Suppose, without loss of generality, that $AB < AC$. We have $\angle PQH = 90^\circ - \angle QAB = 90^\circ - \angle OAB = \frac{1}{2} \angle AOB = \angle ACB$, and similarly $\angle QPH = \angle ABC$. Thus triangles $ABC$ and $HPQ$ are similar. Let $\Omega$ and $\omega$ be the circumcircles of $ABC$ and $HPQ$, respectively. Since $\angle AHP = 90^\circ - \angle HAC = \angle ACB = \angle HQP$, line $AH$ is tangent to $\omega$. ! Let $T$ be the center of $\omega$ and let lines $AT$ and $BC$ meet at $M$. We will take advantage of the similarity between $ABC$ and $HPQ$ and the fact that $AH$ is tangent to $\omega$ at $H$, with $A$ on line $PQ$. Consider the corresponding tangent $AS$ to $\Omega$, with $S \in BC$. Then $S$ and $A$ correspond to each other in $\triangle ABC \sim \triangle HPQ$, and therefore $\angle OSM = \angle OAT = \angle OAM$. Hence quadrilateral $SAOM$ is cyclic, and since the tangent line $AS$ is perpendicular to $AO$, $\angle OMS = 180^\circ - \angle OAS = 90^\circ$. This means that $M$ is the orthogonal projection of $O$ onto $BC$, which is its midpoint. So $T$ lies on median $AM$ of triangle $ABC$.