DE ​
​   AC ​
​ , and the length of DE ​
​
1—
is ​ —
​ ​ AC ​
.
2
3 − 4.5 −1.5
—
​ = —
​
​ ≈ −0.43
slope of DE ​
​  = —
​

b. Sample answer: The slopes of DE ​
​  and AC ​
​  are equal
c.The segment connecting the midpoints of two sides of a
5 − 1.5
3.5
——
—
length of DE ​
​  = √
​
(1.5 − 5)2 + (4.5 − 3)2 ​ ≈ 3.8
−3
1−4
—
​   ​ ≈ −0.43
slope of AC ​
​  = —
​
​ = —
5 − (−2)
7
——
—
5 − (−2) )2​ + (1 − 4)2 ​ ≈ 7.6
length of AC ​
​  = √
​ (​
of the triangle and are (0, 4).
12.
4
y
Z(7, 4)
triangle is parallel to the third side and is half as long as
that side.
2.a.Check students’ work.
b. Sample answer: The triangle formed by the midsegments
2
4
6
of a triangle, △EFD, is similar to the original triangle,
△ABC. The side lengths of △EFD are —
​ 12 ​the side lengths
of △ABC.
x
−2
X(−1, −4)
Y(7, −4)
△ XYZ is a right triangle. Therefore, the orthocenter is on the
triangle at the intersection of the legs, which is (7,−4).
3. The midsegment that is formed by joining the midpoints of
two sides is parallel to the third side and is —
​ 12 ​the length of the
third side.
1
4. UV = ​ —
​RT
2
T
12 = —
​ 12 ​RT
2
⋅ 12 = 2 ⋅ — ​  ​RT
V
1

2
12
24 = RT
In △RST, if UV = 12,
R
U
S
then RT = 24.
Geometry
Worked-Out Solutions
215
Chapter 6
(
6.4 Monitoring Progress (pp. 334–336)
2 − (−1) 2 + 1 3
—
​
= ​ —
​
= ​ —   ​
1. Slope of DE ​
​ :  ​ —
2−0
2
2
0 − (−6) 6 3
—
​
= ​ —   ​ = —
​   ​
Slope of AC ​
​ :  ​ —
5−1
4 2
—
—
—
—
Because the slope of DE ​
​  equals the slope of AC ​
​ , DE ​
​   AC ​
​
because parallel lines have equal slopes.
——
2
2

—
2
2
—
—

——
2
2

—
2
2
—

—

—
—

—

) (  )
0 + 2p 0 + 0
2p
​, ​ —
​  ​ = ​ —
3. The midpoint F is ​ —
​
​   ​,  0  ​ = ( p, 0).
2
2
2
r
r−0
—
Slope of FE ​
​    ​
​ :  ​ —
​ = —
p+q−p q
r
2r − 0
—
Slope of OB ​
​ = —
​    ​
​ :  ​ —

2q − 0 q
—
—
—
—
Because the slopes of FE ​
​  and OB ​
​  are equal, FE ​
​   OB ​
​ .
——
——
2
2

—
2
2
—
—
2
2
2
2
DE = √
​ (2 − 0) + (​  2 − (−1) )​  ​
= ​√(2) + (3)  ​
OB = √
​ (2q− 0) + (2r − 0)  ​
​ 13 ​
= ​√4 + 9 ​ = √
=√
​ 4q + 4r  ​
AC = √
​ (5 − 1) + (​  0 − (−6) )​  ​
=√
​ 4(q + r ) ​ = 2​√ q + r  ​
= ​√(4) + (6)  ​
OB = 2FE
= ​√16 + 36 ​
FE = —
​   ​ OB
= ​√52 ​
= ​√4 13 ​ = 2​√ 13 ​
AC = 2​√ 13 ​
AC = 2DE
DE = —
​   ​ AC
V
R
(
) ( 26 −62 )
−1 + 5 4 + 0
4 4
—
​,
​   ​
The midpoint of BC ​
​  is (​  ​
​ = ​  ​   ​ , ​   ​   ​ = (2, 2).
2
2 ) (2 2)
5 + 1 0 + (−6)
2
2
— —
—  —
—
—
The midpoint F of AC ​
​  is (3, −3) and the midpoint E of BC ​
​
is (2, 2).
—
5
2 − (−3)
2−3
−1
−10
−6 − 4
—
​ = —
​   ​ = −5
Slope of AB ​
​ :  ​ —

1 − (−1)
2
—
—
EF ​
​   AB ​
​  because the slopes are equal.
AB = √
​
​( 1 − (−1) )2​ + (−6 − 4)2 ​
⋅
S
U
​
​   ​  ​ = (3, −3).
​ —
​   ​, ​ —
​ = ​ —
​   ​ , —
⋅
1
2
—
​  is
2. The midpoint of AC ​
1
2
—
​ . If UW = 81, then
4. The third midsegment is UW ​
ST = 2 81 = 162; therefore, VS = —
​ 12 ​ 162 = 81 inches.
⋅
​
= ​ —
Slope of EF ​
​ :  ​ —
​ = −5
———
——
2
2

—

—

—
—

——
2
2

——
2
2

—
—

—

—
FE = √
​
( p + q − p)2 + (r − 0)2 ​ = √
​ q2 + r2 ​
W
T
—
—
—
5. In Example 4, DF ​
​  is a midsegment; therefore, DF ​
​   AB ​
​  and ​
1—
—
DF ​ = —
​ 2 AB ​
​​ .
6. From Peach Street to Plum Street is 2.25 miles; from
Plum Street to Cherry Street is 1.4 miles; from Cherry Street
to Pear Street is 1.3 miles; from Pear Street to Peach Street
is ​ —
​  12 ​ 1.4  ​is 0.7 mile; from Pear Street back home is
(  ⋅ )
​(  ⋅ 2.25 )​is 1.125 miles. The total distance is
​ 12 ​
—
2.25 + 1.4 + 1.3 + 0.7 + 1.125 = 6.775 miles.
This route was less than that taken in Example 5.
6.4 Exercises (pp. 337–338)
= ​√(2) + (−10)  ​
Vocabulary and Core Concept Check
= ​√4 + 100 ​
= ​√104 ​
1. The midsegment of a triangle is a segment that connects the
midpoints of two sides of a triangle.
= ​√4 26 ​ = 2​√ 26 ​
⋅
EF = √
​ (3 − 2) + (−3 − 2)  ​
=√
​ (1) + (−5)  ​
= ​√1 + 25 ​ = √
​ 26 ​
AB = 2​√ 26 ​
AB = 2EF
EF = —
​   ​ AB
216
—
—
2. If DE ​
​  is the midsegment opposite AC ​
​  in △ABC, then
1 —
—
—
—
​   AC ​
DE ​
​  and DE ​
​  = —
​ 2 ​ AC ​
​  by the Triangle Midsegment
Theorem (Thm. 6.8).
Monitoring Progress and Modeling with Mathematics
3. The coordinates are D(−4, −2), E(−2, 0), and F(−1, −4).
1
2
Geometry
Worked-Out Solutions
Chapter 6
2
0 − (−2)
—
​ = —
​   ​ = 1
4. Slope of DE ​
​ :  ​ —

−2 − (−4) 2
−2 − (−6) −2 + 6 4
—
Slope of CB ​
​
= ​ —
​
= —
​   ​ = 1
​ :  ​ —
1 − (−3)
4
4
—
—— —
​ 12 ​BC
7. DE = —
1
8. DE = —
​ 2 ​AB
x = —
​ 12 ​(26)
5 = —
​ 12 ​(AB)
x = 13
x = 10
Because the slope of DE ​
​  equals the slope of CB ​
​ , DE ​
​   CB ​
​ .
DE = √
​ ​(
−2 − (−4) )​2 + ​( 0 − (−2) )​2 ​
=√
​ (2)2 + (2)2 ​
=√
​ 4 + 4 ​
=√
​ 8 ​ = √
​ 4 2 ​ = 2​√ 2 ​
CB = √
​ ​( 1 − (−3) )​ + (​  −2 − (−6) )​  ​
=√
​ (4) + (4)  ​
=√
​ 16 + 16 ​
17.
=√
​ 16 2 ​ = 4​√ 2 ​
3x + 8 = —
​ 2 ​(2x + 24)
3y − 5 = —
​ 2 ​(4y + 2)
Because 2​√ 2 ​ = —
​   ​​ ( 4√
​ 2 ​  )​, DE = —
​   ​ CB.
3x + 8 = x + 12
3y − 5 = 2y + 1
2x + 8 = 12
y−5=1
2x = 4
x = 2
GL = 2
AB = —
​ 2 ​(28) = 14
———
—
—
—
⋅
—
—
———
2
2

—
2
2
—

—
—

⋅
1
2
—
—
1
2
−4
−4
−4 − 0
—
​ = —
​
​   ​ = −4
5. Slope of EF ​
​ :  ​ —

​ = —
−1 − (−2) −1 + 2
1
−8
−8
−6 − 2
—
​ = —
​
​   ​ = −4
Slope of AC ​
​ :  ​ —

​ = —
−3 − (−5)
−3 + 5
2
—
—
—
—
Because the slope of EF ​
​  equals the slope of AC ​
​ , EF ​
​   AC ​
​ .
———
2
2

