Question 67
Let $ABCDEF$ be a cyclic hexagon, such that $AD$ is the diameter of its circumcircle and the lines $BF$ and $CE$ are parallel. Let $K$ be the intersection of the segments $AC$ and $BD$, let $L$ be the intersection of the segments $AE$ and $DF$ and let $M$ be the midpoint of the segment $KL$. Prove that $|MC| = |ME|$.