MathNet 2017-part-04 — 176 problems (EN–VI) Question 21

Let $I$ be the incenter of triangle $ABC$. Let $D$ be a point on side $BC$, and $E$ be a point on ray $BC$ such that $C$ lies between $E$ and $D$ and $\frac{BD}{DC} = \frac{BE}{EC}$. Let $H$ be the foot of the perpendicular from $D$ to line $IE$. Prove that $\angle AHE = \angle IDE$.
(Proposed by B. Battsengel, G. Batzaya)