<s>[INST] 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. 
 [/INST] 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 $$1/2 \cdot bc \cdot \sin(60)=\sqrt3/4 \cdot b \cdot \sqrt{b^2+60} $$
$$\Delta^2 = 3/16 \cdot b^2 \cdot (b^2 + 60) $$
</s>