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

(N/A) Let $ABCD$ and $ABFE$ be two parallelograms on the same base $AB$ and between the same parallel lines $AB$ and $DF$.
Let $\overrightarrow{AB} = \vec{a}$ and $\overrightarrow{AD} = \vec{b}$.
The area of parallelogram $ABCD$ is given by the magnitude of the cross product of its adjacent sides:
$\text{Area}(ABCD) = |\overrightarrow{AB} \times \overrightarrow{AD}| = |\vec{a} \times \vec{b}|$.
Now,for parallelogram $ABFE$,the adjacent sides are $\overrightarrow{AB}$ and $\overrightarrow{AE}$.
Since $D, E, C, F$ lie on the same line parallel to $AB$,we can write $\overrightarrow{AE} = \overrightarrow{AD} + \overrightarrow{DE}$.
Since $\overrightarrow{DE}$ is parallel to $\overrightarrow{AB}$,we have $\overrightarrow{DE} = k\vec{a}$ for some scalar $k$.
Thus,$\overrightarrow{AE} = \vec{b} + k\vec{a}$.
The area of parallelogram $ABFE$ is:
$\text{Area}(ABFE) = |\overrightarrow{AB} \times \overrightarrow{AE}|$
$= |\vec{a} \times (\vec{b} + k\vec{a})|$
$= |(\vec{a} \times \vec{b}) + (\vec{a} \times k\vec{a})|$
$= |(\vec{a} \times \vec{b}) + k(\vec{a} \times \vec{a})|$.
Since the cross product of any vector with itself is zero $(\vec{a} \times \vec{a} = \vec{0})$:
$\text{Area}(ABFE) = |\vec{a} \times \vec{b} + \vec{0}| = |\vec{a} \times \vec{b}|$.
Therefore,$\text{Area}(ABFE) = \text{Area}(ABCD)$.
Hence,it is proved that parallelograms on the same base and between the same parallels are equal in area.