Consider quadrilateral $ABCD$ with $|DC| > |AD|$. Let $P$ be a point on $DC$ such that $PC = AD$ and let $Q$ be the midpoint of $DP$. Let $L_1$ be the line perpendicular on $DC$ passing through $Q$ and let $L_2$ be the bisector of the angle $ \angle ABC$. Let us call $X = L_1 \cap L_2$. Show that if quadrilateral is cyclic then $X$ lies on the circumcircle of $ABCD.$