Solution. Without loss of generality, we may and will assume $AB < AC$. Begin by noticing that the triangles $ABC$ and $HPQ$ are similar. Indeed, $\angle HQP = 90^\circ - \angle QAB = 90^\circ - \angle OAB = \frac{1}{2}\angle AOB = \angle ACB$, and similarly $\angle HPQ = \angle ABC$.

Let $\Omega$ and $\omega$ be the circles $ABC$ and $HPQ$, respectively. The line $AH$ is tangent to $\omega$, since $\angle AHP = 90^\circ - \angle HAC = \angle ACB = \angle HQP$.

Let $T$ be the center of $\omega$ and let the lines $AT$ and $BC$ cross at $M$. We will show that $M$ is the midpoint of the segment $BC$, so $T$ lies on the median $AM$ of the triangle $ABC$.

Consider the similarity of the triangles $ABC$ and $HPQ$ along with the fact that $A$ is the point where the tangent of $\omega$ at $H$ crosses the line $PQ$. Letting the tangent of $\Omega$ at $A$ cross the line $BC$ at $S$, it follows that $S$ and $A$ correspond to one another under the similarity, so $\angle OSM = \angle OAT = \angle OAM$. The quadrangle $AOMS$ is therefore cyclic, and since the tangent $AS$ of $\Omega$ is perpendicular to the radius $OA$, it follows that $\angle OMS = 180^\circ - \angle OAS = 90^\circ$. Consequently, $M$ is the orthogonal projection of $O$ on the line $BC$, which is precisely the midpoint of the segment $BC$.