Given $\triangle ABC$. Let $D$ be the point where the altitude form the $A$ vertex intersect $\overline{BC}$ and the point $E$ is the intersect between the bisector of $\angle ABC$ with $\overline{AC}$. Let $P$ be the point of intersect of $\overline{AD}$ with $\overline{BE}$.
Prove that if $AP=2PD$ and $BP=2PE$, then $\triangle ABC$ is equilateral.
This is essentially what I've tried. But I don't know how to continue, I can't find any useful congruences.