Problem

Source: Iran MO second round P4

Tags: geometry



In the triangle $ABC$, $M$ is the midpoint of $AB$ and $B'$ is the foot of $B$-altitude. $CB'M$ intersects the line $BC$ for the second time at $D$. Circumcircles of $CB'M$ and $ABD$ intersect each other again at $K$. The parallel to $AB$ through $C$ intersects the $CB'M$ circle again at $L$. Prove that $KL$ cuts $CM$ in half.