Find sides of triangle, and radius of inscribed circle

Question

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

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?

Answer 1

In the standard notation: $$BD=\frac{a+c-b}{2}$$ and $$CF=\frac{a+b-c}{2}.$$ Thus, $$BC=BD+CF=6.$$ Also, $$\frac{S_{\Delta ADF}}{AG\cdot AE}=\frac{\frac{1}{2}AD\cdot AF\cdot\sin60^{\circ}}{AF^2}=\frac{\sqrt3}{4}.$$

Answer 2

Tangents to a circle from the same point, in this case, $BD$ and $BE$ from the point $B$, and $CE$ and $CF$ from the point $C$ have the same lengths. Therefore,

$$BC = BE + EC = BD + CF = 4 + 2 = 6$$

This means the question is giving more than enough information.