Korean Mathematical Olympiad

Let $Q^{\prime }$ be the intersection of the line passing $A$ and parallel to $BC$ and the line passing $P$ and perpendicular to $AD$. Let $U$ be the intersection of $DI$ and $A{Q}^{\prime }$, and $V$ be the intersection of $P{Q}^{\prime }$ and the circle $I$. ($V\ne P$) Since $\angle VPD=90^{\circ }$, four points $D,I,V,U$ lie on a line. Since $\angle BDI=90^{\circ }$ we have $\angle AUI=90^{\circ }$ and since $\angle AFI=\angle AEI=90^{\circ }$ we have that five points $A,F,I,E,U$ lie on a circle, say, $C_1$. Four points $A,P,V,U$ lie on a circle, say, $C_2$, because $\angle APV=\angle AUV=90^{\circ }$. For given three circles, three perpendicular bisectors of the line segments joining two centers of two circles meet at one point. Considering three circles $C_1,C_2$, and the incircle $I$, we have that three lines $AU,PV,EF$ meet at one point $Q^{\prime }$. Thus we have $Q^{\prime }=Q$. Since $△AXE\sim △CDE$, we have $\frac{AX}{AE}=\frac{CD}{CE}$. So $$AX=\frac{CD}{CE}\times AE=AE.$$ Since $△AYF\sim △BDF$, we have $\frac{AY}{AF}=\frac{BD}{BF}$. So $$AY=\frac{BD}{BF}\times AF=AF.$$ Since $AE=AF$ we have $AX=AY$, which completes the proof. $\square$