Balkan Mathematical Olympiad Shortlist

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 $E^{\prime }$ 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 $△ADF$ opposite $A$) $=\angle MDC$ (as $DM$ is the external angle bisector of $\angle ADF$), the triangles $△BMF$ and $△DMC$ are similar. Therefore, we have $FH:HB=MP:PC$ (as $MH$ and $DP$ are corresponding altitudes in similar triangles) $=AD:DC$ (as $AM\parallel DP$) $=FD:D{E}^{\prime }$ (as $AF\parallel C{E}^{\prime }$). It follows that $HD\parallel B{E}^{\prime }$ and $E\equiv E^{\prime }$, as needed.