IMO Team Selection Test 2

Let $A^{\prime }$ be the reflection of $H$ across the midpoint of $BC$ and $A^{\prime \prime }$ 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 $\angle B{A}^{\prime }C$ and $\angle B{A}^{\prime \prime }C$. Therefore, both $A^{\prime }$ and $A^{\prime \prime }$ lie on the circle. Moreover, $A^{\prime }{A}^{\prime \prime }$ is parallel to $BC$, which in turn is perpendicular to $A{A}^{\prime \prime }$. Hence, $\angle A^{\prime }{A}^{\prime \prime }A=90^{\circ }$ and $A^{\prime }A$ is a diameter of the circle. Therefore $A^{\prime }A$ is perpendicular to the tangent to $\Gamma$ at $A$. Since $A$ is the reflection of $M$ across the midpoint of $BC$, we note that $A^{\prime }A$ 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 $A$. $\square$