IR-TST 2021 Problem 11

Point $X$ is chosen inside the non trapezoid quadrilateral $ABCD$ such that $\angle AXD +\angle BXC=180$ . Suppose the angle bisector of $\angle ABX$ meets the $D$ -altitude of triangle $ADX$ in $K$ , and the angle bisector of $\angle DCX$ meets the $A$ -altitude of triangle $ADX$ in $L$ .We know $BK \perp CX$ and $CL \perp BX$ . If the circumcenter of $ADX$ is on the line $KL$ prove that $KL \perp AD$ .

Proposed by Alireza Dadgarnia