median of trapezoid

The segment connecting the midpoints of the legs (http://planetmath.org/Trapezoid) of a trapezoid, i.e. the median of the trapezoid, is parallel to the bases and its length equals the arithmetic mean of the legs.

Proof.  Let $AB$ and $CD$ be the bases of a trapezoid $ABCD$ and $E$ the midpoint of the leg $AD$ and $F$ the midpoint of the leg $BC$.  Then the median $EF$ may be determined as vector as follows:

$\overrightarrow{EF}$$=\overrightarrow{ED}+\overrightarrow{DC}+\overrightarrow{CF}$

$={\displaystyle \frac{1}{2}}\overrightarrow{AD}+\overrightarrow{DC}+{\displaystyle \frac{1}{2}}\overrightarrow{CB}$

$={\displaystyle \frac{1}{2}}(\overrightarrow{AD}+\overrightarrow{DC}+\overrightarrow{CB})+{\displaystyle \frac{1}{2}}\overrightarrow{DC}$

$={\displaystyle \frac{1}{2}}\overrightarrow{AB}+{\displaystyle \frac{1}{2}}\overrightarrow{DC}$

$={\displaystyle \frac{1}{2}}(\overrightarrow{AB}+\overrightarrow{DC})$

The last expression tells that $\overrightarrow{EF}\parallel \overrightarrow{AB}+\overrightarrow{DC}\parallel \overrightarrow{AB}$  and $|\overrightarrow{EF}|={\displaystyle \frac{|\overrightarrow{AB}+\overrightarrow{DC}|}{2}}={\displaystyle \frac{|\overrightarrow{AB}|+|\overrightarrow{DC}|}{2}}$.  Q.E.D.