Balti Tee 2023 · Ülesanne 13

Let $H'$ be the reflection of $H$ in $BC$. It is well-known (and easy to prove) that $H'$ lies on the circumcircle of $\triangle ABC$. Let $O$ be the circumcenter of $\triangle ABC$. We have

$$\begin{align*} \angle OH'A &= \angle HAO = \angle BAC - \angle BAH - \angle OAC \\ &= \angle BAC - 2(90^\circ - \angle CBA) = \angle CBA - \angle ACB \\ &= \angle IDH = \angle H'HD = \angle DH'A, \end{align*}$$

hence $O, D, H'$ are collinear. Also note that $\angle HAO = \angle H'HD$ implies that $AO \parallel HD$.

Since $OM \parallel AH$, the above equality gives that $A, O, E$ are collinear. Let $D'$ be the reflection of $D$ in $I$. It is well-known (and easy to prove) that $D'$ lies on $AE$. Since $AH \parallel OM$ and $AD' \parallel HD$, quadrilateral $AHDD'$ is a parallelogram. Therefore $|AH| = |DD'| = 2 \cdot |ID|$.