Prove that this triangle is equilateral? 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.

By the Angle Bisector Theorem in $\triangle ABD$, 
$$\frac{|BA|}{|BD|} = \frac{|PA|}{|PD|} = \frac{2}{1}$$
Therefore, $\triangle ABD$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle; and, then, so is $\triangle BPD$. This implies that your single-tick-mark segments are congruent to your double-tick-mark segments, so that $\triangle APE \cong \triangle BPD$ (SAS). The conclusion follows.
