Let $A^{\prime}$ be the reflection of $H$ in the midpoint of $B C$. Since $A$ is the reflection of $M$ in the midpoint of $B C$, we note that $A^{\prime} A$ is parallel to $H M$ (since these line segments are point reflections of each other). Furthermore, $A^{\prime}$ is one of the so-called orthocentric points: $A^{\prime}$ lies on $\Gamma$ and is the antipode of $A$. Indeed, the first follows from the cyclic quadrilateral theorem, for which we calculate that

$$\angle B A^{\prime} C=\angle C H B=180^{\circ}-\angle B C H-\angle H B C=\angle A B C+\angle B C A=\angle 180^{\circ}-\angle C A B .$$

Similarly, the reflection $A^{\prime \prime}$ of $H$ in the line $B C$ lies on $\Gamma$. Since $A H$ and $H A^{\prime \prime}$ are both perpendicular to $B C$, $A, H$ and $A^{\prime \prime}$ are collinear and since $A^{\prime} A^{\prime \prime}$ is parallel to $B C$ we find that $A^{\prime} A^{\prime \prime} \perp A A^{\prime \prime}$. By Thales, this means that $A^{\prime} A$ is a diameter of $\Gamma$ and thus perpendicular to the tangent at $A$. Therefore, $H M$ is also perpendicular to this.