Question

Prove the theorem : Parallelogram on the same base and between the same parallels are equal in area.

Answer 1

Given two parallelogram ABCD and EFCD that have the same base CD and lie between same parallel AF and CD.

We have prove that $ar\left(ABCD\right)=ar\left(EFCD\right)$

Since opposite sides of $\parallel gm$ are parallel $AB\parallel CD$ and $ED\parallel FC$ with transversal AB

$\Rightarrow \angle DAB=\angle CBF$ [ Corresponding angles ]

with transversal EF

$\Rightarrow \angle DEA=\angle CFE$ [ Corresponding angles ]

$\Rightarrow AD=BC$ [ Opposite sides of parallelogram are equal ]

In $△AED$ $\xi$ $△BFC$

$\Rightarrow \angle DEA=\angle CFE$

$\angle DAB=\angle CBF$

$\therefore AD=BC$

$\Rightarrow △AED\cong △BFC$ [ AAS congruency ]

Hence, $ar\left(△AED\right)=ar\left(△BFC\right)$

( Areas of congruent figures are equal )

$\Rightarrow ar\left(ABCD\right)=ar\left(△ADE\right)+ar\left(EBCD\right)$

$=ar\left(△BFC\right)+ar\left(EBCD\right)$

$=ar\left(EBCD\right)$

$\therefore ar\left(ABCD\right)=ar\left(EBCD\right)$

Hence, the answer is proved.