Question 75
Let $ABCD$ be a cyclic quadruple. Let us denote the midpoints of $AB$, $BC$, $CD$ and $DA$ by $M$, $L$, $N$ and $K$ respectively. It is known, that $\angle BMN = \angle MNC$. Prove that:
a) $\angle DKL = \angle CLK$;
b) $ABCD$ has a pair of parallel sides.
a) $\angle DKL = \angle CLK$;
b) $ABCD$ has a pair of parallel sides.