——
2
2

—
—

———
2
2

——
2
2

—

—
—

EF = √
​ ​( −1 − (−2) )​ + (−4 − 0)  ​
= ​√(1) + (−4)  ​
= ​√1 + 16 ​ = √
​ 17 ​
AC = √
​ ​( −3 − (−5) )​ + (−6 − 2)  ​
10. BE = EC
6 = x
12.JL ​
​   XZ ​
​
— —
—
—
13.XY ​
​   KL ​
​
14.JY
​ ​  ≅ JX
​ ​  ≅ KL
​ ​
—
—
—
15.JL ​
​  ≅ XK ​
​  ≅ KZ ​
​
16.JK ​
​  ≅ YL ​
​  ≅ LZ ​
​
— — —
⋅
Because √
​ 17 ​ = —
​   ​​ ( 2​√ 17 ​  )​, EF = —
​   ​ AC.
1
2
1
2
—
2
−4 − (−2) −4 + 2
—
6. Slope of DF ​
​ = —
​
​ = −​ —  ​
​ :  ​ —

3
−1 − (−4) −1 + 4
2
−4
−4
−2 − 2
—
Slope of AB ​
​ = —
​
​   ​ = −​ —  ​
​ :  ​ —

​ = —
3
1 − (−5) 1 + 5
6
—
—— —
Because the slope of DF ​
​  equals the slope of ​AB ​, DF ​
​   AB ​
​ .
DF = √
​
​( −1 − (−4) )​2 + (​  −4 − (−2) )​2 ​
AB = √
​
​( 1 − (−5) )​2 + (−2 − 2)2 ​
2 + (−4)2 ​
=√
​ (6)

=√
​ 36 + 16 ​
=√
​ 52 ​ = √
​ 4 13 ​
= 2​√ 13 ​
———
—
——
—
2 + (−2)2 ​ = √
=√
​ (3)

​ 9 + 4 ​
=√
​ 13 ​
———
——
⋅
—

—
⋅
=3 6−5
= 18 − 5 = 13
⋅ — ​  ​(7z − 1)
1

2
8z − 6 = 7z − 1
z − 6 = −1
z=5
GA = CB
CB = 4z − 3 = 4
GA = 17
⋅ 5 − 3 = 20 − 3 = 17
—
—
—
20.DE
​ ​ is not parallel to BC
​ ​. So, DE
​ ​ is not a midsegment. So,
according to the contrapositive of the Triangle Midsegment
Theorem (Thm. 6.8), DE
​—​ does not connect the midpoints of ​
—
—
AC ​ and AB
​ ​.
21. The distance between first base and second base is 90 feet.
Because the shortstop is halfway between second and third
bases, and the pitcher is halfway between first and third
bases, using the Triangle Midsegment Theorem, the distance
between the shortstop and the pitcher is —
​ 12 ​ 90 = 45. So, the
distance between the shortstop and the pitcher is 45 feet.
⋅
—
—
HB = AC
HB = 3y − 5
4z − 3 = —
​ 12 ​(7z − 1)
—
y=6
CB = —
​ 12 ​(GA)
⋅
1
= ​√(2) + (−8)  ​
= ​√4 17 ​ = 2​√ 17 ​
= 4 + 24 = 28
2(4z − 3) = 2
AC = —
​ 12 ​(HJ)
1
2 + 24
18.
1
= ​√4 + 64 ​
— — —
AB = —
​ 12 ​(GL)
x=8
—
—
11.JK ​
​   YZ ​
​
19.
9. AE = EC
Because √
​ 13 ​ = —
​ 2 (​​  2​√13 ​  )​, DF = —
​ 2 ​AB.
—
1
—
1
Geometry
Worked-Out Solutions
217
Chapter 6
—
22. Given F is the midpoint of OC ​
​ .
1
—
—
DF ​  BC ​
​  and DF = —
​   ​ (BC)
Prove ​
2
y
24. Sample answer: ends of the swing set
are midsegments.
B(2q, 2r)
D(q, r)
O(0, 0)
The crossbars on the
E
C(2p, 0) x
F
—
( 0 +2 2p ) ( 2p2 )
​, 0  ​ = ​ —
midpoint of OC ​
​  = ​ —
​
​   ​,  0  ​ = F(p, 0)
25.a.The perimeter of the shaded triangle is
8 + 8 + 8 = 24 units.
slope of DF ​
​  = —
​     ​
r
q−p
2r
r
2r − 0
—
​ = —
​
​     ​
slope of BC ​
​  = —
​

