I thought this would be easy but I can't seem to find the answer.
Edit: I did my best to draw the diagram:
$\overline{EC}=\frac{1}{3} \overline{AC}, \overline{AF}=\frac{1}{3} \overline{AB}, \overline{BD}=\frac{1}{3} \overline{BC}$
I drew a line parallel to $\overline{BE}$ through $R$. I called the intersection of that line and $\overline{AB}$ point $M$. I drew a line parallel to $\overline{AD}$ through $B$, and called the intersection of that line and my previous line $G$. Then I want to prove $BGRP = 2\triangle PRQ$. This is for the ultimate goal of solving the $1/7$ area triangle problem. In addition I need to prove $\triangle BGM = \triangle ARM$, and to prove that I think I need to prove $M$ is the midpoint. I think I'm close with similar triangles $\triangle ARM$ and $\triangle APB$ but I can't get the relationship. Thanks if you managed to read all this.
Edit: A full proof is given here, I just need someone to clarify the dialations and transformations.
Proving $\triangle BGM ~ \triangle ARM$ is easy.
$\angle GMB = \angle AMR$ because they are vertically opposite angles.
$\angle GBM = \angle MAR$ because line $BG$ is parallel to line $AR$, and the angles are corresponding angles made by intercept $GR$.
For proving $BGRP=2PRQ$, you need to show $BP=PQ$ which is not clear to me using this information.