62nd Ukrainian National Mathematical Olympiad

From the fact that $XY$ is tangent to the circumcircle of $△BXC$, we have $\angle BXY=\angle BCX$. Combining this with the fact that $\angle YBX=\angle XBC$, we have the similarity $△BXY\sim △BCX$. Therefore, we have $\frac{BX}{BC}=\frac{BY}{BX}$, so $BY=\frac{B{X}^2}{BC}$. Notice that $\angle IBX=90^{\circ }$, so (Fig. 9)



$$BY=\frac{B{X}^2}{BC}=\frac{I{X}^2-B{I}^2}{BC}=\frac{C{I}^2-B{I}^2}{BC}.$$

Let $K$ be the point of tangency of the incircle of $△ABC$ with side $BC$. Then $$BY=\frac{C{I}^2-B{I}^2}{BC}=\frac{(I{K}^2+C{K}^2)-(I{K}^2+B{K}^2)}{BC}=\frac{C{K}^2-B{K}^2}{BC}=CK-BK.$$

$AY=AB+BY=AB+CK-BK=c+(p-c)-(p-b)=b=AC$,

as desired.