Question 136
Given an acute $\triangle ABC$ with $AB \gt AC$, let $M$ be the midpoint of the side $BC$. Let $I$ be the incentre of $\triangle ABC$, and let $D$ be the foot of the perpendicular from $I$ to $BC$. The line passing through $A$ and perpendicular to $BC$ meets the circumcircle of $\triangle ABC$ again at $E \neq A$. Suppose $\angle MID = \angle ACB - \angle ABC$. Prove that $DE = DI$.