Question 81
Point $X$ is chosen inside of a non-trapezoid quadrilateral $ABCD$ such that $\angle AXD + \angle BXC = 180^\circ$. Suppose the angle bisector of $\angle ABX$ meets the $D$-altitude of triangle $ADX$ at $K$, and the angle bisector of $\angle DCX$ meets the $A$-altitude of triangle $ADX$ at $L$. Let $BK \perp CX$ and $CL \perp BX$. If the circumcenter of $ADX$ lies on the line $KL$, prove that $KL \perp AD$.