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LESSON
5.3
Answers for the lesson “Use Angle Bisectors
of Triangles”
Skill Practice
16. No; you do not know that the
perpendicular segment bisects the
angle.
1. bisector
2. Perpendicular bisectors bisect
line segments while angle
bisectors bisect angles; both
divide the segment or angle into
two equal parts, and both have
special points of intersection.
3. 208
4. 12
5. 9
6. Yes; ŽBAD > ŽCAD, }
DB > }
AB
and }
DC > }
AC so by the Angle
Bisector Theorem DB 5 DC.
7. No; you do not know that
Ž BAD > ŽCAD.
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8. No; you do not know that
}
DB > }
AB or }
DC > }
AC.
9. No; you don’t know that
]›, or
}
HG > }
HF, }
HF > EF
]›.
}
HG > EG
17. Yes; x 5 7 using the Angle
Bisector Theorem.
18. B
19. 9
20. 8
21. GD is not the perpendicular
}. The same
distance from G to CE
is true about GF; the distance
from G to each side of the triangle
is the same.
22. T is not the incenter of nUWY.
Sample answer: }
UZ > }
ZY,
}
}
}
}
WX > XY, and UV > VW
23. C
24. 6
25. 0.5
26. They all have the same length;
Concurrency of Angle Bisectors
of a Triangle Theorem.
Sample:
B
10. Yes; Converse of Angle Bisector
Theorem
11. No; you don’t know that
]›.
]› or }
}
HF > EF
HG > EG
12. 5
13. 4
14. 8
A
C
15. No; the segments with length x
and 3 are not perpendicular to
their respective rays.
Geometry
Answer Transparencies for Checking Homework
146
27. Sample answer: Since nABC
30. AAS; HL;
is a right triangle, its area is
B
1
2
}(AB + AC). The area of nABC
is also the sum of the areas of
nABD, nADC, and nDBC.
D
A
This sum is
C
1
2
1
2
1
2
}x(AB) 1 }x(AC) 1 }x(BC), or
1
2
}x(AC 1 AB 1 BC). Setting
B
D
A
1
2
}(AB + AC) equal to
1
}x(AC 1 AB 1 BC) and solving
2
C
31. a. Equilateral; 3; the angle
bisector would also be the
perpendicular bisector.
AB + AC
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for x gives x 5 }}
.
AC 1 AB 1 BC
b. Scalene; 6; each angle
Problem Solving
bisector would be different
than the corresponding
perpendicular bisector.
28. No; G is on the angle bisector of
ŽLBR.
29. At the incenter of the pond;
A
32. Angle bisector; more; no; the
diameter of the inscribed circle
is greater than 5 inches.
33. Perpendicular bisectors; (10, 10);
I
100 yd; about 628 yd;
y
B
C
P(10, 20)
T(2, 4)
N(16, 2)
5
2
x
Geometry
Answer Transparencies for Checking Homework
147
34. Statements (Reasons)
1. ŽBAC is bisected by }
AD,
}
}
}
}
DB > AB, DC > AC. (Given)
2. ŽBAD > ŽCAD (Definition of
angle bisector)
3. ŽDBA and ŽDCA are right
angles.
(Definition of
perpendicular lines)
6. nABD > nACD
(HL)
7. ŽBAD > ŽCAD (Corr. parts
s are >.)
of > n
]› bisects ŽABC.
8. AD
(Definiton of angle bisector)
36. Statements (Reasons)
4. ŽDBA > ŽDCA
(Right
Angles Congruence Theorem)
} bisects ŽCAB,
1. nABC, AD
},
} bisects ŽCBA, DE
} >AB
BD
}, DG
} > CA
}. (Given)
} > BC
DF
6. nABD > nACD
2. ŽDGC, ŽDFC, ŽDFB, and
ŽDEB are right angles.
(Def. of perpendicular lines)
} > DA
} (Reflexive Property
5. DA
of Segment Congruence)
7. }
DB > }
DC
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5. }
AD > }
AD (Reflexive Property
of Segment Congruence)
8. DB 5 DC
(AAS)
(Corr. parts
s are >.)
of > n
(Definition of
congruent segments)
35. Statements (Reasons)
1. ŽBAC with D in its interior,
}
DB > }
AB, }
DC > }
AC,
DB 5 DC.
(Given)
2. ŽABD and ŽACD are right
angles.
(Definition of
perpendicular lines)
3. nABD and nACD are right
triangles.
(Definition of
right triangle)
4. }
BD > }
CD
(Definition of
congruent segments)
3. nCGD, nCFD, nBED, and
nBFD are right triangles.
(Definition of right triangle)
BD > }
BD, }
CD > }
CD
4. }
(Reflexive Property
of Segment Congruence)
5. ŽEBD > ŽFBD (Definition of
angle bisector)
6. The angle bisector of ŽACB
passes through point D, the
incenter of nABC.
(Definition of incenter)
7. ŽGCD > ŽFCD
(Definition of angle bisector)
8. nCGD > nCFD,
nDEB > nDFB
(AAS)
Geometry
Answer Transparencies for Checking Homework
148
36. (cont.)
9. }
DG > }
DF, }
DE > }
DF
s are >.)
(Corr. parts of > n
10. }
DG > }
DE > }
DF
Mixed Review of
Problem Solving
1. Sample answer: The park would
be located outside of the county.
(Transitive Property of
Segment Congruence)
Elk County
37. a. Use the Concurrency of Angle
Bisectors of Triangle Theorem;
if you move the circle to any
other spot it will extend into
the walkway.
Q
Elm
County
Z
Deer
County
X
Y
Bear
County
R
Forest County
I
2. a. incenter; angle bisectors
Q
R
Q
R
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P
b. HL
c. 3.9 cm; (AE)2 1 (EG)2 5
P
P
b. Yes; the incenter will allow the
largest tent possible.
38. Sample answer: Construct three
circles exterior to the triangle,
each one tangent to one side of
the triangle and the other two
lines. The centers of the circles
are the three points.
(GA)2 or 72 1 (EG)2 5 82
‹]›
3. AC ; y 5 2x 1 9; the y-intercepts
are 26, 4, and 9. The slope of
the line with 9 as the y-intercept
is 21 so the equation of that line
is y 5 21x 1 9.
y
H(6, 8)
A(4, 5)
B(8, 6)
2
22
J(10, 4)
G(2, 2) C(6, 3)
x
Geometry
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149
4. 9 ft;
9
5. a. 262 ft
b. 840 ft2
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6. Equilateral triangle
} i PR
}
7. ŽQPR, ŽSTU, ŽTUR; ST
} a transversal, so ŽQPR
with QP
and ŽQST are corresponding
} i UT
} a trans} with ST
angles. PQ
versal, so ŽQST and ŽSTU are
} i PR
}
alternate interior angles. ST
} a transversal, so ŽSTU
with TU
and ŽTUR are alternate interior
angles and ŽQST > ŽTUR by the
Transitive Property.
Geometry
Answer Transparencies for Checking Homework
150