Question 56
The bisector of the angle on vertex $A$ of triangle $ABC$ intersects the circumcircle of triangle $ABC$ at point $F$ ($F \neq A$). Points $D$ and $E$ are chosen on the sides $AB$ and $AC$, respectively, in such a way that the lines $DE$ and $BC$ are parallel. Let $G$ and $H$ be the points of intersection of the rays $FD$ and $FE$, respectively, with the circumcircle of triangle $ABC$ ($G \neq F, H \neq F$). The circumcircles of triangles $AGD$ and $AHE$ intersect at point $P$ ($P \neq A$). Prove that point $P$ lies on the line $AF$.