Question

Prove that the two parallelogram on the same base and between two parallels are equal in area

Answer 1

To prove :ar (||gm ABCD)=ar (||gm EBCF)
Proof : In $\Delta ABE$ and $\Delta DCF$
$\angle EAB=\angle FDC\left[\begin{array}{c}AB||DC,AFisthetransversal\\ \Rightarrow Pairofcorrespondinganglesareequal\end{array}\right]...\left(1\right)$
$\angle BEA=\angle CFD.\left[\begin{array}{c}BE||CF,AFisthetransversal\\ \Rightarrow Pairofcorrespondinganglesareequal\end{array}\right]....\left(2\right)$
Now, $\angle EAB+\angle ABE+\angle BEA=\angle FDC+\angle DCF+\angle CFD={180}^{\circ }$
[Sum of measures of the interior or angles of a triangle is ${180}^{\circ }$
$\therefore \angle ABE=\angle DCF...\left(3\right)$ [Using (1)and (2)]
Also, AB =DC [Opposite sides of a ||gm are equal]...(4)
Therefore, using (1),(3)and (4); $\Delta ABE\cong \Delta DCF$ [By ASA congruency axioms]
$\therefore ar\left(\Delta ABE\right)\cong \left(\Delta DCF\right)$
$\therefore$ ar (||gm ABCD)=$ar\left(\Delta ABE\right)+ar$ (trapezium EBCD)
=ar $\left(\Delta DCF\right)$=ar (trapezium EBCD)
=ar (||gm EBCF)
Hence, ar (||gm ABCD)=ar (||gm EBCF)