Given $\Omega ABC$ with $AB<AC$, let $AD$ be the bisector of $\angle BAC$ with $D$ on the side $BC$. Let $\Gamma$ be a circle passing through $A$ and $D$ which is tangent to $BC$ at $D$. Suppose $\Gamma$ cuts the side $AB$ again at $E\ne A$. The tangent to the circumcircle of $\Delta BDE$ at $D$ intersects $\Gamma$ again at $F\ne D$. Let $P$ be the intersection point of the segments $EF$ and $AC$. Prove that $PD$ is perpendicular to $BC$.