Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

First solution. We show that the angles $APD$ and $AQM$ are either equal or one is the supplement of the other.
To begin, recall that the reflection of $H$ across $M$ is the antipode of $A$ in the circle $ABC$, to infer that the angle $APM$ is right, so $A$, $D$, $M$, $P$ are concyclic. Hence the angles $APD$ and $AMD$ are either equal or one is the supplement of the other.
The circle $ABC$ is the image of the nine-point circle under a homothety of scale factor $-2$ centred at $G$. This homothety sends $M$ to $A$ and $D$ to $Q$, so $AQ$ and $BC$ are parallel.
Hence $\angle AMD = \angle MAQ$, on the one hand; and $MA = MQ$, on the other, so $\angle MAQ = \angle AQM$. Consequently, $\angle AMD = \angle AQM$, and the conclusion follows by the preceding paragraph.

Second solution. As in the previous solution, $A$ and $Q$ are reflections of one another in the perpendicular bisectrix of the segment $BC$.
Let the line $MP$ cross the circle $ABC$ again at $P'$, and let the lines $QP'$ and $BC$ cross at $S$. Since $P'$ is the antipode of $A$ in the circle $ABC$, the angle $AQS = AQP'$ is right, so the lines $QS$ and $BC$ are perpendicular.
Let the line $MQ$ cross the circle $ABC$ again at $Q'$, let the lines $PQ'$ and $BC$ cross at $R$, and refer to the butterfly theorem to infer that $R$ and $S$ are reflections of one another across $M$.
The lines $AR$ and $QS$ are therefore reflections of one another in the perpendicular bisectrix of the segment $BC$.
Finally, since $QS$ is perpendicular to $BC$, so is $AR$, and, consequently, the points $R$ and $D$ coincide. The conclusion follows.