Question

Use a vector argument to prove that the segment connecting the midpoints of two sides of a triangle is parallel to the third side of the triangle and half of its length.

Short answer

It is proved that the segment connecting the midpoints of two sides of a triangle is parallel to the third side of the triangle and half of its length.

Step-by-step solution

Given 

A triangle $ABC$, $D\mathrm{and}E$is the mid point of sides $AB\mathrm{and}AC$respectively.

Proof

From figure,

$AB=\overrightarrow{c},BC=\overrightarrow{a,}AC=\overrightarrow{b}$

In $∆ABC,$

role="math" localid="1663731958913" $$\begin{gathered} \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0 \\ \overrightarrow{a}=-(\overrightarrow{b}+\overrightarrow{c})...\left(1\right) \\ \mathrm{In}∆\mathrm{ADE}, \\ \frac{1}{2}\overrightarrow{c}+\overrightarrow{d}+\frac{1}{2}\overrightarrow{b}=0 \\ \overrightarrow{d}=-\frac{1}{2}\left(\overrightarrow{b}+\overrightarrow{c}\right) \\ \overrightarrow{d}=\frac{1}{2}\overrightarrow{a}(\mathrm{From}\mathrm{equation}1) \end{gathered}$$

Thus, the segment connecting the midpoints of two sides of a triangle is parallel to the third side of the triangle and half of its length.

Hence, proved.