Question 53
Let $\triangle ABC$ be an isosceles triangle with $AB = AC$. The incircle $\Gamma$ of $\triangle ABC$ has centre $I$, and it is tangent to the sides $AB$ and $AC$ at $F$ and $E$ respectively. Let $\Omega$ be the circumcircle of $\triangle AFE$. The two external common tangents of $\Gamma$ and $\Omega$ intersect at a point $P$. If one of these external common tangents is parallel to $AC$, prove that $\angle PBI = 90^\circ$.