Question 137
Let $ABCD$ be a cyclic quadrilateral with obtuse angle at $C$. Let $H$ be such that $BCDH$ is a parallelogram. Let line $AH$ meet $CD$ at $M$ and $BC$ at $N$. A line through $H$ parallel to $AC$ meets $CD$ at $K$ and $BC$ at $L$. If the points $K, L, M, N$ lie on a common circle, prove that $\angle ABC = 90^\circ$.
(Khulan Tumenbayar)
(Khulan Tumenbayar)