Triangle Area Problem with intersecting lines I recently saw this geometry problem and it's been killing me. Given $\overline{AB}=\overline{AC}=20$, $\overline{AD}=\overline{AE}=12$, and $[ADFE]=24$ with $[\cdot]$ denoting area, find the area of $\triangle BCF$ (see diagram)

I started by considering line $\overline{DE}$. It's pretty clear to see that $\triangle DEF \sim \triangle CBF$. We also have $\triangle {ADE}\sim\triangle{ABC}$, with $9[ABC]=25[ADE]$. But once i've figured out these parts, I get stuck. What would be the next steps in this situation? I assume I'd gave to take advantage of the area ratios I've gotten, but I can't workout the details of that argument.
 A: Connect $AF$. By symmetry, $[ADF] = [AEF]= 12$.
Notice that $\triangle DEF \sim \triangle CBF$ in the same ratio $3:5$, so we have $\dfrac {EF}{FC} = \dfrac {DF}{FB} = \dfrac35$.
Hence $\dfrac {[AEF]}{[AFC]} = \dfrac35 = \dfrac {12}{12+ [DFC]}$, giving $[DFC] = 8$.
We also have $\dfrac {[CDF]}{[FBC]} = \dfrac 35$, giving $[FBC] = \dfrac{40}3$.
A: $CD = BE = 8$
Join $AF$. $[ADF] = [AEF] = 12$
Now
$$ \dfrac{[AFD]}{[FDC]} = \dfrac{AD}{CD} = \dfrac{12}{8}$$
$$ \Rightarrow [FDC] = 8 = [FEB]$$
Also
$$ \dfrac{AD}{CD} = \dfrac{[ABD]}{[BDC]} =  \dfrac{[ADFE]+[FEB]}{[FDC]+[BFC]} = \dfrac{24+8}{8+[BFC]}$$
$$ \Rightarrow [BFC] = \dfrac{40}{3}$$
A: 
$\triangle ADF = \frac{1}{2} \times 24 = 12$
$\triangle ADF$ shares the same altitude with $\triangle CDF$ (from point $F$ to $AC$). So their area is in the ratio of their bases $12:8$.
$\triangle CDF = 8 \implies \triangle ACE = (8 + 24) = \frac{1}{2} \times AC \times EH$
As $EH = \frac{12}{20} BG$
$\frac{1}{2} \times AC \times BG = \frac{160}{3} = \triangle ABC$
So $\triangle BCF = \frac{160}{3} - 24 - 8 -8 = \frac{40}{3}$
