Question 118
Let $ABCD$ be a convex quadrilateral with $\angle ABC = \angle ADC = 90^\circ$, and diagonals intersecting at $P$. A line through $P$ parallel to $AD$ intersects $BC$ at $K$, and a line through $P$ parallel to $AB$ intersects $DC$ at $L$. Let $\omega$ be the circumcircle of triangle $KLC$. Let $AC$ intersect $\omega$ again at $E$. Let line $KL$ intersect $AB$ at $M$ and $AD$ at $N$. Let $\omega$ intersect $EM$ again at $Q$ and $EN$ again at $R$. Prove that $QR = KL$.
(Khulan Tumenbayar)
(Khulan Tumenbayar)