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.
 A: If AP = 2PD and BP = 2PE, then P is the centroid. ….(#)
From (#), AD is then served both as the median and altitude. This further implies AB = AC. …. (%)
From (#), BE is then served both as the median and angle bisector.
By bisector theorem BA : BC = AE : EC = 1 : 1, meaning that BA = BC. ….(@)
(%) + (@) implies  △ABC is equilateral.
A: 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.
A: 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 $C=C^*$.
