, 2020

Let $M$ be the circumcentre of $\triangle ABC$ and let $\alpha = \angle BAC$. We first show that $\triangle BYA \sim \triangle BMP$ and then that $\triangle YBM \sim \triangle ABP$.

By the inscribed angle theorem we have $\angle BMC = 2\angle BAC = 2\alpha$. Quadrilateral $PBMC$ is a kite with axis of symmetry $PM$ (by the equality of radii $|MB| = |MC|$ and equality of tangent segments $|PB| = |PC|$), so $MP$ bisects angle $\angle BMC$. Therefore $\angle BMP = \frac{1}{2}\angle BMC = \alpha$.

Moreover, we have $\angle PBM = 90^\circ$ (tangent to a circle is perpendicular to its radius), so by the sum of angles of a triangle we have $\angle MPB = 90^\circ - \alpha$.

On the other hand, we are given that $\angle ABY = 90^\circ$ and we also have $\angle YAB = \angle YAC - \angle BAC = 90^\circ - \alpha$. Therefore $\angle ABY = \angle PBM$ and $\angle YAB = \angle MPB$, from which follows that $\triangle BYA \sim \triangle BMP$.

From this similarity it follows that $\frac{|YB|}{|AB|} = \frac{|MB|}{|PB|}$. Combining this with the equality of angles
$$\angle YBM = \angle YBA + \angle ABM = 90^\circ + \angle ABM = \angle ABM + \angle MBP = \angle ABP,$$
we see that $\triangle YBM \sim \triangle ABP$.

Let $T$ now be the intersection of $AP$ and $YM$, then we have
$$\angle BYT = \angle BYM = \angle BAP = \angle BAT,$$
from which it follows that $BYAT$ is a cyclic quadrilateral. Therefore $\angle ATY = \angle ABY = 90^\circ$, so $AT \perp YM$. Analogously, $AP \perp XM$. But from this it now follows that $YM$ and $XM$ coincide and we get that $AP \perp XY$. $\square$