You should use that theorem
Given that
triangle
A
B
C
is an irregular triangle,
A
D
is the median which bisects
B
C
, and the green, blue and red regions are squares. If the total area of green region and blue region is 9, find the area of the red region.
The answer is 18.
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4 solutions
Stewart’s theorem with usual notation,
b
2
∗
m
+
c
2
∗
n
=
a
∗
(
d
2
+
m
∗
n
)
.
B
u
t
h
e
r
e
m
=
n
=
2
1
∗
a
.
⟹
b
2
∗
2
1
∗
a
+
c
2
∗
2
1
∗
a
=
a
∗
(
d
2
+
2
1
∗
a
∗
2
1
∗
a
)
.
a
=
0
,
∴
d
i
v
i
d
i
n
g
b
o
t
h
s
i
d
e
s
b
y
a
,
b
2
∗
2
1
+
c
2
∗
2
1
=
(
d
2
+
2
1
∗
a
∗
2
1
∗
a
)
.
⟹
2
1
(
A
C
2
∗
+
A
B
2
)
=
(
A
D
2
+
B
D
∗
D
C
)
B
u
t
B
D
=
D
C
,
∴
A
C
2
∗
+
A
B
2
=
2
∗
(
A
D
2
+
B
D
2
)
=
2
∗
9
=
1
8
.
Its a simple application of appolonius theorem
Using Apollonius' Theorem , we get
A B 2 + A C 2 = 2 ( A D 2 + B D 2 )
The area of red region = 2 × 9 = 1 8
And it is done!