Let $ABC$ be a triangle, and let $P$ be the midpoint of $AC$. A circle intersects $AP, CP, BC, AB$ sequentially at their inner points $K, L, M, N$. Let $S$ be the midpoint of $KL$. Let also $2 \cdot | AN |\cdot |AB |\cdot |CL | = 2 \cdot | CM | \cdot| BC | \cdot| AK| = | AC | \cdot| AK |\cdot |CL |.$ Prove that if $P\ne S$, then the intersection of $KN$ and $ML$ lies on the perpendicular bisector of the $PS$. (Jan Mazák)