triangle mid-segment theorem

Theorem.  The segment connecting the midpoints of any two sides of a triangle is parallel to the third side and is half as long.

Proof.  In the triangle $ABC$, let $A^{\prime }$ be the midpoint of $AC$ and $B^{\prime }$ the midpoint of $BC$.  Using the side-vectors $\overrightarrow{AC}$ and $\overrightarrow{CB}$ as a basis (http://planetmath.org/Basis) of the plane, we calculate the mid-segment $A^{\prime }{B}^{\prime }$ as a vector:

$$\overrightarrow{A^{\prime }{B}^{\prime }}=\overrightarrow{A^{\prime }C}+\overrightarrow{C{B}^{\prime }}=\frac{1}{2}\overrightarrow{AC}+\frac{1}{2}\overrightarrow{CB}=\frac{1}{2}(\overrightarrow{AC}+\overrightarrow{CB})=\frac{1}{2}\overrightarrow{AB}$$

The last expression indicates that the segment $A^{\prime }{B}^{\prime }$ is such as asserted.

Corollary (Varignon’s theorem).  If one connects the midpoints of the of a quadrilateral, one obtains a parallelogram.