My solution in the exam. The condition gives $\angle AYP = \angle BXQ$, so $ABXY$ is cyclic, and it follows that $CD$ is parallel to $XY$ since $ABCD$ is also cyclic after easy angle chasing.

This is an easy consequence of Reim Theorem after proving $ABXY$ is cyclic.

It is easy to see that $ABYX$ is cyclic. So $\angle AYX = \angle BAX + \angle AXB = 180 - \angle ABX = \angle ADC$. So $XY || CD$...

I know that it might be easy to see that $ABYX$ is cyclic, however I believe in the competition you would have to show your reasoning. I am going to show my reasoning, however should you have a more elegant approach, feel free to improve upon my solution. Cmiiw Let $\measuredangle{XYZ}$ denote the directed angle of $\angle{XYZ}$. (Written in a counter-clockwise order.) Let M be the midpoint of arc CD. So AM and BM are bisectors of angles facing the same arc CD. Let $AX \cap BM = K$, $BY \cap AM = J$ and $AX \cap BY = L$. Clearly $JMKL$ is cyclic, since we have two right angles on opposite sides. This results in $\measuredangle{KLB} = \measuredangle{JLA} = \measuredangle{JMK}$. Since $\measuredangle{LJA} = \measuredangle{BKL}$, from the two equal angles in triangles $ALJ$ and $KLB$, $\measuredangle{LAJ} = \measuredangle{KBL}$ so adding with $\measuredangle{JAY} = \measuredangle{XBK}$ on both sides respectively we obtain $\measuredangle{XBY} = \measuredangle{XAY}$ so $ABXY$ is cyclic. Hence $\measuredangle{YAB} = \measuredangle{BXY} = \measuredangle{BCD}$, hence since $B, X, C$ are defined to be collinear, $XY \parallel CD$, as desired.