Problem - 3104
Let $ABCD$ be a trapezoid. Points $M$ and $N$ are the mid points of its diagonal $AC$ and $BD$, respectively. Show that $MN \parallel AB$ and $MN = \frac{1}{2}\mid AB - CD\mid$.
This problem can be solved in several different ways. A vector based solution is one of the simplest.
$$2\overrightarrow{MN} = (\overrightarrow{MA}+\overrightarrow{AB}+\overrightarrow{BN}) + (\overrightarrow{MC}+\overrightarrow{CD}+\overrightarrow{DN}) = \overrightarrow{AB} + \overrightarrow{CD}$$