​ = —
2q − 2p 2(q − p) q − p
—
—
8
— — —
——
——
2
2
2
2

——
2
2

——
2
2

——
2
2

——
2
2

——
2
2

DF = √
​ (q − p) + (r − 0)  ​ = ​√ (q − p) + r
​
BC = √
​ (2q − 2p) + (2r − 0)  ​
=√
​ 4(q − p) + 4r  ​
=√
​ 4(q − p) + r
​
= 2​√ (q − p) + r  ​
16
16
8
Because BC = 2DF, DF =
b.The perimeter of all the shaded triangles is
⋅
24 + 3(3 4) = 24 + 36 = 60 units.
​ 12 ​BC.
—
4
—
—
23. An eighth segment, FG
​ ​, would connect the midpoints of DL
​ ​
—
and EN
​ ​.
DE = —
​ 2 ​(XY + LN)
1 1
= ​ —
​ ​ ​ —  ​(LN) + LN  ​
2 2
FG
= —
​ 14 ​LN + —
​ 12 ​LN
3
= ​ —
​LN
4
3
= —
​ 4 ​LN + LN = —
​ 78 ​LN
[
use 8p, 8q, and 8r. So, L(0, 0), M(8q, 8r), and N(8p, 0).
Find the coordinates of X, Y, D, E, F, and G.
—
Because X is the midpoint of LM ​
​ , the coordinates are X(4q, 4r).
—
Because Y is the midpoint of MN ​
​  , the coordinates are
Y(4q + 4p, 4r).
—
Because D is the midpoint of XL ​
​ , the coordinates are D(2q, 2r).
—
Because E is the midpoint of YN ​
​ , the coordinates are
E(2q + 6p, 2r).
—
Because F is the midpoint of DL ​
​ , the coordinates are F(q, r).
—
Because G is the midpoint of EN ​
​ , the coordinates are
G(q + 7p, r).
—
The y-coordinates of D and E are the same, so DE
​ ​ has a
slope of 0. The y-coordinates of F and G are also the same,
—
so FG
​—​ has a slope of 0. LN
​ ​ is on the x-axis, so its slope is 0.
—
—
—
Because their slopes are the same, DE
​ ​   LN
​ ​   FG
​ ​.
Use the Ruler Postulate (Post. 1.1) to find DE, FG, and LN.
DE = 6p, FG = 7p, LN = 8p
Because 6p = —
​ 34 ​(8p), DE = —
​ 34 ​LN. Because 7p = —
​ 78 ​(8p),
7
FG = ​ —
​LN.
8
218
4
8
Geometry
Worked-Out Solutions
8
8
16
4
]
Because you are finding quarter segments and eighth segments,
16
4
4
1
8
=√
​ ​( 2(q − p) )​ + 4r  ​
16
Because the slope of DF ​
​  equals the slope of BC ​
​ ,  ​DF ​   BC ​
​ .
4
4
4
16
4
c.The perimeter of all the shaded triangles is
⋅
⋅
24 + 3 12 + 9 6 = 24 + 36 + 54 = 114 units.
16
6
12
6
24
6
16
6
6
12
12
6
6
6
16
6
26. Two sides of the red triangle have a length of 4 tile widths. A
yellow segment connects the midpoints, where there are 2 tile
lengths on either side. The third red side has a length of
4 tile diagonals, and the other two yellow segments meet at the
midpoint, where there are 2 tile diagonals on either side.
Chapter 6
—
27. Construct a line (​ DF ​
​  )​with midpoint P parallel to and twice
—
the length of QR ​
​ . Construct a line (​ EF ​
—
​  )​with midpoint R
—
parallel to and twice the length of QP ​
​ . Construct a line (​ DE ​
—
​  )​
—
with midpoint Q parallel to and twice the length of PR ​
​ . The vertices
of the original large triangle are D(−1, 2), E(9, 8), and F(5, 0).
10
y
E
8
6
Q
4
R
2.a.Check students’ work. Using the sample in the text:
AB ≈ 3.61
AC ≈ 5.10
BC = 5
b.Check students’ work. Using the sample in the text:
BC = 5 < 8.71 = AC + AB
AC = 5.10 < 8.61 = AB + BC
AB = 3.61 < 10.10 = BC + AC
D
P
−2
F
2
4
8
10 x
−2
A(x, y)
B(x, y)
C(x, y)
AB
AC + BC
AC
A(5, 1)
B(7, 4)
C(2, 4)
3.61
9.24
4.24
A(2, 4)
B(4, −2)
C(7, 6)
6.32
13.93
5.39
A(1, 0)
B(7, 0)
C(1, 7)
6
16.22
7
A(1, 0)
B(7, 0)
C(5, 1)
6
6.36
4.12
AB + BC
BC
AB + AC
8.61
5
7.85
14.86
8.54
11.71
Maintaining Mathematical Proficiency
28. Sample answer: A counterexample to show that the
difference of two numbers is not always less than the greater
of the two numbers is 6 and −2: 6 − (−2) = 8, which is not
less than 6, so the original conjecture is false.
29. Sample answer: An isosceles triangle has at least two sides
that are congruent. An isoseles triangle whose sides are 5
centimeters, 5 centimeters, and 3 centimeters is not equilateral.
6.5 Explorations (p. 339)
1.a.Check students’ work. Using the sample in the text:
AC ≈ 6.08, AB ≈ 4.47, BC ≈ 3.61, m∠ A ≈ 36.03°,
m∠ B ≈ 97.13°, m∠ C ≈ 46.85°
b.Check students’ work. Using the sample in the text,
BC < AB < AC and m∠ A < m∠ C < m∠ B. The shortest
side is opposite the smallest angle, and the longest side is
opposite the largest angle.
A(x, y)
B(x, y)
15.22
9.22
13
8.24
2.24
10.12
The length of each side is less than the sum of the other two.
3. The largest angle is opposite the longest side, and the
smallest angle is opposite the shortest side; The sum of any
two side lengths is greater than the third side length.
4. no; The sum 3 + 4 is not greater than 10, and it is not
possible to form a triangle when the sum of the lengths
of the two sides is less than the length of the third side.
6.5 Monitoring Progress (pp. 340–343)
1. Given △ABC is a scalene triangle.
C(x, y)
AB
AC
BC
Prove △ABC does not have two congruent angles.
5
Assume temporarily that △ABC is a scalene triangle with
A(5, 1)
B(7, 4)
C(2, 4)
3.61 4.24
A(2, 4)
B(4, −2)
C(7, 6)
6.32 5.39 8.54
A(1, 0)
B(7, 0)
C(1, 7)
m∠ A
m∠ B
m∠ C
78.69°
56.31°
45°
93.37°
38.99°
47.64°
90°
49.4°
40.6°
6
7
9.22
If one side of a triangle is longer than another side, then
the angle opposite the longer side is larger than the angle
opposite the shorter side. Similarly, if one angle of a triangle
is larger than another angle, then the side opposite the larger
angle is longer than the side opposite the smaller angle.
∠ A ≅ ∠ B. By the Converse of Base Angles Theorem
(Thm. 5.7), if ∠ A ≅ ∠ B, then the opposite sides are
—​  ≅ AC
—
congruent: ​
BC
​ ​. A scalene triangle cannot have two
congruent sides. So, this contradicts the given information.
So, the assumption that △ABC is a scalene triangle with two
congruent angles must be false, which proves that a scalene
triangle cannot have two congruent angles.
—
—
2. The sides of △PQR from smallest to largest are PR ​
​ , RQ ​
​ , and ​
—
PQ ​. So, by the Triangle Longer Side Theorem, the angles
from smallest to largest are ∠Q, ∠P, and ∠R.
3. The angles of △RST from smallest to largest are ∠R, ∠T,
and ∠S. So, by the Triangle Larger Angle Theorem, the sides
—
—
—
from shortest to longest are ST
​ ​, RS
​ ​,  and RT
​ ​.
Geometry
Worked-Out Solutions
219
Chapter 6
4. Let x represent the length of the third side. By the Triangle
Inequality Theorem: x + 12 > 20 and 12 + 20 > x.
x + 12 > 20
The length of the third side must be greater than 8 inches and
and
12 + 20 > x
x > 8
32 > x, or x < 32
less than 32 inches.
5. 4 + 9 > 10 → 13 > 10
Yes
4 + 10 > 9 → 14 > 9
Yes
9 + 10 > 4 → 19 > 4
Yes
yes; The sum of any two side lengths of a triangle is greater
—
​ . Using point A as the center, draw an
10. Draw a segment AB ​
—
​ , then draw an arc with B as the center
—
​— to intersect the first arc on both sides of AB ​
​ .
Construct a segment from one arc intersection toward the
other but stop at AB ​
​—. Label the arc-segment intersection as C
—
and the intersection with AB ​
​  as G. △BGC is a right scalene
triangle.
C
than the length of the third side.
6. no; The sum 8 + 9 = 17 is not greater than 18.
A
G
B
7. no; The sum 5 + 7 = 12 is not greater than 12.
6.5 Exercises (pp. 344–346)
Vocabulary and Core Concept Check
1. In an indirect proof, rather than proving a statement directly,
you show that when the statement is false, it leads to a
2. The longest side of a triangle is opposite the largest angle
and the shortest side is opposite the smallest angle.
Monitoring Progress and Modeling with Mathematics
3. Assume temporarily that WV = 7 inches.
4. Assume temporarily that xy is even.
5. Assume temporarily that ∠ B is a right angle.
—
6. Assume temporarily that JM
​ ​ is not a median.
7. A and C; The angles of an equilateral triangle are always 60°.
