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.


$\triangle ADF = \frac{1}{2} \times 24 = 12$
$\triangle ADF$ 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 $\triangle BCF = \frac{160}{3} - 24 - 8 -8 = \frac{40}{3}$
