Mongolian Mathematical Olympiad

Let $A^{\prime },D^{\prime },C^{\prime },P^{\prime },Q^{\prime },X^{\prime },Y^{\prime }$ be images of points $A,D,C,P,Q,X,Y$ transformed by inversion $I_{\omega (B,1)}$ respectively.

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Then points $A^{\prime },P^{\prime },D^{\prime },Q^{\prime },C^{\prime }$ are colinear and $A^{\prime }{P}^{\prime }=P^{\prime }{D}^{\prime }$, $D^{\prime }{Q}^{\prime }=Q^{\prime }{C}^{\prime }$. This implies from $△B{A}^{\prime }{D}^{\prime }\sim △BAD$ and fact that $BP$ is simedian. Consequently $X^{\prime }=\omega (B{P}^{\prime }{C}^{\prime })\cap A^{\prime }B$, $Y^{\prime }=\omega (A^{\prime }{Q}^{\prime }B)\cap B{C}^{\prime }$. Now it is sufficient to prove that points $B,X^{\prime },Y^{\prime },D^{\prime }$ are cyclic. Furthermore

$$\angle X^{\prime }{A}^{\prime }{P}^{\prime }=\angle Q^{\prime }{Y}^{\prime }{C}^{\prime }\angle A^{\prime }{X}^{\prime }{P}^{\prime }=\angle Q^{\prime }{C}^{\prime }{Y}^{\prime }\Rightarrow \frac{A^{\prime }{P}^{\prime }}{P^{\prime }{X}^{\prime }}=\frac{Q^{\prime }{Y}^{\prime }}{Q^{\prime }{C}^{\prime }}\Rightarrow \frac{P^{\prime }{D}^{\prime }}{P^{\prime }{X}^{\prime }}=\frac{Q{Y}^{\prime }}{D^{\prime }{Q}^{\prime }}\text{and}\angle X^{\prime }{P}^{\prime }{D}^{\prime }=\angle D^{\prime }{Q}^{\prime }{Y}^{\prime }\Rightarrow △P^{\prime }{D}^{\prime }{X}^{\prime }\sim △Q^{\prime }{Y}^{\prime }{D}^{\prime }\Rightarrow \angle X^{\prime }{D}^{\prime }{P}^{\prime }+\angle Y^{\prime }{D}^{\prime }{Q}^{\prime }=\angle X^{\prime }{P}^{\prime }{A}^{\prime }=180^{\circ }-\angle X^{\prime }B{Y}^{\prime }\Rightarrow \angle X^{\prime }B{Y}^{\prime }+\angle X^{\prime }{D}^{\prime }{Y}^{\prime }=180^{\circ }.$$ This completes the proof.