Point $H$ is the incenter of triangle $A_1B_1C_1$, while $A$ is the excenter of the same triangle, corresponding to the side $B_1C_1$. Let $D$, $E$, and $F$ be the orthogonal projections of points $P$, $Q$, and $H$, respectively, onto the line $B_1C_1$. Let $S$ be the intersection of lines $PH$ and $A_1C_1$, and let $T$ be the intersection of lines $QH$ and $A_1B_1$. Then it follows that $C_1D = C_1S = C_1F$ (1), and $B_1E = B_1T = B_1F$ (2). If $M$ is the projection of $A$ onto $B_1C_1$, it follows that $FC_1 = MB_1$ (3). From (1), (2), and (3) we obtain that $M$ is the midpoint of the line segment $DE$. Thus, $AM$ is the perpendicular bisector of the line segment $DE$ and, in the right trapezoid $DEQP$, it cuts side $PQ$ at its midpoint.

Let $A'$, $B'$ and $C'$ be the reflections of point $H$ with respect to the sides $BC$, $CA$, and $AB$, respectively. Triangle $A'B'C'$ is the image of triangle $A_1B_1C_1$ through a homothety centered at $H$, and has the same circumcircle as $ABC$. Moreover, $A'H$, $B'H$, and $C'H$ are the angle bisectors of triangle $A'B'C'$. Triangle $QHB'$ is isosceles with $QH = QB'$, therefore $\angle HB'Q = \angle QHB'$. It follows that $\angle C'B'Q = \angle HB'Q + \angle C'B'H = \angle QHB_1 + \angle C_1B_1H = \angle QHB_1 + \angle HB_1A_1 = 90^\circ$. Similarly, $\angle B'C'P = 90^\circ$. As $AB' = AC'$, the perpendicular from $A$ to $B_1C_1$ is the perpendicular bisector of the line segment $B'C'$, hence parallel with $QB'$ and $PC'$, so it will cut the side $PQ$ of the trapezoid $B'C'PQ$ at its midpoint.