51st Ukrainian National Mathematical Olympiad, 3rd Round

Since $ABCD$ is cyclic, then $\angle ABD=\angle ACD$. Moreover, $\angle ABD=\angle ABY=\angle ABY$. Hence, $CXYD$ is cyclic because $\angle XCD=\angle XYB$ (fig. 12).



This implies that $\angle XCY=\angle XDY$, from what it follows that $\sin \angle ACY=\sin \angle BDX$. Moreover, $BX=AY$, and applying sine Law for triangles $BXD$ and $CAY$ we get: $$R_{BDX}=\frac{BX}{2\sin \angle BDX}=\frac{AY}{2\sin \angle ACY}=R_{CYA}$$