Points $A, B$ and $P$ lie on the circumference of a circle $\Omega_1$ such that $\angle APB$ is an obtuse angle. Let $Q$ be the foot of the perpendicular from $P$ on $AB$. A second circle $\Omega_2$ is drawn with centre $P$ and radius $PQ$. The tangents from $A$ and $B$ to $\Omega_2$ intersect $\Omega_1$ at $F$ and $H$ respectively. Prove that $FH$ is tangent to $\Omega_2$.