triangle mid-segment theorem

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:

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

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

 Title triangle mid-segment theorem Canonical name TriangleMidsegmentTheorem Date of creation 2013-03-22 17:46:35 Last modified on 2013-03-22 17:46:35 Owner pahio (2872) Last modified by pahio (2872) Numerical id 12 Author pahio (2872) Entry type Theorem Classification msc 51M04 Classification msc 51M25 Synonym mid-segment theorem Related topic MutualPositionsOfVectors Related topic ParallelogramTheorems Related topic MedianOfTrapezoid Related topic CommonPointOfTriangleMedians Related topic Grafix Related topic SimonStevin Related topic InterceptTheorem