Let $AA_1$ be the altitude of triangle $ABC$. Consider the case when the lines $BQ$ and $CP$ intersect at some point $T$. We need to prove that the point $T$ lies on the line $AA_1$. Suppose first that points $A$ and $P$ lie on the same side of line $BC$, and points $A$ and $Q$ lie on different sides of line $BC$ (see figure).

As is known, the points symmetric to the orthocenter of a triangle with respect to the midpoints of its sides lie on the circumcircle of this triangle: indeed, if in triangle $ABC$ the point $F$ is symmetric to the orthocenter $H$ with respect to the midpoint $N$ of side $AC$, then $\angle AFC = \angle AHC = 180^\circ - \angle ABC$. Further, since $FC \parallel AH$, we have $\angle FCB = 90^\circ$, and points $B$ and $F$ are diametrically opposite. Hence, $HQ \perp BT$. Similarly, $HP \perp CT$.

The quadrilaterals $HQT P$ and $BQ A_1 H$ are cyclic. Therefore, we have: $\angle QBC = \angle QH A_1$, $\angle QHT = \angle QPC$. Since $\angle QPC = \angle QBC$, it follows that $\angle QHT = \angle QH A_1$, which completes the proof.

If the chord $PQ$ and point $A$ lie on different sides of line $BC$, then $\angle QBC = \angle QH A_1$, $\angle QHT = \angle QPT$, $\angle QPT = \frac{\angle QP}{2} + \frac{\angle PC}{2} = \angle QBC$. Other cases of the arrangement of points $P$ and $Q$ on the circumcircle of triangle $ABC$ are considered similarly.

If the lines $BQ$ and $CP$ are parallel, it is not difficult to prove that the point $H$ lies on the segment $MN$, points $P$ and $F$ coincide, and therefore $AH \parallel PC$.

Let $\omega$ be the circumcircle of triangle $ABC$, and $\omega_B$ and $\omega_C$ be the circles constructed on diameters $HB$ and $HC$ respectively. Since $\angle HPC = 90^\circ$, we have $P \in \omega_C$, and the line $CP$ is the radical axis of circles $\omega$ and $\omega_C$. Similarly, $BQ$ is the radical axis of circles $\omega$ and $\omega_B$. Further, the line $AA_1$ is the radical axis of circles $\omega_B$ and $\omega_C$. Therefore, the statement of the problem follows from the theorem on the radical axes of three circles.

(This solution was proposed by olympiad participant Anna Mitrushchenkova.)

The case when points $P$ and $F$ coincide, i.e., $AH \parallel PC$, was considered in Solution 1. Let $Y$ be such a point that $H$ is the midpoint of segment $AY$. Then $BY \parallel PH$, and $CP \perp BY$. Similarly, $BQ \perp CY$. Moreover, $AH \perp BC$. The required result follows from the concurrency of the altitudes of triangle $BCY$.