Question 59
Let $P$ be a point inside the triangle $ABC$. The lines $AP$, $BP$ and $CP$ intersect the sides $BC$, $CA$ and $AB$ at points $D$, $E$ and $F$, respectively. Let $Q$ be a point on the ray $[BE$ such that $E \in [BQ]$ and $\angle EDQ = \angle FDB$. Suppose that $BE \perp AD$ and $|DQ| = 2|BD|$. Prove that $\angle FDE = 60^\circ$.