Prove that the area of a parallelogram formed by taking as its sides the diagonals of a parallelogram ABCD is twice the area of parallelogram ABCD.
Consider a parallelogram $ABCD$.
Let $\overrightarrow{AB} = \overrightarrow{a}$ $\;$ and $\;$ $\overrightarrow{BC} = \overrightarrow{b}$
Then, $\overrightarrow{DC} = \overrightarrow{a}$ $\;$ and $\;$ $\overrightarrow{AD} = \overrightarrow{b}$
Diagonal $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{a} + \overrightarrow{b}$
Diagonal $\overrightarrow{BD} = \overrightarrow{BA} + \overrightarrow{AD} = - \overrightarrow{a} + \overrightarrow{b}$
Area of parallelogram $ABCD = A_1 = \left|\overrightarrow{a} \times \overrightarrow{b}\right|$ $\;\;\; \cdots \; (1)$
Consider parallelogram $PQRS$ where $\overrightarrow{PQ} = \overrightarrow{a} + \overrightarrow{b}$ $\;$ and $\;$ $\overrightarrow{QR} = \overrightarrow{b} - \overrightarrow{a}$
$\begin{aligned}
\text{Area of parallelogram PQRS } = A_2 & = \left|\overrightarrow{PQ} \times \overrightarrow{QR}\right| \\\\
& = \left|\left(\overrightarrow{a} + \overrightarrow{b}\right) \times \left(\overrightarrow{b} - \overrightarrow{a}\right)\right| \\\\
& = \left|\overrightarrow{a} \times \overrightarrow{b} - \overrightarrow{a} \times \overrightarrow{a} + \overrightarrow{b} \times \overrightarrow{b} - \overrightarrow{b} \times \overrightarrow{a}\right| \\\\
& = \left|\overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{a} \times \overrightarrow{b}\right| \\\\
& = \left|2 \left(\overrightarrow{a} \times \overrightarrow{b}\right)\right| \\\\
& = 2 A_1
\end{aligned}$
Hence proved.