# median of trapezoid

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:

 $\displaystyle\overrightarrow{EF}$ $\displaystyle=\overrightarrow{ED}+\overrightarrow{DC}+\overrightarrow{CF}$ $\displaystyle=\frac{1}{2}\overrightarrow{AD}+\overrightarrow{DC}+\frac{1}{2}% \overrightarrow{CB}$ $\displaystyle=\frac{1}{2}(\overrightarrow{AD}+\overrightarrow{DC}+% \overrightarrow{CB})+\frac{1}{2}\overrightarrow{DC}$ $\displaystyle=\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{DC}$ $\displaystyle=\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{DC})$

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

Title median of trapezoid MedianOfTrapezoid 2013-03-22 17:46:44 2013-03-22 17:46:44 pahio (2872) pahio (2872) 6 pahio (2872) Theorem msc 51M04 msc 51M25 MutualPositionsOfVectors MidSegmentTheorem TriangleMidSegmentTheorem HarmonicMeanInTrapezoid