Using vectors, prove that the parallelogram on the same base and between the same parallels are equal in area.
Let $ABCD$ and $ABFE$ are parallelograms on the same base $AB$ and between the same parallel lines $AB$ and $DF$.

Let $\overrightarrow {{\rm{AB}}} = \overrightarrow {\rm{a}}$ and $\overrightarrow {{\rm{AD}}} = \overrightarrow {\rm{b}}$

$\therefore$ Area of parallelogram $ABCD = \overrightarrow {\rm{a}} \times \overrightarrow {\rm{b}}$

Now, area of parallelogram $ABFF = \overrightarrow {AB} \times \overrightarrow {AE}$

$= \overrightarrow {{\rm{AB}}} \times (\overrightarrow {{\rm{AD}}} + \overrightarrow {{\rm{DE}}} )$

$= \overrightarrow {AB} \times (\vec b + k\vec a)$,

[let $\overrightarrow {DE} = k\vec a$where $k$ is a scalar ]

$= \overrightarrow {\rm{a}} \times (\overrightarrow {\rm{b}} + k\overrightarrow {\rm{a}} )$

$= (\overrightarrow {\rm{a}} \times \overrightarrow {\rm{b}} ) + (\overrightarrow {\rm{a}} \times k\overrightarrow {\rm{a}} )$

$= (\overrightarrow {\rm{a}} \times \overrightarrow {\rm{b}} ) + k(\overrightarrow {\rm{a}} \times \overrightarrow {\rm{a}} )$

$= (\overrightarrow {\rm{a}} \times \overrightarrow {\rm{b}} )$
= Area of parallelogram $ABCD$

LONG ANSWER