55rd Ukrainian National Mathematical Olympiad - Fourth Round

By Varignon's Theorem quadrangle $MPNQ$ is a parallelogram and its sides are parallel to the diagonals of trapezoid $ABCD$, so it is a rectangle (Fig. 38). Denote $O$ the intersection point of $AC$ and $BD$. Then from the properties of rectangular $△MNL$ we get that: $OL=OM=ON\Rightarrow OL=OP=OQ$, so $△PQL$ is rectangular. That means $PL\perp LQ$, and since $AB\parallel CD$ then $PL\perp AB$.