Question 59
Let $\omega$ be the circumcircle of triangle $ABC$. Points $K$, $L$, $M$ lie on sides $AB$, $BC$, $CA$, respectively, and satisfy $CM \cdot CL = AM \cdot BL$. The ray $LK$ meets the line $AC$ at $P$. The common chord of $\omega$ and the circumcircle of triangle $KMP$ meets the segment $AM$ at $S$. Prove that $SK \parallel BC$.