The 36th KOREAN MATHEMATICAL OLYMPIAD Final Round

Let $Z$ be the intersection of the circumcircles of $ABC$ and $ADE$. Then we have $\angle ZDA=\angle ZEA=\angle ZYA$ and $\angle ZBA=\angle ZCA=\angle ZXA$. Hence the triangles $ZBD$, $ZCE$ and $ZXY$ are all similar.

Now let $L,M,N$ be the midpoints of $BD$, $CE$, $XY$, respectively. Then by the similarity of the triangles $ZBD$, $ZCE$, $ZXY$ we have $\angle ZLA=\angle ZMA=\angle ZNA$. Hence $Z,L,M,N,A$ are concyclic. Since $\angle ALP=\angle AMP=90^{\circ }$, $P$ lies on this circle. Hence $\angle ANP=90^{\circ }$. As $N$ is the midpoint of $XY$, we have $PX=PY$. $\square$