Let $D'$ be the orthogonal projection of $H$ onto $AM$, $E'$ the intersection point of the lines $BD'$ and $AH$, and let $F'$ be the intersection point of the lines $D'H$ and $AB$. We prove that $D' = D$, $E' = E$, $F' = F$, hence $HD \perp AM$.
Let $H_0$ and $D'_0$ be the reflections across the point $M$, of the points $H$ and $D'$, respectively. It is well known that $H_0$ is the antipode of $A$ on the circumcircle of triangle $ABC$ and, as $\angle H_0D'_0A = \angle HD'M = 90^\circ$, point $D'_0$ also lies on the circumcircle of $ABC$. Thus, points $B, C, H, D'$ all lie on the reflection of this circle across $M$, i.e. the quadrilateral $BHD'C$ is cyclic. It follows that $\angle HD'E' = \angle HD'B = \angle HCB = 90^\circ - \angle B = \angle F'AH$, which means that the quadrilateral $AF'E'D'$ is cyclic. We obtain that $\angle F'E'A = \angle F'D'A = 90^\circ$, which means that lines $E'F'$ and $BC$ are parallel. If $D$ is between $A$ and $D'$, then $E$ is between $A$ and $E'$, and $F$ is between $B$ and $F'$ (where $F$ is the intersection point of the lines $DH$ and $AB$), hence $EF$ can not be parallel to $BC$. Similarly in the case when $D$ is between $D'$ and $M$. It follows that it is necessary to have $D = D'$, and then $E = E'$, $F = F'$.
As $ABD'_0C$ is cyclic, we have $\angle ABD'_0 = 180^\circ - \angle ACD'_0$, and therefore $\sin(\angle ABD'_0) = \sin(\angle ACD'_0)$. But since $M$ is the midpoint of $BC$, triangles $ABD'_0$ and $ACD'_0$ have the same area surface, i.e. $AB \cdot BD'_0 \sin(\angle ABD'_0) = AC \cdot CD'_0 \sin(\angle ABD'_0)$, hence $AB \cdot BD'_0 = AC \cdot CD'_0$, or $AB \cdot CD = AC \cdot BD$, which means $\frac{AB}{AC} = \frac{DB}{DC}$. The converse of the Angle Bisector Theorem proves the conclusion.