Prove using vectors, the mid-points of two opposite sides of a quadrilateral and the mid-points of the diagonals are the vertices of a parallelogram.
Let $ABCD$ be a quadrilateral.
$E$ and $F$ are the midpoints of sides $AB$ and $CD$ respectively.
$AC$ and $BD$ are the diagonals of the quadrilateral $ABCD$.
$G$ and $H$ are the mid-points of the diagonals $AC$ and $BD$ respectively.
Position vectors of points $A$, $B$, $C$ and $D$ are $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ and $\overrightarrow{d}$ respectively.
Position vectors of points $E$, $H$, $F$ and $G$ are $\dfrac{\overrightarrow{a} + \overrightarrow{b}}{2}$, $\dfrac{\overrightarrow{b} + \overrightarrow{d}}{2}$, $\dfrac{\overrightarrow{c} + \overrightarrow{d}}{2}$ and $\dfrac{\overrightarrow{a} + \overrightarrow{c}}{2}$ respectively.
To prove that $EHFG$ is a parallelogram, it is sufficient to show that $\overrightarrow{GE} = \overrightarrow{FH}$ and $\overrightarrow{EH} = \overrightarrow{GF}$.
Now,
$\begin{aligned}
\overrightarrow{GE} & = \text{position vector of E} - \text{position vector of G} \\\\
& = \left(\dfrac{\overrightarrow{a} + \overrightarrow{b}}{2}\right) - \left(\dfrac{\overrightarrow{a} + \overrightarrow{c}}{2}\right) \\\\
& = \dfrac{\overrightarrow{b} - \overrightarrow{c}}{2}
\end{aligned}$
and
$\begin{aligned}
\overrightarrow{FH} & = \text{position vector of H} - \text{position vector of F} \\\\
& = \left(\dfrac{\overrightarrow{b} + \overrightarrow{d}}{2}\right) - \left(\dfrac{\overrightarrow{c} + \overrightarrow{d}}{2}\right) \\\\
& = \dfrac{\overrightarrow{b} - \overrightarrow{c}}{2}
\end{aligned}$
$\therefore$ $\;$ $\overrightarrow{GE} = \overrightarrow{FH}$
$\implies$ $GE \parallel FH$ and $GE = FH$ $\;\;\; \cdots \; (1)$
Now,
$\begin{aligned}
\overrightarrow{EH} & = \text{position vector of H} - \text{position vector of E} \\\\
& = \left(\dfrac{\overrightarrow{b} + \overrightarrow{d}}{2}\right) - \left(\dfrac{\overrightarrow{a} + \overrightarrow{b}}{2}\right) \\\\
& = \dfrac{\overrightarrow{d} - \overrightarrow{a}}{2}
\end{aligned}$
and
$\begin{aligned}
\overrightarrow{GF} & = \text{position vector of F} - \text{position vector of G} \\\\
& = \left(\dfrac{\overrightarrow{c} + \overrightarrow{d}}{2}\right) - \left(\dfrac{\overrightarrow{a} + \overrightarrow{c}}{2}\right) \\\\
& = \dfrac{\overrightarrow{d} - \overrightarrow{a}}{2}
\end{aligned}$
$\therefore$ $\;$ $\overrightarrow{EH} = \overrightarrow{GF}$
$\implies$ $EH \parallel GF$ and $EH = GF$ $\;\;\; \cdots \; (2)$
$\therefore$ $\;$ From $(1)$ and $(2)$, $EHFG$ is a parallelogram.
Hence proved.