Bisectors
Let
A
B
C
D
be a trapezoid with
A
B
/
/
C
D
. The bisectors of
∠
C
D
A
and
∠
D
A
B
meet at
E
, the bisectors of
∠
A
B
C
and
∠
B
C
D
meet at
F
, the bisectors of
∠
B
C
D
and
∠
C
D
A
meet at G, and the bisectors of
∠
D
A
B
and
∠
A
B
C
meet at
H
. Quadrilaterals
E
A
B
F
and
E
D
C
F
have areas
2
4
and
3
6
, respectively, and triangle
A
B
H
has area
2
5
.
The area of triangle C D G can be expressed as n m , where m , n are coprime positive integers.
Find m + n .
The answer is 263.
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2 solutions
Because of the four angle bisectors, perpendiculars dropped on each side
are equal to h.
∴
x
=
E
F
=
a
n
d
∣
∣
A
B
a
n
d
C
D
.
⟹
A
E
F
B
a
n
d
E
D
C
F
a
r
e
t
r
a
p
e
z
o
i
d
s
.
A
r
e
a
Δ
E
H
F
=
2
5
−
2
4
=
1
A
r
e
a
s
Δ
E
H
F
Δ
A
H
B
=
1
2
5
=
(
x
A
B
)
2
.
∴
A
B
=
5
x
.
.
.
.
.
.
.
.
.
.
.
.
(
∗
)
A
r
e
a
s
A
E
F
B
E
D
C
F
=
2
4
3
6
=
2
3
=
2
1
∗
h
∗
(
x
+
5
x
)
2
1
∗
h
∗
(
x
+
D
C
)
.
.
.
.
(
∗
∗
)
∴
D
C
=
8
x
.
.
.
.
(
∗
∗
∗
)
Δ
s
G
D
C
=
E
D
C
F
+
G
E
F
.
∴
A
r
e
a
s
G
E
F
G
D
C
=
G
E
F
3
6
+
G
F
F
=
(
x
2
(
8
x
)
2
)
S
o
l
v
i
n
g
a
r
e
a
G
E
F
=
7
4
.
⟹
A
r
e
a
G
E
C
=
3
6
+
7
4
.
=
7
2
5
6
=
n
m
.
m
+
n
=
.
.
.
.
.
.
.
2
6
3