Let $ABC$ be an obtuse triangle with $AB = AC$, and let $\Gamma$ be the circle that is tangent to $AB$ at $B$ and to $AC$ at $C$. Let $D$ be the point on $\Gamma$ furthest from $A$ such that $AD$ is perpendicular to $BC$. Point $E$ is the intersection of $AB$ and $DC$, and point $F$ lies on line $AB$ such that $BC = BF$ and $B$ lies on segment $AF$. Finally, let $P$ be the intersection of lines $AC$ and $DB$. Show that $PE = PF$.