Given triangle $ABC$ with $CE=6 cm $ , $BD=9 cm $ , $\angle A = 60$ , CE , BD are medians. Find the area of $ ABC $ 
Given  triangle  $ABC$ with  $CE=6 cm $ , $BD=9 cm $ , $\angle A = 60$ , CE , BD are medians. 
  Find the area of $ ABC $  


This problem is too difficult for me , the teacher said it is a challenge  problem.
Any help will be appreciated , thank you. 
 A: Let $AB=c$, $AC=b$ and $BC=a$.
Thus, $a^2=b^2+c^2-bc$, $\frac{1}{2}\sqrt{2a^2+2c^2-b^2}=9$ and $\frac{1}{2}\sqrt{2a^2+2b^2-c^2}=6$, which gives
$$\frac{2(b^2+c^2-bc)+2c^2-b^2}{2(b^2+c^2-bc)+2b^2-c^2}=\frac{9}{4}$$
A: Well let's use the following names:$AC=2x$, $AB=2y$, and $\angle BAC=\alpha$. 
Now write the cosine law for both the triangles $ADB$ and $AEC$ both centred at the angle $\alpha$:
$$
\begin{cases}
 ABD: x^2 + 4y^2-4xy\cos\alpha = |DB|^2 \\
 AEC: 4x^2 + y^2 - 4xy\cos\alpha= |EC|^2
\end{cases}
$$
Now replacing your numerical values gives:
$$
(*)
 \begin{cases} 
x^2 + 4y^2- 2xy = 81 \quad(1)  \\
4x^2+y^2-2xy=36      \quad(2)
\end{cases}
$$
Now there probably should be some slick way to get $xy$ out of this. Since I am  not that bright, going to wolfram alpha gives:
$$
xy = \frac{1}{7}\left(39 + 5 \sqrt{249} \right)
$$
And your area, A, according to a simple trigonometry is:
$$
A= 2xy\sin\alpha = \frac{2}{7}\left(39 + 5 \sqrt{249} \right) \frac{\sqrt3}{2}
$$

Update:
According to @Lozenges, System $(*)$ can be solved as follows:
Denote $xy=p$, and $x+y=s$. Then, add $(1)$ and $(2)$ to get:
$$ 5s^2 - 14 p - 117 = 0. \quad (3)$$
Subtract $(2)$ from $(1)$ to get:
$$ s^4 - 4ps^2 -225= 0. \quad (4)$$
Eliminating $s$ from $(3)$ and $(4)$ gives:
$$ 7p^2-78p-672=0, $$
whose solution is:
$$ p = \frac{1}{7}\left(39 + 5 \sqrt{249} \right). $$
A: I tried to find an elegant geometric solution but couldn't. Here is the gung-ho algebraic version.
First by cosine rule about $A$  in $\Delta ADB$, and $\Delta AEC$, we get $$b^2 + c^2/4 -bc/2 = 36$$ and  $$b^2/4 + c^2 -bc/2 = 81$$ From these we get by subtracting $$c^2=b^2 + 60 $$ Substituting this in the first equation we get
$$5b^2/4 -21 = b\sqrt{b^2+60}  $$ Squaring both sides 
$$25b^4/16 + 221 -105b^2/2 = b^4 + 60b^2 $$ $$b^4 - 200b^2 + 3536/9 = 0$$
You can solve this to get $b^2$. Since $c = \sqrt{b^2+60}$, and therfore the area is $$1/2 \cdot bc \cdot \sin(60)=\sqrt3/4 \cdot b \cdot \sqrt{b^2+60} $$
$$\Delta^2 = 3/16 \cdot b^2 \cdot (b^2 + 60) $$
A: 
Let's express coordinates 
of the vertices of $\triangle ABC$ 
in some scaled units ($\mathrm{su}$)
such that $\triangle ABC$ is inscribed in the unit circle
(with the radius $R=1\,\mathrm{su}$).
Since $\angle BAC=\tfrac\pi3$,
let's start with an equilateral $\triangle A_0BC$, inscribed in a unit circle
with the center $O$.
In order to get the location 
of the vertex $A$,
we need to rotate $A_0$ by the angle $\phi\in(0,\tfrac{2\pi}3)$
such that $|BD|:|CE|=9:6=3:2$,
 or $|BD|^2:|CE|^2=9:4$.
Thus the coordinates of vertices in ($\mathrm{su}$) are:
\begin{align}
A&=(\cos\phi,\sqrt{1-\cos^2\phi}),\,
B=(-\tfrac12,-\tfrac{\sqrt3}2),\,
C=(-\tfrac12,\tfrac{\sqrt3}2)
,\\
D&=(\tfrac12\cos\phi-\tfrac14,\tfrac12\sin\phi+\tfrac14\sqrt3))
,\\
E&=(\tfrac12\cos\phi-\tfrac14,\tfrac12\sin\phi-\tfrac14\sqrt3))
.
\end{align}
Condition
\begin{align}
4\cdot|B-D|^2&=9\cdot|C-E|^2
\end{align}
simplifies to a quadratic equation in $\cos\phi$:
\begin{align}
4588\cos^2\phi+400\cos\phi-2963
,
\end{align}
which provides one valid root
\begin{align}
\cos\phi&=\tfrac{117}{2294}\sqrt{249}-\tfrac{50}{1147}
.
\end{align}
The area of the triangle $ABC$
\begin{align}
S_{\triangle ABC}&=|CF|\cdot|FG|
, 
\end{align}
expressed
in scaled units (su)
\begin{align}
 S'_{\triangle ABC}&= 
 (\tfrac{\sqrt3}2 \cdot(\tfrac12+\cos\phi))\,\mathrm{su}^2
 =
(\tfrac{1047}{4588}\sqrt3+\tfrac{351}{4588}\sqrt{83})
\,\mathrm{su}^2.
\end{align}
We can convert this value to $\mathrm{cm}^2$,
using one of known medians, for example, $CE$:
\begin{align}
|CE|^2&=36\,\mathrm{cm}^2
=\left(2+\tfrac14\cos\phi-\tfrac34\sqrt3\sqrt{1-\cos^2\phi}\right)
\,\mathrm{su}^2
=
(\tfrac{1404}{1147}+\tfrac9{1147}\sqrt{83}\sqrt3)
\,\mathrm{su}^2
,\\
1\,\mathrm{su}^2&=
(\tfrac{208}{7}-\tfrac4{21}\sqrt{83}\sqrt{3}) \,\mathrm{cm}^2
,\\
S_{\triangle ABC}&=
(\tfrac{1047}{4588}\sqrt3+\tfrac{351}{4588}\sqrt{83}
)\,\mathrm{su}^2
\cdot
(\tfrac{208}{7}-\tfrac4{21}\sqrt{83}\sqrt{3}) \,\frac{\mathrm{cm}^2}{\mathrm{su}^2}
=
\tfrac{39}7\sqrt3+\tfrac{15}7\sqrt{83}
\approx
29.172355
.
\end{align}
