Let $\omega$ be the circle of center $M$ and radius $MH$ and let the tangent to $\omega$ through $D$ different from $DC$ meets the line $AB$ at $F$. It suffices to show that $E$ lies on $DF$.
Let $E'$ be the intersection of $DF$ and the line through $C$ parallel to $AB$ and let $P$ be the projection of $D$ onto $MC$.
Let $T$ be the tangency point of $AC$ and $\omega$. Consider the case when $D$ lies in the segment $AT$; the case when $D$ lies in the segment $TC$ is treated analogously.

Since $\angle FBM = \angle MCD$ and $\angle FMB = 90^\circ - \angle AMF = 90^\circ - \frac{1}{2}\angle ADF$ (as $M$ is the excenter of $\triangle ADF$ opposite $A$) $= \angle MDC$ (as $DM$ is the external angle bisector of $\angle ADF$), the triangles $\triangle BMF$ and $\triangle DMC$ are similar.
Therefore, we have $FH : HB = MP : PC$ (as $MH$ and $DP$ are corresponding altitudes in similar triangles) $= AD : DC$ (as $AM \parallel DP$) $= FD : DE'$ (as $AF \parallel CE'$). It follows that $HD \parallel BE'$ and $E \equiv E'$, as needed.