Question 163
Let $ABC$ be a triangle whose three angles are acute, with $AB \gt AC$, and let $\Omega$ be its circumcircle. Let $M$ be the midpoint of $[BC]$. The tangents to $\Omega$ at $B$ and $C$ meet at $P$, and the lines $(AP)$ and $(BC)$ meet at $S$. Let $D$ be the foot of the altitude from $B$ in $ABP$, and $\omega$ the circumcircle of $CSD$. Finally, let $K$ be the second point of intersection (other than $C$) of $\omega$ and $\Omega$.
Prove that $\widehat{CKM} = 90^{\circ}$.
Prove that $\widehat{CKM} = 90^{\circ}$.