So, an equilateral triangle cannot be a right triangle.
8. B and C; If both ∠ X and ∠ Y have measures less than 30°,
then their sum is less than 60°. Therefore, the sum of their
measures cannot be 62°.
9. To construct a scalene triangle, draw a segment and label it ​
—. Ensuring that AB ​
—
—
—
AC ​
​ , BC ​
​ , and AC ​
​  are all different lengths,
—
draw an arc with center A and radius AB ​
​  and an arc with
—
center C with radius CB ​
​ . Where the two arcs intersect place
point B.
B
A
C
—
The largest angle is ∠ ABC and the opposite side, AC ​
​ , is the
longest side. The smallest angle is ∠ ACB and the opposite
side, AB ​
​—, is the shortest side.
The largest angle is ∠CGB because it is the right angle and
—
the opposite side, CB ​
​ , is the longest side. The smallest angle
—
is ∠GCB and the opposite side, GB ​
​ , is the shortest side.
—
—
11. The sides of △RST from smallest to largest are RT ​
​ , TS ​
​ ,  and ​
—
RS ​.  So, by the Triangle Longer Side Theorem, the angles
from smallest to largest are ∠ S, ∠ R, and ∠ T.
—
—
12. The sides of △JKL from smallest to largest are KL ​
​  , JL ​
​ , and ​
—
JK ​.  So, by the Triangle Longer Side Theorem, the angles
from smallest to largest are ∠ J, ∠ K, and ∠ L.
13. The angles of △ABC from smallest to largest are ∠ C, ∠ A,
and ∠ B. So, by the Triangle Larger Angle Theorem, the
—
—
—
sides from shortest to longest are AB ​
​ , BC ​
​ , and AC ​
​ .
14. The angles of △XYZ from smallest to largest are ∠ Z, ∠ X,
and ∠ Y. So, by the Triangle Larger Angle Theorem, the
—
—
—
sides from shortest to longest are XY ​
​ , ZY ​
​ ,  and ZX ​
​ .
15. m∠ M = 180° − (127° + 29°) = 24°
The angles of △MNP from smallest to largest are ∠ M, ∠ P,
and ∠ N. So, by the Triangle Larger Angle Theorem, the
—
—
—
sides from shortest to longest are PN ​
​ , MN ​
​ , and MP ​
​ .
16. m∠ D = 180° − (90° + 33°) = 57°
The angles of △DFG from smallest to largest are ∠ G, ∠ D,
and ∠ F. So, by the Triangle Larger Angle Theorem, the
—
—
—
sides from shortest to longest are DF ​
​  , GF ​
​ , and GD ​
​ .
17. x + 5 > 12
x>7
5 + 12 > x
17 > x or x < 17
The possible lengths of the third side are greater than
7 inches and less than 17 inches.
220
Geometry
Worked-Out Solutions
Chapter 6
28. Assume temporarily that the second group has 15 or more
18. x + 12 > 18
x>6
12 + 18 > x
30 > x or x < 30
The possible lengths of the third side are greater than
6 feet and less than 30 feet.
students. Because the first group has 15 students, the total
number of students in the class would be 30 students or
more. Because the class has fewer than 30 students, the
assumption must be false, and the second group must have
fewer than 15 students.
29. C; m∠ U = 180° − (84° + 48°) = 180° − 132° = 48°,
which indicates that △UTV is isosceles. By the Triangle
Longer Side Theorem, UV > TV.
19. x + 24 > 40
x > 16
30. C and D; m∠ R = 180° − (65° + 56°) = 180° − 121° = 59°;
24 + 40 > x
64 > x or x < 64
The possible lengths of the third side are greater than
16 inches and less than 64 inches.
20. x + 25 > 25
By the Triangle Inequality Theorem, the order of the
angles from smallest to largest is ∠ T, ∠ R, and ∠ S.
The order of the sides from shortest to longest is
RS < ST < RT ⇒ 8 < ST < RT. ST could possibly be
9 or 10, but not 7 or 8.
31. Given
x>0
An odd number
Prove An odd number is not divisible by 4.
25 + 25 > x
50 > x or x < 50
The possible lengths of the third side are greater than
0 meters and less than 50 meters.
21. 6 + 7 = 13 → 13 > 11
Yes
7 + 11 = 18 → 18 > 6
Yes
11 + 6 = 17 → 17 > 7
Yes
Assume temporarily that an odd number is divisible by 4.
Let the odd number be represented by 2y + 1 where y is a
positive integer. Then there must be a positive integer x such
that 4x = 2y + 1. However, when you divide each side of the
equation by 4, you get x = —12 y + —14 , which is not an integer.
So, the assumption must be false, and an odd number is not
divisible by 4.
32. Given
yes; The sum of any two side lengths of a triangle is greater
than the length of the third side.
22. no; The sum 3 + 6 = 9 is not greater than 9.
23. no; The sum 28 + 17 = 45 is not greater than 46.
24. 35 + 120 = 155 → 155 > 125
Yes
120 + 125 = 255 → 255 > 35
Yes
125 + 35 = 160 → 160 > 120
Yes
yes; The sum of any two side lengths of a triangle is greater
than the length of the third side.
25. An angle that is not obtuse could be acute or right. Assume
temporarily that ∠ A is not obtuse.
—
26. Because 30° < 60° < 90° and 1 < √ 3 < 2, the longest side,
which is 2 units long, should be across from the largest
angle, which is the right angle.
Prove
△QRS, m∠ Q + m∠ R = 90°
m∠ S = 90°
Assume temporarily that in △QRS, m∠ Q + m∠ R = 90°
and m∠ S ≠ 90°. By the Triangle Sum Theorem (Thm. 5.1),
m∠ Q + m∠ R + m∠ S = 180°. Using the Substitution
Property of Equality, 90° + m∠ S = 180°. So, m∠ S = 90°
by the Subtraction Property of Equality, but this contradicts
the given information. So, the assumption must be false,
which proves that in △QRS, if m∠ Q + m∠ R = 90°,
then m∠ S = 90°.
33. The right angle of a right triangle must always be the largest
angle because the other two will have a sum of 90°. So,
according to the Triangle Larger Angle Theorem (Thm. 6.10),
because the right angle is larger than either of the other
angles, the side opposite the right angle, which is the
hypotenuse, will always have to be longer than either of
the legs.
34. yes; If the sum of the lengths of the two shortest sides is
greater than the length of the longest side, then the other
two inequalities will also be true.
60°
1
2
——
30°
√3
27. Assume temporarily that your client committed the crime.
Then your client had to be in Los Angeles, California, at the
time of the crime. Security footage shows that your client
was in New York at the time of the crime. Therefore, the
assumption must be false, and the client must be innocent.
Geometry
Worked-Out Solutions
221
Chapter 6
35.a.The width of the river must be greater than 35 yards and
—
less than 50 yards. In △BCA, the width of the river, BA
​ ​,
—
must be less than the length of CA
​ ​, which is 50 yards,
—
because the measure of the angle opposite BA
​ ​ is less than
—
the measure of the angle opposite CA
​ ​, which must be 50°. In
△BDA, the width of the river, BA
​—​, must be greater than the
—
length of DA
​ ​, which is 35 yards, because the measure of
—
the angle opposite BA
​ ​ is greater than the measure of the
—
angle opposite DA
​ ​, which must be 40°.
b.You could measure from distances that are closer together.
In order to do this, you would have to use angle measures
that are closer to 45°.
36.a.By the side length requirements for a triangle,
x < 489 + 565 = 1054 kilometers and
x > 565 − 489 = 76 kilometers.
b.Because ∠2 is the smallest angle, the distance between
Granite Peak and Fort Peck Lake must be the shortest side
of the triangle. So, the second inequality becomes
x < 489 kilometers.
37. ∠WXY, ∠ Z, ∠ YXZ, ∠WYX and ∠ XYZ, ∠W;
In △WXY, because WY < WX < YX, by the Triangle Longer
Side Theorem (Thm. 6.9), m∠WXY < m∠WYX < m∠W.
Similarly, in △XYZ, because XY < YZ < XZ, by the Triangle
Longer Side Theorem (Thm. 6.9), m∠ Z < m∠ YXZ < m∠ XYZ.
Because m∠WYX = m∠ XYZ and ∠W is the only angle
greater than either of them, you know that ∠W is the largest
angle. Because △WXY has the largest angle and one of
