Define $M_C$ and $M_B$ as midpoints of $AB$ and $AC$ respectively. Then $H$ lies on the line $M_B M_C$ due to its initial position. Note that projections of $O$ on the lines $AB$, $AC$ and $PQ$ are points $M_C$, $M_B$ and $H$ which lie on the Simson line (Fig 33). From this can be concluded that $O$ lies on the circumscribed circle of the $\triangle PAQ$.

Intersect a ray $HO$ with the line $BC$ in a point $S$. Observe that $\triangle OM_B M_C \sim \triangle HBC$ as they have the same angles. Also, $\angle OHM_B = \angle HSB$, so $S$ and $H$ are respective vertices of the similar triangles. Since $BC = 2M_B M_C$, we have that $SC = 2M_C H = BD$. Hence, points $S$ and $D$ are symmetric over $BC$.

Name midpoints of $BP$ and $CQ$ as $P_1$ and $Q_1$ respectively. Define $C_1$ as a symmetric point of $C$ over $D$. Therefore, can deduce that
$$\angle OQC = \angle OHM_B = \angle OSC_1 = \angle HSC_1 \text{ and } \angle OCA = 90^\circ - \angle B = \angle HCB = \angle HC_1S,$$
which means that $\triangle HSC_1 \sim \triangle OQC$. Let $X$ be a midpoint of the $SC_1$. From $C_1D = DC = BS$, we have that the midpoints of the $SC_1$ and $BD$ coincide, so $X$ is also the midpoint of the $BD$. From the previously described similarity $\triangle HSC_1 \sim \triangle OQC$ we get that $\angle OQ_1A = \angle HXS$ (an angle between a median and a side) and $\angle OQ_1A = \angle HXS = \angle B$ (midline). Similarly,

$\angle OP_1A = \angle C$. From the angle sum $\angle A + \angle B + \angle C = 180^\circ$ we conclude that $P_1, O$ and $Q_1$