RMO 2025 K3

Let $ABCDE$ be a convex pentagon in which $AB=AE$ , $CB=CD$ , and $\angle AED =\angle CDE= 90^\circ$ . Let the internal bisectors of $\angle EAB$ and $\angle DCB$ intersect at $I$ , and let $M$ be the midpoint of $AC$ . Prove that $\angle MIC = \angle AIB$ .

Proposed by - Rijul Saini