Question 117
Let $ABC$ be an isosceles triangle with $AB = BC$. Let $CC'$ be the diameter of its circumcircle $\Omega$. The line through $C'$ parallel to $BC$ intersects segments $AB$ and $AC$ at $M$ and $P$, respectively. Prove that $M$ is the midpoint of $C'P$. (B. Obukhov)