the congruent angles, the remaining angle, ∠WXY, is the
smallest.
m∠ D + m∠ E + m∠ F = 180°
38.
(x + 25)° + (2x − 4)° + 63° = 180°
3x = 96
x = 32
m∠D = 32 + 25 = 57°
m∠E = 2
m∠F = 63°
The order of the angles from least to greatest is
3x + 84 = 180
⋅ 32 − 4 = 60°
m∠ D < m∠ E < m∠ F. The order of the sides from
least to greatest is EF < DF < DE.
39. By the Exterior Angle Theorem (Thm. 5.2),
m∠1 = m∠ A + m∠B. Then by the Subtraction Property of
Equality, m∠1 − m∠B = m∠ A. If you assume temporarily
that m∠1 ≤ m∠B, then m∠ A ≤ 0. Because the measure
of any angle in a triangle must be a positive number, the
assumption must be false. So, m∠1 > m∠B. Similarly, by the
Subtraction Property of Equality, m∠1 − m∠ A = m∠B. If
you assume temporarily that m∠1 ≤ m∠ A, then m∠B ≤ 0.
Because the measure of any angle in a triangle must be a
positive number, the assumption must be false. So,
m∠1 > m∠ A.
222
Geometry
Worked-Out Solutions
40.
JK + KL > JL
x + 11 + 2x + 10 > 5x − 9
3x + 21 > 5x − 9
−2x > −30
x < 15
x + 11 + 5x − 9 > 2x + 10
4x > 8
x>2
2x + 10 + 5x − 9 > x + 11
7x + 1 > x + 11
6x > 10
The possible values for x are x > 2 and x < 15.
41.
JK + JL > KL
6x + 2 > 2x + 10
KL + JL > JK
x > —
​ 10
​ = —
​ 53 ​≈ 1.667
6
UV + VT > TU
3x − 1 + 2x + 3 > 6x − 11
5x + 2 > 6x − 11
−x > −13
x < 13
6x − 11 + 2x + 3 > 3x − 1
TU + TV > UV
8x − 8 > 3x − 1
5x > 7
3x − 1 + 6x − 11 > 2x + 3
The possible values for x are x > 2​ —
​ and x < 13.
7
2
x > —
​ 75 ​= 1​ —
​
5
UV + TU > TV
9x − 12 > 2x + 3
7x > 15
1
x > —
​ 15
​= 2​ —  ​
7
7
1
42. The shortest route is along Washington Avenue. By the
Triangle Inequality Theorem (Thm. 6.11), the length of
Washington Avenue must be shorter than the sum of the
lengths of Eighth Street and View Street, as well as the sum
of the lengths of Hill Street and Seventh Street.
Chapter 6
Prove m∠ BAC > m∠ C
by the constructions in parts (a)–(c) because a triangle is
formed only when the two sides formed by arc radii have
—
a sum that is greater than the length of side QR
​ ​.
B
1
A
2
3
D
46. Given △ABC
C
It is given that BC > AB and BD = BA. By the Base Angles
Theorem (Thm. 5.6), m∠ 1 = m∠ 2. By the Angle Addition
Postulate (Post. 1.4), m∠ BAC = m∠ 1 + m∠ 3. So,
m∠ BAC > m∠ 1. Substituting m∠ 2 for m∠ 1 produces
m∠ BAC > m∠ 2. By the Exterior Angle Theorem (Thm. 5.2),
m∠ 2 = m∠ 3 + m∠ C. So, m∠ 2 > m∠ C. Finally, because
m∠ BAC > m∠ 2 and m∠ 2 > m∠ C, you can conclude that
m∠ BAC > m∠ C.
44. As an example, if the 24-centimeter string is divided into
10 centimeters, 10 centimeters, and 4 centimeters, the
triangle is an acute isosceles triangle.
10 cm
10 cm
4 cm
For a right scalene triangle, the sides could be 6 centimeters,
8 centimeters, and 10 centimeters.
10 cm
8 cm
6 cm
45.a.
a
b
M
Q
R
b.
c
d
M
R
—
yes; The arcs intersect in one point that lies on QR
​ ​.
They do not form a triangle because all three points
are collinear.
Prove AB + BC > AC, AC + BC > AB, and
AB + AC > BC
B
2 3
D
1
C
A
—
Assume BC
​ ​ is longer than or the same length as each of
—
—
the other sides, AB
​ ​ and AC
​ ​. Then AB + BC > AC and
AC + BC > AB. The proof for AB + AC > BC follows.
STATEMENTS
REASONS
1. △ABC
1. Given
2. Extend —
​AC ​ to D so
that —
​AB ​ ≅ —
2. Ruler Postulate
(Post. 1.1)
3. Definition of segment
congruence
4. AD + AC = DC
Postulate (Post. 1.2)
5. ∠ 1 ≅ ∠ 2
5. Base Angles Theorem
(Thm. 5.6)
6. m∠ 1 = m∠ 2
6. Definition of angle
congruence
7. m∠ DBC > m∠ 2
7. Protractor Postulate
(Post. 1.3)
8. m∠ DBC > m∠ 1
8. Substitution Property
9. DC > BC
9. Triangle Larger Angle
Theorem (Thm. 6.10)
10. AD + AC > BC
10. Substitution Property
11. AB + AC > BC
11. Substitution Property
no; The arcs do not intersect.
Q
d. The Triangle Inequality Theorem (Thm. 6.11) is verified
43. Given BC > AB, BD = BA
47. The perimeter of △HGF must be greater than 4 and less
than 24; Because of the Triangle Inequality Theorem
(Thm. 6.11), FG must be greater than 2 and less than 8,
GH must be greater than 1 and less than 7, and FH must
be greater than 1 and less than 9. So, the perimeter must be
greater than 2 + 1 + 1 = 4 and less than 8 + 7 + 9 = 24.
c.
f
Q
M
R
g
yes; The arcs intersect in two points. Each point is a
—
vertex of a triangle with side QR
​ ​.
Geometry
Worked-Out Solutions
223
Chapter 6
6.6 Monitoring Progress (pp. 349–350)
Maintaining Mathematical Proficiency
—
—
​  and BE ​
​  is ∠ AEB.
48. The included angle between AE ​
1.
PR = PSGiven
—
—
PQ ​
​  ≅ PQ ​
​
Reflexive Property of Congruence
(Thm. 2.1)
—
—
49. The included angle between AC ​
​  and DC ​
​  is ∠ ACD.
—
—
50. The included angle between AD ​
​  and DC ​
m∠ QPR > m∠ QPSGiven
—
—
51. The included angle between CE ​
​  and BE ​
​  is ∠ BEC.
6.6 Explorations (p. 347)
2.
1.a.Check students’ work.
b.Check students’ work.
c.Check students’ work.
d.AC
​ ​ ≅ DC
​ ​, because all points on a circle are equidistant
— —
—
—
from the center; BC
​ ​ ≅ BC
​ ​ by the Reflexive Property of
Congruence (Thm. 2.1).
—
—
DB ​ is 2.7 units, so AB > DB; m∠ ACB = 90° and
e.As drawn, the length of AB ​
​  is 3.6 units and the length of ​
f.
D
AC
BC
1. (4.75, 2.03)
2
3
2. (4.94, 2.5)
2
3. (5, 3)
m∠ ACB
m∠ BCD
3.61 2.68
90°
61.13°
3
3.61 3.16
90°
75.6°
2
3
3.61 3.61
90°
90°
4. (4.94, 3.5)
2
3
3.61
90°
104.45°
5. (3.85, 4.81)
2
3
3.61 4.89
90°
154.93°
AB
BD
4
g.If two sides of one triangle are congruent to two sides
of another triangle, and the included angle of the first is
larger than the included angle of the second, then the third
side of the first is longer than the third side of the second.
2. If the included angle of one is larger than the included angle
of the other, then the third side of the first is longer than the
third side of the second. If the included angles are congruent,
then you already know that the triangles are congruent by
the SAS Congruence Theorem (Thm. 5.5). Therefore, the
third sides are congruent because corresponding parts of
congruent triangles are congruent.
3. Because the sides of the hinge do not change in length, the
angle of the hinge can model the included angle and the
distance between the opposite ends of the hinge can model
the third side. When the hinge is open wider, the angle is
larger and the ends of the hinge are farther apart. If the hinge
is open less, the ends are closer together.
224
Hinge Theorem (Thm. 6.12)
PR = PSGiven
—
—
PQ ​
​  ≅ PQ ​
​
Reflexive Property of Congruence
RQ < SQGiven
m∠ RPQ < m∠ SPQConverse of the Hinge Theorem
(Thm. 2.1)
m∠ DCB = 61°, so m∠ ACB > m∠ DCB; yes; the results
are as expected because the triangle with the longer third
side has the larger angle opposite the third side.
RQ > SQ
—
RQ ​
​  is the longer segment.
Geometry
Worked-Out Solutions
(Thm. 6.13)
∠ SPQ is the larger angle.
3. Assume temporarily that the third side of the first triangle
with the larger included angle is not longer than the third
side of the second triangle with the smaller included angle.
This means the third side of the first triangle is equal to or
shorter than the third side of the second triangle.
4. Group A: 135°
Group B: 150°
Group C: 180° − 40° = 140°
Because 135° < 140° < 150°, Group C is closer to the camp
than Group B, but not as close as Group A.
6.6 Exercises (pp. 351–352)
Vocabulary and Core Concept Check
1. Theorem 6.12 refers to two angles with two pairs of sides
that have the same measure, just like two hinges whose sides
are the same length. Then the angle whose measure is greater
