Prove that the sum of the vectors directed from the vertices to the mid-points of opposite sides of a triangle is zero.
Let $O$ be the origin.
Let the position vectors of points $A$, $B$ and $C$ be
$\overrightarrow{OA} = \overrightarrow{a}$, $\overrightarrow{OB} = \overrightarrow{b}$ and $\overrightarrow{OC} = \overrightarrow{c}$ respectively.
Let $D$, $E$ and $F$ be the midpoints of $AB$, $AC$ and $BC$ respectively.
Then,
position vector of point $D = \overrightarrow{OD} = \dfrac{\overrightarrow{a} + \overrightarrow{b}}{2}$
position vector of point $E = \overrightarrow{OE} = \dfrac{\overrightarrow{a} + \overrightarrow{c}}{2}$
position vector of point $F = \overrightarrow{OF} = \dfrac{\overrightarrow{b} + \overrightarrow{c}}{2}$
To prove that: $\;$ $\overrightarrow{CD} + \overrightarrow{BE} + \overrightarrow{AF} = \overrightarrow{O}$
Now,
$\overrightarrow{CD} = \overrightarrow{OD} - \overrightarrow{OC} = \dfrac{\overrightarrow{a} + \overrightarrow{b}}{2} - \overrightarrow{c}$
$\overrightarrow{BE} = \overrightarrow{OE} - \overrightarrow{OB} = \dfrac{\overrightarrow{a} + \overrightarrow{c}}{2} - \overrightarrow{b}$
$\overrightarrow{AF} = \overrightarrow{OF} - \overrightarrow{OA} = \dfrac{\overrightarrow{b} + \overrightarrow{c}}{2} - \overrightarrow{a}$
$\begin{aligned}
\therefore \; \overrightarrow{CD} + \overrightarrow{BE} + \overrightarrow{AF} & = \dfrac{\overrightarrow{a} + \overrightarrow{b}}{2} - \overrightarrow{c} + \dfrac{\overrightarrow{a} + \overrightarrow{c}}{2} - \overrightarrow{b} + \dfrac{\overrightarrow{b} + \overrightarrow{c}}{2} - \overrightarrow{a} \\\\
& = \dfrac{2 \overrightarrow{a} + 2 \overrightarrow{b} + 2 \overrightarrow{c}}{2} - \left(\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}\right) \\\\
& = \left(\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}\right) - \left(\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}\right) \\\\
& = 0 \\\\
& = \overrightarrow{O}
\end{aligned}$
Hence proved.