Through the midpoint $M$ of the side $BC$ of the triangle $ABC$ passes a line which intersects the rays $AB$ and $AC$ at $D$ and $E$, respectively, such that $AD=AE$. Let $F$ be the foot of the perpendicular from $A$ onto $BC$ and $P{}$ the circumcenter of triangle $ADE$. Prove that $PF=PM$.
