Question 133
Let $L$ be a point on the side $AC$ of a triangle $ABC$ such that $BL$ is the bisector of the angle $ABC$. Let $M$ be a point on the segment $CL$. The tangent through $B$ to the circumcircle $\Omega$ of $ABC$ meets the ray $CA$ at $P$. Tangents through $B$ and $M$ to the circumcircle $\Gamma$ of $\triangle BLM$ meet at $Q$. Prove that $PQ \parallel BL$.
(A. Kuznetsov)
(A. Kuznetsov)