is opposite a longer side, just like the ends of a hinge are
farther apart when the hinge is open wider.
—
—
—
—
2. In △ABC and △DEF, AB ​
​  ≅ DE ​
​ , BC ​
​  ≅ EF ​
​ , and AC < DF.
So, m∠ E > m∠ B by the Converse of the Hinge Theorem
(Theorem 6.13).
Monitoring Progress and Modeling with Mathematics
3. m∠ 1 > m∠ 2; By the Converse of the Hinge Theorem
(Thm. 6.13), because ∠ 1 is the included angle in the triangle
with the longer third side, its measure is greater than that
of ∠ 2.
4. m∠ 1 < m∠ 2; By the Converse of the Hinge Theorem
(Thm. 6.13), because ∠ 1 is the included angle in the triangle
with the shorter third side, its measure is less than that of ∠ 2.
5. m∠ 1 = m∠ 2; The triangles are congruent by the SSS
Congruence Theorem (Thm. 5.8). So, ∠ 1 ≅ ∠ 2 because
corresponding parts of congruent triangles are congruent.
Chapter 6
6. m∠ 1 > m∠ 2; By the Converse of the Hinge Theorem
(Thm. 6.12), because ∠ 1 is the included angle in the triangle
with the longer third side, its measure is greater than that of ∠ 2.
—
7. AD > CD; By the Hinge Theorem (Thm. 6.12), because AD
is the third side of the triangle with the larger included angle,
—.
it is longer than CD
50 miles
160°
H
20°
100 miles
Friend’s flight:
—
8. MN < LK; By the Hinge Theorem (Thm. 6.12), because MN
is the third side of the triangle with the smaller included
—.
angle, it is shorter than LK
50 miles
30°
—
9. TR < UR; By the Hinge Theorem (Thm. 6.12), because TR is
the third side of the triangle with the smaller included angle,
—.
it is shorter than UR
—
150°
100 miles
Because 160° >150°, the distance you flew is a greater distance
than your friend flew by the Hinge Theorem (Thm. 6.12).
10. AC > DC; By the Hinge Theorem (Thm. 6.12), because AC is
the third side of the triangle with the larger included angle, it
—.
is longer than DC
11. Given
Prove
— ≅ YZ
—, m∠ WYZ > m∠ WYX
XY
W
WZ > WX
Z
X
STATEMENTS
REASONS
1.
1. Given
—
XY ≅ —
YZ
2. —
WY ≅ —
WY
210 miles
Y
2. Reflexive Property of
Congruence (Thm. 2.1)
110°
70°
80 miles
3. m∠ WYZ > m∠ WYX 3. Given
4. WZ > WX
4. Hinge Theorem (Thm. 6.12)
12. Given
— ≅ DA
—, DC < AB
BC
Prove
m∠ BCA > m∠ DAC
B
A
210 miles
D
STATEMENTS
BC ≅ —
DA
1. —
AC ≅ —
AC
2. —
Friend’s flight:
C
50°
REASONS
130°
1. Given
2. Reflexive Property of
Congruence (Thm. 2.1)
3. DC < AB
3. Given
4. m∠ BCA > m∠ DAC
4. Converse of the Hinge
Theorem (Thm. 6.13)
80 miles
Because 130° > 110°, the distance your friend flew is a
greater distance than the distance you flew by the Hinge
Theorem (Thm. 6.12).
15. The measure of the included angle in △PSQ is greater than
the measure of the included angle in △SQR; By the Hinge
Theorem (Thm. 6.12), PQ > SR.
16. Given EF > ED and GD = GF, then m∠ EGF > m∠ DGE by
the Converse of the Hinge Theorem (Thm. 6.13).
Geometry
Worked-Out Solutions
225
Chapter 6
17. The angle bisector of ∠ FEG will also pass through incenter H.
180°
​   ​ = 90°,
Then, m∠ HEG + m∠ HFG + m∠ HGF = —
2
because they are each half of the measure of an angle of
a triangle. By subtracting m∠ HEG from each side, you
can conclude that m∠ HFG + m∠ HGF < 90°. Also,
m∠ FHG + m∠ HFG + m∠ HGF = 180° by the Triangle
Sum Theorem (Thm. 5.1). So, m∠ FHG > 90°, which means
that m∠ FHG > m∠ HFG and m∠ FHG > m∠ HGF. So,
FG > FH and FG > HG.
—
—
—
—
—
18. Because NR
​ ​ is a median, PR
​ ​ ≅ QR
​ ​. NR
​ ​ ≅ NR
​ ​ by the
Reflexive Property of Congruence (Thm. 2.1). So, by
the Converse of the Hinge Theorem (Thm. 6.13),
m∠ NRQ > m∠ NRP. Because ∠ NRQ and ∠ NRP form a
linear pair, they are supplementary. So, ∠ NRQ must be
obtuse and ∠ NRP must be acute.
19. 180° − (27° + 102°) = 180° − 129° = 51°
110° > 51°
3x + 2 > x + 3
2x > 1
x > —
​ 12 ​
4x − 3 > 2x
2x > 3
x > —
​ 32 ​
—
​  is the perpendicular bisector of AB
​ ​.
21. Given CP​
Prove CA = CB
C
A
P
24. The sum of the measures of the angles of a triangle in
spherical geometry must be greater than 180°; The area of
πr2
spherical △ABC = —
​
​(m∠ A + m∠B + m∠C − 180°),
180°
where r is the radius of the sphere.
Maintaining Mathematical Proficiency
25. x° + 115° + 27° = 180°
x + 142 = 180
x = 38
27. 3x° = 180°
2x = 144
x = 72
28. x° = 44°+ 64°
x = 60
x = 108
B
1. Let n be the stage, then the side length of the new triangles in
each stage is 24 − n. So, the perimeter of each new triangle is
(​ 3  24 − n )​. The number of new triangles is given by (​  3n − 1) ​.
So, to find the perimeter of all the shaded triangles in each
​( 3 24 − n )(​​  3n − 1 )​. The perimeter of the new triangles in
stage 4 will be ​( 3 24 − 4 )(​​  34 − 1 )​ = 81. The total perimeter
of the new triangles and old triangles is 81 + 114 = 195 units.
⋅
⋅
or CA < CB. By the Reflexive Property of Congruence
—
—
(Thm. 2.1), CP
​ ​ ≅ CP
​ ​. Also, by the definition of
—
—
perpendicular bisector, AP
​ ​ ≅ BP
​ ​, ∠ APC is a right angle,
and ∠ BPC is a right angle. So, by the Right Angles
Congruence Theorem (Thm 2.3), ∠ APC ≅ ∠ BPC. If
CA > CB, then by the Converse of the Hinge Theorem
(Thm. 6.13), m∠ APC > m∠ BPC, which contradicts the
fact that ∠ APC ≅ ∠ BPC. Also, if CA < CB, then by the
Converse of the Hinge Theorem (Thm. 6.13),
m∠ APC < m∠ BPC, which also contradicts the fact that
∠ APC ≅ ∠ BPC. So, the assumption must be false, and
CA = CB.
22. By the Converse of the Hinge Theorem (Thm. 6.13), because
28 ft > 22 ft → AD > AB, then m∠ ACD > m∠ ACB.
Geometry
Worked-Out Solutions
⋅
2. x + 5 > 12, x + 12 > 5, 5 + 12 > x; Because the length of
the third side has to be a positive value, the inequality
x + 12 > 5 will always be true. So, you do not have to
consider this inequality in determining the possible values
of x. Solve the other two inequalities to find that the length
of the third side must be greater than 7 and less than 19.
3. If △ABC is an acute triangle, then m∠ BAC < m∠ BDC and
the orthocenter D is inside the triangle.
B
Assume temporarily that CA ≠ CB. Then either CA > CB
226
26. 2x° + 36° = 180°
6.4–6.6 What Did You Learn? (p. 353)
20. By the Exterior Angle Theorem, m∠ ABD = m∠ BDC + m∠ C.
So, m∠ ABD > m∠ BDC.
23. △ ABC is an obtuse triangle; If the altitudes intersect inside
the triangle, then m∠ BAC will always be less than m∠ BDC
—
because they both intercept the same segment, CD
​ ​. However,
because m∠ BAC > m∠ BDC, ∠ A must be obtuse, and the
altitudes must intersect outside of the triangle.
D
A
C
If △ABC is a right triangle, then m∠ BAC = m∠ BDC
because the orthocenter D is on vertex A, where ∠ A is the
right angle.
B
A
D
C
Chapter 6
If △ABC is an obtuse triangle, then m∠ BAC > m∠ BDC and
the orthocenter D is outside the triangle, where ∠ A is the
obtuse angle.
5. Graph △LMN.
4
X(−2, 1)
B
y
D(4, 3)
2
−2
A
C
−4
Chapter 6 Review (pp. 354–356)
—
​  ⊥ AC
1. DC = 20; Point B is equidistant from A and C, and BD​
​ ​.
So, by the Converse of the Perpendicular Bisector Theorem
(Thm. 6.2), DC = AD = 20.
—
—
​  , and SP
​  . So,
2. RS = 23; ∠ PQS ≅ ∠ RQS, SR
​ ​ ⊥ QR​
​ ​  ⊥ QP​
by the Angle Bisector Theorem (Thm. 6.3), SR = SP. This
means that 6x + 5 = 9x − 4, and the solution is x = 3. So,
RS = 9(3) − 4 = 23.
​  and FH​
3. m∠ JFH = 47°; Point J is equidistant from FG​
​  . So,
by the Converse of the Angle Bisector Theorem (Thm. 6.4),
m∠ JFH = m∠ JFG = 47°.
−6
−4
y
U(0, −1) x
−2
y=3
Z(6, −3)
—
​
midpoint of YZ
​ ​  = ​ —
​   ​, ​ —
​
The equation of the line perpendicular to YZ
​ ​  is x = 4.
​ = —
​
​   ​ = −1
Slope of XY
​ ​:  m = —
​

