$ \angle IDA=\angle IEA=\angle IAB=\angle IAC $.Also that $ \angle IED=\angle IAD $ $\Rightarrow \angle DAE=\angle DIE=2\angle EIC=2\angle ICD-2\angle IED $.Then $ IC $ is the bisector of triangle $ IDE $ $\Rightarrow \angle DIC=\angle CIE=\angle MED=\angle MDE $

My solution: Let $ P=AC \cap IE $ . From $ \angle AEI=\angle BAI=\angle IAP \Longrightarrow \triangle IAP \sim \triangle IEA $ , so $ \angle IPC=180^{\circ}-\angle API=180^{\circ}-\angle EAI=\angle IDC $ , hence $ P $ and $ D $ are symmetry WRT $ CI \Longrightarrow CI $ is the bisector of $ \angle DIE $ , so we get $ M $ is the midpoint of arc $ DE $ in circle $ \omega \Longrightarrow MD=ME $ . Q.E.D