<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] 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
$$\frac{2(b^2+c^2-bc)+2c^2-b^2}{2(b^2+c^2-bc)+2b^2-c^2}=\frac{9}{4}$$
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