Let $ABC$ be an acute triangle and $D$ be the foot of the altiutude from $A$ on $BC$, $E$ and $F$ are the midpoints of $BD$ and $DC$ respectively. $O$ and $Q$ are the circumcenters of the triangles $\vartriangle BF$ and $\vartriangle ACE$ respectively. $P$ is the intersection point of $OE$ and $QF$, show that $PB = PC$.