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.

Hint: Show that $AB \parallel DE$.
Hint: Show that $BD=DE= \frac{AB}{2}$. 
Hint: Show that $\angle ABD = 60^\circ$. 
Define $C^*$ on $BC$ such that $ABC^*$ is an equilateral triangle.
Hint: Show that $P$ is the centroid of $ABC^*$.
Use the fact that $BP = 2 PE$ to conclude that $C=C^*$.
