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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.
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.
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.
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 $C=C^*$.
