Question

Prove that if the diagonals of a parallelogram are equal then it is a rectangle.

Answer 1

Let $ABCD$ be a parallelogram. To show that $ABCD$ is a rectangle, we have to prove that one of its interior angles is ${90}^0$.

In $\Delta ABC$ and $\Delta DCB$,

$AB=DC$ (Opposite sides of a parallelogram are equal)

$BC=BC$ (Common)

$AC=DB$ (Given)

$\therefore \Delta ABC\cong \Delta DCB$ (By SSS Congruence rule)

$\Rightarrow \angle ABC=\angle DCB$

It is known that the sum of the measures of angles on the same side of transversal is ${180}^0$.

$\angle ABC+\angle DCB={180}^0$ (AB || CD)

$\Rightarrow \angle ABC+\angle ABC={180}^0$

$\Rightarrow 2\angle ABC={180}^0$

$\Rightarrow \angle ABC={90}^0$

Since $ABCD$ is a parallelogram and one of its interior angles is ${90}^0$.

Hence, $ABCD$ is a rectangle.