Let $P$ be a point in the interior of the triangle $ABC$. The lines $AP$, $BP$, and $CP$ intersect the sides $BC$, $CA$, and $AB$ at $D,E$, and $F$, respectively. A point $Q$ is taken on the ray $[BE$ such that $E\in [BQ]$ and $m(\widehat{EDQ})=m(\widehat{BDF})$. If $BE$ and $AD$ are perpendicular, and $|DQ|=2|BD|$, prove that $m(\widehat{FDE})=60^\circ$.