​ = —
​
= —
​
​
= —
​   ​ = 0
Slope of YZ
​ ​:  m = —
​
( 2 +2 6 −3 +2 (−3) )
8 −6
= ​( ​   ​ , ​   ​  )​ = (4, −3)
2 2
—
—  —
—
—
−4
−4
−3 − 1
2 − (−2) 2 + 2
4
—
The slope of the line perpendicular to XY
​ ​  is m = 1.
​,
​
​
midpoint of XY
​ ​  = ​ —
​
​
—
−1 = 1
−1 = b
( −22+ 2 1 + 2(−3) )
0 −2
= ​( ​   ​ , ​   ​  )​ = (0, −1)
2 2
—
—  —
y = mx + b
⋅0 + b
—
The equation of the line perpendicular to XY
​ ​ through (0, −1)
Intersection of x = 4 and y = x − 1:
y=4−1=3
So, the coordinates of the circumcenter of △XYZ are (4, 3).
is y = x − 1.
2
x = −3
Y(2, −3)
x
4. Graph △TUV.
6
−3 − (−3) −3 + 3 0
6−2
4
4
—
The slope of the line perpendicular to YZ
​ ​ is undefined.
D
2
6. By the Incenter Theorem (Thm. 6.6), x = 5.
−2
−4
T(−6, −5)
−6
V(0, −5)
( 02 −1 +2 (−5) ) ( 20 −62 )
−6 + 0 −5 + (−5)
—
​,
​
​
midpoint of TV
​ ​  = (​  ​
)​ = (​  ​ −62 ​,  ​ −102 ​  )​
2
2
—
​
​
​  ​ = (0, −3)
midpoint of UV
​ ​  = ​ ​ —   ​, —
​ = ​ —
​   ​ , ​ —
The equation of the perpendicular bisector of UV
​ ​  through
— —
= (−3, −5)
—  —
—
its midpoint (0, −3) is y = −3, and the equation of the
perpendicular bisector of TV
​—​ through its midpoint
(−3, −5) is x = −3. The point of intersection of the two
perpendicular bisectors is (−3, −3). So, the coordinates of
the circumcenter of △TVU are (−3, −3).
Geometry
Worked-Out Solutions
227
Chapter 6
7.
8.
y
A(−10, 3)
F(−6, 3)
B(−4, 5)
y
4
2
6
−2
6
8
x
E(2, −2) G(5, −2) F(8, −2)
4
D(−4, 3)
−4
2
E(−7, 2)
−10
−8
−6
—
−4
−2
C(4, −4)
H(2, −5)
−6
C(−4, 1)
x
−8
( −4 +2 (−4) 5 +2 1 )
−8 6
= (​  ​   ​,  ​    ​  )​ = (−4, 3)
2 2
D(2, −8)
—
( 2 +2 8 −2 +2 (−2) )
10 −4
= (​  ​   ​,  ​   ​  )​ = (5, −2)
2 2
​,
​ —
​  ​
midpoint of BC
​ ​  = ​ ​ —
​
midpoint of EF
​ ​  = ​ —
​   ​, ​ —
​
​   ​ = 0.
​ ​  is ——
​
​ = —
​
= —
​
​
= —
​   ​ = 2.
The slope of DG
​ ​  is —
​
​ ​ through (−4, 3) is y = 3.
​,
​ —
​  ​
midpoint of AC
​ ​  = ​ ——
​
3
3
5−2
—
​
​   ​ = 1.
The slope of BE
​ ​  is —
​
​ = —
​ = —
−4 − (−7) −4 + 7 3
y = mx + b
2=1
2 = −7 + b
9=b
—  —
—
3−3
−4 − (−10)
—
—
0
6
( −10 +2 (−4) 3 +2 1 )
−14 4
= (​  ​   ​, ​    ​  )​ = (−7, 2)
2 2
—  —
⋅ (−7) + b
—
—  —
—
−2 = 2
−2 = 10 + b
−12 = b
—
The equation of DG
​ ​ through (5, −2) is y = 2x − 12.
​
midpoint of ED
​  ​ = ​ —
​   ​, ​ —
​
​
= —
​
​
= —
​   ​ = —
​   ​ .
The slope of FH
​ ​  is —
​
—
( 2 +2 2 −2 +2 (−8) )
4 −10
= (​  ​   ​ , ​   ​
​ = (2, −5)
2 2 )
—  —
—
The centroid has the coordinates of the intersection of y = 3
−5 = —
​   ​  2 + b
y=x+9
3=x+9
−6 = x
So, the coordinates of the centroid are (−6, 3).
228
Geometry
Worked-Out Solutions
−2 − (−5)
8−2
−2 + 5
6
3
6
1
2
y = mx + b
6
3
⋅5 + b
The equation of BE
​ ​ through (−7, 2) is y = x + 9.
−2 + 8
3
y = mx + b
and y = x + 9.
−2 − (−8)
5−2
⋅
1
2
−5 = 1 + b
−6 = b
—
The equation of FH ​
​  through (2, −5) is y = —
​ 2 ​x − 6.
The centroid has the coordinates of the intersection of
2x − 12 = —
​   ​ x − 6
3x = 12
x=4
y = —
​   ​  4 − 6 = 2 − 6 = −4
So, the coordinates of the centroid are (4, −4).
1
1
y = 2x − 12 and y = —
​ 2 ​x − 6.
1
2
1
2x = —
​   ​ x + 6
2
1
2 2x = 2 —
​   ​ x + 2 6
2
4x = x + 12
⋅
1
2
⋅
⋅
⋅
Chapter 6
9.
10.
2
5
y= x+4
y
6
y
K(−8, 5)
4
G(1, 6)
H(5, 6)
L(−6, 3)
y=x+5
2
J(3, 1)
4
x=3
6
x
—
6−6 0
5−1 4
—
The slope of the line perpendicular to GH
​ ​  through J(3, 1)
is x = 3.
​ = —
​   ​ = 0
slope of GH
​ ​  = —
​

