Let $A'$ be the reflection of $H$ across the midpoint of $BC$ and $A''$ the reflection of $H$ across $BC$. Then by angle chasing, we find that $\angle BHC = \angle ABC + \angle BCA = 180^\circ - \angle CAB$. This angle is also equal to $\angle BA'C$ and $\angle BA''C$. Therefore, both $A'$ and $A''$ lie on the circle. Moreover, $A'A''$ is parallel to $BC$, which in turn is perpendicular to $AA''$. Hence, $\angle A'A''A = 90^\circ$ and $A'A$ is a diameter of the circle. Therefore $A'A$ is perpendicular to the tangent to $\Gamma$ at $A$. Since $A$ is the reflection of $M$ across the midpoint of $BC$, we note that $A'A$ is parallel to $HM$ (because these line segments are transformed to each other under the reflection). Therefore $HM$ is also perpendicular to the tangent to $\Gamma$ at $A$. $\square$