# Prove that twice the area of parallelogram is equal to the area of another parallelogram formed by taking as its adjacent sides the diagonals of the former parallelogram.

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• The area of a parallelogram whose sides are $\overrightarrow a \: and \overrightarrow b \: is \: |\overrightarrow a \times \overrightarrow b|$
Step 1
Let $ABCD$ be a parallelogram .
Consider the parallelogram formed by its diagonals $\overrightarrow {AC} \: and \: \overrightarrow {BD}$ as adjacent sides
Step 2
The vector are of such a parallelogram would be $\overrightarrow {AC} \times \overrightarrow {BD} = (\overrightarrow {AB} +\overrightarrow {BC}) \times (\overrightarrow {BA}+\overrightarrow {AD})$
$= \overrightarrow {AB} \times \overrightarrow {BA}+\overrightarrow {AB} \times \overrightarrow {AD}+\overrightarrow {BC} \times \overrightarrow {BA}+\overrightarrow {BC} \times \overrightarrow {AD}$
$= \overrightarrow 0 + \overrightarrow {AB} \times \overrightarrow {AD}-\overrightarrow {BC} \times \overrightarrow {AB} + \overrightarrow {AD} \times \overrightarrow {AD}$
$= \overrightarrow {AB} \times \overrightarrow {AD} - \overrightarrow {AD} \times \overrightarrow {AB}$
$= \overrightarrow {AB} \times \overrightarrow {AD} +\overrightarrow {AB} \times \overrightarrow {AD}$
$= 2\overrightarrow {AB} \times \overrightarrow {AD} = 2$ ( vector area of parallelogram $ABCD$ )
Hence proved

edited Jun 23, 2013