—
6−1
1−3
5
2
​
= −​ —  ​
slope of GJ
​ ​  = —
​
The slope of the line perpendicular to GJ
​ ​  is —
​   ​ .
y = mx + b
4=b
—
2
5
⋅
2
6 = —
​    ​ 5 + b
5
6=2+b
—
The equation of the line perpendicular to GJ
​ ​ that passes
through H(5, 6) is y = —
​   ​ x + 4.
2
5
2
x = −6
O(3, 5.2)
M(0, 5)
2
The orthocenter is the intersection of x = 3 and y = —
​   ​ x + 4.
5
2
y = —
​   ​  3 + 4
5
6 20
​   ​
y = —
​   ​ + —
5
5
26
y = —
​   ​
5
The orthocenter of △GHJ is inside the triangle with
26
coordinates ​ 3, —
​   ​   ​.
5
⋅
(  )
−4
x
O(−6, −1)
—
0
5−5
0 − (−8) 8
—
The equation of the line perpendicular to KM
​ ​  through
L(−3, 3) is x = −6.
​   ​ = 0
slope of KM
​
​ = —
​
​ = —
—
5−3
−8 − (−6)
2
−8 + 6
2
−2
​
​     ​ = −1
slope of KL
​ ​  = —
​
​ = —
​ = —
The slope of the line perpendicular to KL
​ ​ is 1.
y = mx + b
—
⋅0 + b
5 =1
5=b
y=x+5
—
The equation of the line perpendicular to KL
​ ​ that passes
The orthocenter is the intersection of x = −6 and y = x + 5.
through M(0, 5) is y = x + 5.
y=x+5
y = −6 + 5
y = −1
The orthocenter of △KLM is outside the triangle with
coordinates (−6, −1).
(
)
−12 12
= (​  ​   ​, ​   ​  )​ = (−6, 6)
2 2
−6 + 0 8 + 4
−6 12
—
​,
​   ​
midpoint of AC
​ ​  = (​  ​
​ = ​  ​   ​,  ​   ​   ​ = (−3, 6)
2
2 ) ( 2 2)
−6 + 0 4 + 4
−6 8
—
​,
​   ​
midpoint of BC
​ ​  = (​  ​
​ = ​  ​   ​,  ​    ​  ​ = (−3, 4)
2
2 ) ( 2 2)
−6 + (−6) 8 + 4
—
11. midpoint of AB
​,
​ —
​  ​
​ ​  = ​ —
​
2
2
The coordinates of the midsegments of △ABC are (−6, 6),
—  —
— —
—  —
— —
—  —
(−3, 6), and (−3, 4).
(
) (  )
−3 + 1 1 + (−5)
—
​,
​
​
midpoint of DF
​ ​  = (​  ​
)​
2
2
−2 −4
= (​  ​   ​,  ​   ​  )​ = (−1, −2)
2 2
3 + 1 5 + (−5)
—
​
midpoint of EF
​ ​  = (​  ​   ​, ​
)​ = (​  ​ 42 ​ , ​ 02 ​  )​ = (2, 0)
2
2
−3 + 3 1 + 5
0 6
—
​,
​   ​  ​ = ​ —
​    ​  ​ = (0, 3)
12. midpoint of DE
​ ​  = ​ —
​
​   ​ , —
—
2
2
2 2
The coordinates of the midsegments of △DEF are (0, 3),
— —
—  —
— —
—  —
(−1, −2), and (2, 0).
Geometry
Worked-Out Solutions
229
Chapter 6
13. x + 4 > 8
4.
WY = WZAngle Bisector Theorem (Thm. 6.3)
x>4
6x + 2 = 9x − 13
−3x = −15
x=5
WY = 6x + 2 = 6
4+8>x
The possible lengths for the third side of the triangle with
12 > x, or x < 12
sides 4 inches and 8 inches are 4 in. < x < 12 in.
14. x + 6 > 9
x>3
6+9>x
The possible lengths for the third side of the triangle with
15 > x, or x < 15
sides 6 meters and 9 meters are 3 m < x < 15 m.
15. x + 11 > 18
x>7
11 + 18 > x
5. By the Incenter Theorem (Thm. 6.6), the incenter point is
equidistant to each side of the triangle. Because WC = 20,
BW = 20.
6. AB > CB; By the Hinge Theorem (Thm. 6.12)
7. m∠ 1 < m∠ 2; By the Converse of the Hinge Theorem
(Thm. 6.13)
8. m∠ MNP < m∠ NPM; By the Triangle Larger Angle
Theorem (Thm. 6.10)
9.
The possible lengths for the third side of the triangle with
y
C(0, 6)
29 > x, or x < 29
y=2
x=2
4
circumcenter
E(2, 2)
D(1.3, 0.7)
centroid
sides 11 feet and 18 feet are 7 ft < x < 29 ft.
16. Given △XYZ, XY = 4, and XZ = 8
Prove YZ > 4
Assume temporarily that YZ > 4. Then it follows that either
YZ < 4 or YZ = 4. If YZ < 4, then XY + YZ < XZ because
4 + YZ < 8 when YZ < 4. If YZ = 4, then XY + YZ = XZ
because 4 + 4 = 8. Both conclusions contradict the Triangle
Inequality Theorem (Thm. 6.11), which says that
XY + YZ > XZ. So, the temporary assumption that YZ > 4
cannot be true. This proves that in △XYZ, if XY = 4 and
XZ = 8, then YZ > 4.
17. Given that m∠ QRT > m∠SRT, by the Hinge Theorem
(Thm. 6.12), QT > ST.
18. Given that QT > ST, by the Converse of the Hinge Theorem
(Thm. 6.13), m∠ QRT > m∠SRT.
Chapter 6 Test (p. 357)
1. By the Triangle Midsegment Theorem (Thm. 6.8),
x = —
​ 12 ​ 12 = 6.
⋅
2. By the definition of midpoint, x = 9.
RS = STPerpendicular Bisector
Theorem (Thm. 6.1)
3x + 8 = 7x − 4
−4x = −12
x=3
ST = 7
3.
⋅ 3 − 4 = 21 − 4 = 17
−2
Geometry
Worked-Out Solutions
4
A(0, −2)
orthocenter
6 x
B(4, −2)
Circumcenter:
​
midpoint of AB
​ ​  = ​ —
​   ​, ​ —
​
The line perpendicular to AB
​ ​ through (2, −2) is x = 2.
​
​
midpoint of AC
​ ​  = ​ —
​    ​, —
​ = ​ 0, —
​   ​  ​ = (0, 2)
The line perpendicular to AC
​ ​ through (0, 2) is y = 2.
The intersection of x = 2 and y = 2 is (2, 2).
The coordinates of the circumcenter are (2, 2).
Orthocenter:
​
= ​ —   ​ = undefined.
The slope of AC
​ ​  is —
​
( 0 +2 4 −2 +2 (−2) )
4 −4
= (​  ​   ​ , ​   ​  )​ = (2, −2)
2 2
—
—  —
—
—
( 02 6 + 2(−2) ) (  −42 )
—
—
6 − (−2) 8
0−0
0
—
The equation of the line perpendicular to AC
​ ​  through
B(4, −2) is y = −2.
−2 − (−2) 0
—
​
= —
​   ​ = 0.
The slope of AB
​ ​  is —
​
4−0
4
—
The equation of the line perpendicular to AB
​ ​  through C(0, 6)
The intersection of x = 0 and y = −2 is (0, −2).
The coordinates of the orthocenter are (0, −2).
Centroid:
The slope of the line that contains B(4, −2) and the
​
= ​ —
midpoint of AC
​ ​, (0, 2), is —
​
​ = −1.
is x = 0.
230
⋅ 5 + 2 = 32
—
4
2 − (−2)
0−4
−4
The equation of the line that contains B(4, −2) and the
—
midpoint of AC
​ ​, (0, 2), is y = −x + 2.
Chapter 6
The slope of the line that contains C(0, 6) and the midpoint
​
= ​ —
​ = −4.
of AB
​ ​, (2, −2), is —
​
—
−2 − 6 −8
2−0
2
The equation of the line that contains C(0, 6) and the
—
midpoint of AB
​ ​, (2, −2), is y = −4x + 6.
The intersection of y = x + 2 and y = −4x + 6 is:
−x + 2 = −4x + 6
3x = 4
4
x = —
​   ​
3
4 6 2
4
​   ​ = —
​   ​
y = −​ —  ​  + 2 = −​ —  ​  + —
3
3 3 3
( 43 23 )
R
Prove ∠R ≅ ∠P ≅ ∠Q
Assume temporarily that △PQR is equilateral and
P
Q
11. By the Triangle Midsegment Theorem (Thm. 6.8),
GH = —
​ 12 ​ FD. By the markings EG = GD. By the Segment
Addition Postulate (Post. 1.2), EG + GD = ED. So, when
you substitute EG for GD, you get EG + EG = ED,
or 2(EG) = ED, which means that EG = —
​ 12 ​ ED.
1
So, the area of △GEH = —
​ 2 ​bh
7 + 9 > x
The possible lengths of Pine Avenue are
1
= ​ —
​(EG)(GH)
2
1 1
= ​ —
​​ —
​   ​ED  ​​ —
​  12 ​FD  ​
2 2
1
= ​ —
​(ED)(FD).
8
Note that the area of △DEF = —
​ 12 ​bh = —
​ 12 ​(ED)(FD).
So, the area of △GEH = —
​ 18 ​(ED)(FD) = —
​ 14 ​​ —
​  12 ​(ED)(FD)
(  )(  )
[
]​ =
12.
1.8 miles
hiker 1
4 miles
140°
128°
Visitor
Center
4 miles
1.8 miles
52°
Hiker one: 180° − 40° = 140°
Hiker two: 180° − 52° = 128°
Because 140° > 128°, the first hiker is farthest from the
beach
movie
theater
Chapter 6 Standards Assessment (pp. 358–359)
1. B;
( 1 +2 5 5 +2 2 ) ( 62 72 )
—
​  ​ = ​ —
​   ​   ​ = (3, 3.5)
midpoint of MN
​ ​  = ​ —
​   ​, ​ —
​   ​ , —
The slope of the equation that contains the midpoint of MN
​ ​
—
through L(3, 8):
8 − 3.5 4.5
​
= —
​   ​ = undefined
slope = —
​
3−3
0
The equation of the line is x = 3.
​  ​ = ​ —
​   ​ = (4, 5)
midpoint of LN
​ ​  = ​ —
​   ​, ​ —
​   ​ , ​ —
The slope of the equation that contains the midpoint of LN
​ ​
​  14 ​ A.
—
your
house
market
—
( 3 +2 5 8 +2 2 ) ( 82 102 )
—
through M(1, 5):
hiker 2
40°
16 > x
15. To determine the placement of the market, construct the
perpendicular bisectors of each side. Where they intersect is
the location of the market.
equiangular. Then it follows that m∠ P ≠ m∠ Q,
m∠ Q ≠ m∠ R, or m∠ P ≠ m∠ R. By the contrapositive of
the Base Angles Theorem (Thm. 5.6), if m∠ P ≠ m∠ Q,
then PR ≠ QR, if m∠ Q ≠ m∠ R, then QP ≠ RP, and
if m∠ P ≠ m∠ R, then PQ ≠ RQ. All three conclusions
contradict the fact that △PQR is equilateral. So, the
temporary conclusion must be false. This proves that if
△PQR is equilateral, it must also be equiangular.
14. 9 miles; Because the path represents the shortest distance
from the beach to Main Street, it must be perpendicular to
Main Street, and you ended up at the midpoint between
your house and the movie theater. So, the trail must be the
perpendicular bisector of the portion of Main Street between
your house and the movie theater. By the Perpendicular
Bisector Theorem (Thm. 6.1), the beach must be the same
distance from your house and the movie theater. So, Pine
Avenue is the same length as the 9-mile portion of Hill Street
between your house and the beach.
x > 2
2 miles < x < 16 miles.
​    ​  ​.
The coordinates of the centroid are ​ —
​   ​ , —
—​  ≅ QR
—
—
PQ
​ ​  ≅ PR
​ ​
10. Given ​
13. 7 + x > 9
5−5
4−1
0
3
​
= —
​   ​ = 0
slope = —
​
The equation of the line is y = 5.
The intersection of x = 3 and y = 5 is (3, 5). So, the
coordinates of the centroid are (3, 5).
visitor center, because the longer side is opposite the
larger angle.
Geometry
Worked-Out Solutions
231
Chapter 6
2. G; By the Converse of the Perpendicular Bisector Theorem
(Thm. 6.2), because point P is equidistant from points A and
—
B, it lies on the perpendicular bisector of AB
​ ​.
3. 16; Because BD = DF and BE = EG, D is a midpoint of ​
—​  and E is a midpoint of BG
—
—
BF
​ ​. So, by definition, DE
​ ​ is a
midsegment of △BFG. So, by the Triangle Midsegment
1
Theorem (Thm. 6.8), DE = ​ —
​FG.
2
1
DE = —
​ 2 ​FG
4 = —
​ 2 ​FG
2(4) = 2​ —
​ 2 ​FG  ​
8 = FG
Also, because BG = GC and BF = FA, G is a midpoint
FG = —
​ 2 ​AC
8 = —
​ 2 ​AC
2(8) = 2​ —
​ 2 ​AC  ​
16 = AC
8. C; In Step 1, the two points where arcs meet are each
equidistant from A and B, so by the Converse of the
Perpendicular Bisector Theorem (Thm. 6.2), the constructed
—
line is the perpendicular bisector of AB
​ ​. Similarly, in Step 2
—
the constructed line is the perpendicular bisector of BC
​ ​. So,
the point where the perpendicular bisectors intersect is the
circumcenter, which is equidistant from each vertex. So, in
Step 3, the circle is circumscribed about △ABC.
9. J; ∠ 4 forms a linear pair with ∠ 3, so ∠ 4 and ∠ 3 are
supplementary. Also, because they are corresponding angles
formed by parallel lines, ∠ 3 ≅ ∠ 7. So, ∠ 4 and ∠ 7 are
supplementary but not necessarily congruent. So, statement
J is not true.
1
(  )
1
—
—
—
of BC
​ ​ and F is a midpoint of BA
​ ​. So, by definition, FG
​ ​  is
a midsegment of △ABC. So, by the Triangle Midsegment
1
Theorem (Thm. 6.8), FG = ​ —
​AC.
2
10. A; By the Converse of the Hinge Theorem (Thm. 6.13),
m∠ JKM < m∠ LKM. Because m∠ LKM = 25°, the only
possible measure for ∠ JKM is 20°.
1
1
(  )
1
—
So, the length of AC
​ ​ is 16 units.
4. A; Reflection in the x-axis
(x, y) → (x, −y) : M(3, −1) → Z(3, 1).
5. G; By the Triangle Inequality Theorem (Thm. 6.11), you
get x + 16 > 24, x + 24 > 16, and 16 + 24 > x. Disregard
x + 24 > 16 because it has a negative solution, and all side
lengths will have to be greater than a negative number. Solve
x + 16 > 24 to obtain x > 8, and solve 16 + 24 > x to obtain
40 > x. So, the combined solution is 8 < x < 40.
—
—
6. A; Because YG
​ ​ is a perpendicular bisector of DF
​ ​, you know
that ∠ DEY and ∠ FEY are congruent right angles, and by
the Perpendicular Bisector Theorem (Thm. 6.1), YD = YF.
So, you can prove △DEY ≅ △FEY by the HL Congruence
Theorem (Thm. 5.9). (Note: You can also use the Reflexive
Property of Congruence (Thm. 2.1) and the SAS Congruence
Theorem (Thm. 5.5).)
7. H; The transformation (x, y) → (3x, 3y) is a dilation because
the image will be similar to the original but dilated by a
factor of 3. (x, y) → (3x, y) is a horizontal stretch of the
original. (x, y) → (−y, x) is a 90° rotation about the origin
and represents a rigid motion, so the image is congruent to
the original. (x, y) → (x + 3, y + 3) is a translation 3 units
right and 3 units up and represents a rigid motion, so the
image is congruent to the original.
232
Geometry
Worked-Out Solutions