Since
LMK
MNK
and
K
K
,
LMK
MNK
by AA Similarity.
M
LK
K
¬
M
NK
K
2(
K
1
P
6
)
9
KP
¬
2(
KP
)
9
KP
3(
2
K
5
P
)
¬
3(
K
9
P
)
25(9)
¬
9(
KP
)
2
25
¬
(
KP
)
2
5
¬
KP
KM
KP
PM
KP
2(
KP
)
3(
KP
)
3(5) or 15
PM
2(
KP
)
2(5) or 10
Since
L
QRM
and
K
QPM
,
LKM
RPM
by AA Similarity.
L
R
K
P
¬
K
PM
M
R
25
P
¬
1
1
5
0
25(10)
¬
15(
RP
)
2
1
5
5
0
¬
RP
5
3
0
¬
RP
Since
KNM
PQM
and
K
QPM
,
KNM
PQM
by AA Similarity.
K
PQ
N
¬
K
PM
M
P
9
Q
¬
1
1
5
0
9(10)
¬
15(
PQ
)
90
¬
15(
PQ
)
6
¬
PQ
RP
RQ
PQ
5
3
0
RQ
6
5
3
0
6
RQ
3
3
2
RQ
m
MQP
m
QPM
m
PMQ
¬
180
90
m
QPM
m
PMQ
¬
180
m
QPM
m
PMQ
¬
90
Since
RMP
is a right angle,
m
RMQ
m
PMQ
90
m
QPM
m
PMQ
m
RMQ
m
PMQ
m
QPM
m
RMQ
, so
QPM
RMQ
.
Therefore,
RQM
MQP
by AA Similarity.
M
RQ
Q
¬
M
QP
Q
¬
M
6
Q
(
MQ
)
2
¬
3
3
2
6
(
MQ
)
2
¬
64
MQ
¬
8
R
M
M
P
¬
Q
Q
M
P
R
1
M
0
¬
8
6
6(
RM
)
¬
10(8)
RM
¬
8
6
0
or
4
3
0
Since
KNM
PQM
,
N
QM
M
¬
K
PQ
N
N
8
M
¬
9
6
6(
NM
)
¬
8(9)
NM
¬
7
6
2
or 12
Since
L
QRM
and
LNM
RQM
,
QRM
NLM
by AA Similarity.
Q
N
R
L
¬
R
LM
M
¬
3
3
2
(
LM
)
¬
16
4
3
0
LM
¬
16
4
3
0
3
3
2
LM
¬
20
30.
X
IJ
J
H
YJ
J
and
J
J
,
so
IJH
XJY
by SAS Similarity.
m
JXY
180
m
WXJ
180
130
50
m
JIH
m
JXY
by corr.
50
m
YIZ
m
JIH
by vert.
50
m
JYX
m
YIZ
+
m
WZG
50
20
70
because exterior angle
sum of remote interior
angles
m
JHI
m
JYX
70 by corr.
m
J
m
JXY
m
JYX
¬
180
m
J
50
70
¬
180
m
J
¬
60
m
JHG
m
JHI
¬
180 Linear pair
m
JHG
70
¬
180
m
JHG
¬
110
31.
RST
is a right angle, so
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 Fall '10
 Dr.Zhan
 Calculus, PreCalculus, Trigraph, AA Similarity

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