<s>[INST] Find sides of triangle, and radius of inscribed circle 

$AD = x, \angle A = 60^{\circ}$

I was asked to find $BC$ when $BD=4, CF=2$
I tried using law of cosines, 
$(4+x)^2 + (2+x)^2 - 2(4+x)(2+x) \cos 60^{\circ} = BC^2$
I get $x^2+6x+12=BC^2$
But the answer is $BC= 6, x^2+6x-24=0$
Also i was asked to find $\frac{\triangle ADF}{AG.AE}$. 
I found $\triangle ADF = \frac{\sqrt3}{4}x^2$. But how to find $AG.AE$? Some hints anyone?
 [/INST] In the standard notation: $$BD=\frac{a+c-b}{2}$$ and $$CF=\frac{a+b-c}{2}.$$
Thus, 
$$BC=BD+CF=6.$$
Also, $$\frac{S_{\Delta ADF}}{AG\cdot AE}=\frac{\frac{1}{2}AD\cdot AF\cdot\sin60^{\circ}}{AF^2}=\frac{\sqrt3}{4}.$$
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