First, note that $\angle ATH = \angle ADH = \angle AQH = 90 \implies ATHDQ$ is cyclic. Call the circumcircle $\omega$. Let the bisector of $\angle BHC$ meet $BC$ and the external bisector of $\angle A$ at $Y$ and $K$, respectively. It's well-known that $\angle KAQ = 90$, since the internal and external bisectors of an angle are perpendicular. Also, $\angle QHY = \angle QHC + \angle CHY = \angle TAQ + \frac{1}{2} \angle DHT = \frac{1}{2} \angle A + \frac{1}{2} (180 - \angle A) = 90$. Thus, $\angle QHK = 90$, so $QAKH$ is a rectangle and $K$ lies on $\omega$. Since $AQ$ bisects $\angle TAD$, $Q$ is the midpoint of arc $\widehat{TD}$ of $\omega$, so $QM \perp TD$. Since $QK$ is a diameter of $\omega$, by symmetry, $QM$ passes through $K$, which solves the problem.

Note first that $Q\in(ADTH)$ and from $\angle QDT=\angle QAT=\angle QAD=\angle QAD=\angle QTD$. we get $QD=QT$. Now that if we let $F=MQ\cap (ADT)$ it is clear that $F$ is the antipode of $Q$ in $(ADT)$ and the midpoint of arc $DT$. From the first one we get that $\angle FAQ=90$ making $AF$ the external angle bisector of $\angle A$ and from the second one we get that $FT$ is the angle bisector of $\angle DHT$ and therefore $\angle BHC$ as well.