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Let $H^{\prime }$ be the reflection of $H$ in $BC$. It is well-known (and easy to prove) that $H^{\prime }$ lies on the circumcircle of $△ABC$. Let $O$ be the circumcenter of $△ABC$. We have $$\begin{array}{rl}\angle O{H}^{\prime }A& =\angle HAO=\angle BAC-\angle BAH-\angle OAC\\ & =\angle BAC-2(90^{\circ }-\angle CBA)=\angle CBA-\angle ACB\\ & =\angle IDH=\angle H^{\prime }HD=\angle D{H}^{\prime }A,\end{array}$$ hence $O,D,H^{\prime }$ are collinear. Also note that $\angle HAO=\angle H^{\prime }HD$ implies that $AO\parallel HD$.

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