PL-OM 2024 Problem 6

Let $ABCD$ be a parallelogram. Let $X \notin AC$ lie inside $ABCD$ so that $\angle AXB = \angle CXD = 90^ {\circ}$ and $\Omega$ denote the circumcircle of $AXC$ . Consider a diameter $EF$ of $\Omega$ and assume neither $E, \ X, \ B$ nor $F, \ X, \ D$ are collinear. Let $K \neq X$ be an intersection point of circumcircles of $BXE$ and $DXF$ and $L \neq X$ be such point on $\Omega$ so that $\angle KXL = 90^{\circ}$ . Prove that $AB = KL$ .