Question 29
Point $M$ is the midpoint of side $AC$ of an acute triangle $ABC$ with $AB\gt BC$. Let $\Omega$ be its circumcircle. The tangents to $\Omega$ at $A,C$ meet at $P$. Segments $BP,AC$ meet at $S$. Let $AD$ be the altitude of triangle $ABP$. The circumcircle $\omega$ of triangle $CSD$ meets $\Omega$ at $K\ne C$. Prove that $\angle CKM=90^\circ$. (V. Shmarov)