MathNet 2024-part-03 — 168 problems (EN–VI) Question 114

The diagonals $AC$ and $BD$ of the cyclic quadrilateral $ABCD$ (a quadrilateral inscribed in a circle) are perpendicular and intersect at the point $P$. Let $X$ be a point on the side $BC$. The circumcircle of triangle $APD$ and the line $XP$ intersect at $Y$ distinct from $P$. Prove that $\angle AXD + \angle BYC = 90^\circ$.

(Khulan Tumenbayar)