Question 131
The circle $\omega$ is circumscribed about an acute-angled triangle $ABC$. Points $D$ and $E$ are chosen on the sides $AB$ and $BC$, respectively, so that $AC \parallel DE$. Points $P$ and $Q$ are chosen on the smaller arc $AC$ of $\omega$ so that $DP \parallel EQ$. Rays $QA$ and $PC$ meet $DE$ at $X$ and $Y$, respectively. Prove that $\angle XBY + \angle PBQ = 180^\circ$. (A. Kuznetsov)