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123
Chapter 5
10. Given:
C
w
D
w
is the perpendicular bisector of
A
w
B
w
.
E
is a point on
C
w
D
w
.
Prove:
EB
5
EA
Proof:
C
w
D
w
is the perpendicular bisector of
A
w
B
w
.By
definition of perpendicular bisector,
D
is the
midpoint of
A
w
B
w
.Thus,
A
w
D
w
>
B
w
D
w
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
C
w
D
w
,
/
EDA
and
/
EDB
are right angles and
are congruent. By the Reflexive Property,
E
w
D
w
>
E
w
D
w
.Thus,
n
EDA
>
n
EDB
by SAS.
E
w
B
w
>
E
w
A
w
because CPCTC, and by definition of
congruence,
EB
5
EA
.
11. Given:
n
UVW
is isosceles with vertex angle
UVW
.
Y
w
V
w
is the bisector of
/
UVW
.
Prove:
Y
w
V
w
is a median.
Proof:
12. Given:
G
w
L
w
is a median of
n
EGH
.
J
w
M
w
is a median of
n
IJK
.
n
EGH
>
n
IJK
Prove:
G
w
L
w
>
J
w
M
w
Proof:
13.
M
w
S
w
is an altitude of
n
MNQ
,so
/
MSQ
is a right
angle.
m
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This note was uploaded on 12/19/2011 for the course MAC 1140 taught by Professor Dr.zhan during the Fall '10 term at UNF.
 Fall '10
 Dr.Zhan
 Calculus, PreCalculus

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