23rd Korean Mathematical Olympiad Final Round

It suffices to show that $X$, $B$, $Y$, $Q$ are cyclic under the assumption that $Q$ is the intersection of the line $BD$ and ${\Gamma }_2$. There are six cases depending on the ordering of $P$, $D$, $B$, $Q$. Although we should consider all those six cases, we here give the proof only for the case where the order is $P$, $D$, $B$, $Q$.

Let $\angle ABP=\alpha$, $\angle DPX=\beta$. Since $AB//CD$, we have $\angle ABP=\angle BDC=\alpha$. For the circle ${\Gamma }_1$, the circumference angle for $PB$ equals $\frac{\pi }{2}-\alpha$; hence we have $(\frac{\pi }{2}-\alpha )+\angle PXB=\pi$. That is, $\angle PXB=\frac{\pi }{2}+\alpha$.

On the other hand, we have $$\angle DYQ=\frac{1}{2}\angle DCQ=\frac{\pi }{2}-\alpha .$$ For the circle $\Gamma$, $\angle DPX=\angle DYX=\beta$. Now it suffices to show that $\angle DBX=\angle XYQ$.

For the triangle $DBX$, we have $$\angle DBX=\pi -\beta -(\frac{\pi }{2}+\alpha )=\frac{\pi }{2}-\alpha -\beta .$$ Meanwhile, since $\angle DYQ=\frac{\pi }{2}-\alpha$ and $\angle DYX=\beta$, we have $$ \angle XYQ = \angle DYQ - \angle DYX = \frac{\pi}{2} - \alpha - \beta. \quad \square