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 $A'$ be the midpoint of $AC$ and $B'$ the midpoint of $BC$ . Using the side-vectors $\overrightarrow{AC}$ and $\overrightarrow{CB}$ as a basis of the plane, we calculate the mid-segment $A'B'$ as a vector: $$\overrightarrow{A'B'} \,=\, \overrightarrow{A'C}+\overrightarrow{CB'} \,=\, \frac{1}{2}\overrightarrow{AC}+\frac{1}{2}\overrightarrow{CB} \,=\, \frac{1}{2}(\overrightarrow{AC}+\overrightarrow{CB}) \,=\, \frac{1}{2}\overrightarrow{AB}$$ The last expression indicates that the segment $A'B'$ is such as asserted.

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