123
Chapter 5
10. Given:
CD
is the perpendicular bisector of
AB
.
E
is a point on
CD
.
Prove:
EB
EA
Proof:
CD
is the perpendicular bisector of
AB
. By
definition of perpendicular bisector,
D
is the
midpoint of
AB
. Thus,
AD
BD
by the Midpoint
Theorem.
CDA
and
CDB
are right angles by
definition of perpendicular. Since all right angles
are congruent,
CDA
CDB
. Since
E
is a point
on
CD
,
EDA
and
EDB
are right angles and
are congruent. By the Reflexive Property,
ED
ED
. Thus,
EDA
EDB
by SAS.
EB
EA
because CPCTC, and by definition of
congruence,
EB
EA
.
11. Given:
UVW
is isosceles with vertex angle
UVW
.
YV
is the bisector of
UVW
.
Prove:
YV
is a median.
Proof:
12. Given:
GL
is a median of
EGH
.
JM
is a median of
IJK
.
EGH
IJK
Prove:
GL
JM
Proof:
13.
MS
is an altitude of
MNQ
, so
MSQ
is a right
angle.
m
1
m
2
¬
90
3
x
11
7
x
9
¬
90
10
x
20
¬
90
10
x
¬
70
x
¬
7
m
2
¬
7(7)
9
¬
58
14.
MS
is a median of
MNQ
, so
QS
SN
.
3
a
14
¬
2
a
1
a
14
¬
1
a
¬
15
m
MSQ
¬
7
a
1
¬
7(15)
1
¬
106
MS
is not an altitude of
MNQ
because
m
MSQ
106.
I
K
M
J
E
H
L
G
U
Y
W
V
A
D
C
E
B
Statements
Reasons
1.
UVW
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 Fall '10
 Dr.Zhan
 Calculus, PreCalculus, Trigraph, perpendicular bisector, eB eA

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