Question 124
Let $\triangle ABC$ be a triangle with an incenter $I$. The line $CI$ intersects for a second time the circumcircle of $\triangle ABC$ at $L$, where $CI = 2 \cdot IL$. Points $M$ and $N$ lie on the segment $AB$, such that $\angle AIM = \angle BIN = 90^\circ$. Prove that $AB = 2 \cdot MN$.