Question 45
Let $M$ be a midpoint of the side $AC$ of an acute-angled triangle $ABC$ with $AB \gt BC$. Let $\Omega$ be the circumcircle of the triangle $ABC$. The tangents to $\Omega$ at points $A$ and $C$ meet at $P$. The segments $BP$ and $AC$ meet at $S$. Let $AD$ be the altitude in the triangle $ABP$. The circumcircle $\omega$ of the triangle $CSD$ intersects $\Omega$ at point $K \neq C$. Prove that $\angle CKM = 90^\circ$.