Question 69
Let $\Gamma$ be the circumcircle of acute triangle $ABC$. The bisector of $\angle ABC$ meets $AC$ at $B_1$ and the minor arc $AC$ of $\Gamma$ at $P$. The line through $B_1$ perpendicular to $BC$ meets the minor arc $BC$ of $\Gamma$ at $K$. The line through $B$ perpendicular to $AK$ meets $AC$ at $L$. Prove that $K,L,P$ are collinear.