Let $BY$ and $CX$ cross at $S$ and let circles $APX$ and $AQY$ cross again at $T$. We first prove that $A, S, T$ are collinear. Invert from $A$ with power $AP \cdot AB = AQ \cdot AC$. As $\angle ABY = \angle AXP$, the circle $APX$ is mapped to line $BY$. Similarly, the circle $AQY$ is mapped to line $CX$, so the inversion switches $S$ and $T$. Hence $A, S, T$ are collinear, as stated.

Next, we show that the lines $AT$, $PX$ and $QY$ are projectively concurrent.
Let $PX$ and $QY$ cross projectively at $R$. Apply Pappus' theorem to the hexagram $BPXCQY$ to deduce that $A = BP \cap CQ$, $R = PX \cap QY$ and $S = XC \cap YB$ are collinear. The desired concurrence now follows by the preceding paragraph.

We now prove that $PQYX$ is cyclic. If $AT$, $PX$ and $QY$ are parallel, then $PQYX$ is an isosceles trapezoid, so it is cyclic. Otherwise, read the power of $R$ from circles $APX$ and $AQY$ to write $RX \cdot RP = RT \cdot RA = RY \cdot RQ$, so $PQYX$ is cyclic.

Finally, read angles from circle $PQYX$ and the given circle through $B$ and $C$ to write $\angle PYX = \angle PQX \equiv \angle PQB = \angle PCB$. Consequently, $XY$ and $BC$ are parallel, as required.