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# Using vectors,prove that the parallelogram on the same base and between the same parallels are equal in area.

Toolbox:
• Vector area of a parallelogram is the cross product of its adjacent sides.
Let us consider a parapollegram $ABCD$
Let $\overrightarrow{AB}=\overrightarrow{a}$ and $\overrightarrow {AD}=\overrightarrow b$
We know that the vector area of paralleogram is the cross product of its adjacent sides.
Therefore vector area of the parallelogram is $\overrightarrow a \times \overrightarrow b.$-----(1)
Now Consider the parallelogarm $ABB'A'$
Here $\overrightarrow {AB}=\overrightarrow a$ and let $A'D=m\overrightarrow a.$ because $A'D$ is parallel to $AB$
Consider the triangle $ADA'$
By triangle law of vectors
$AA'=m \overrightarrow a+\overrightarrow b$
Hence the vector area of parallelogram $ABB'A'=\overrightarrow a \times (m \overrightarrow a+\overrightarrow b)$
On expanding we get
$\overrightarrow a \times m \overrightarrow a+ \overrightarrow a \times \overrightarrow b$
But we know $\overrightarrow a \times \overrightarrow a=0$
Therefore vector area of parallelogram $ABB'A'=\overrightarrow a \times \overrightarrow b$-----(2)
Hence from equ(1) and equ(2)
It is proved that vector area of a parallelogram on the same base and between same paralles are equal