Question

If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Answer 1

Step $1$: Prove that $∆DCB\cong ∆CBA$.

Let $ABCD$ is a parallelogram.

Given : Diagonals of a parallelogram are equal, $AC=BD$

In $∆DCB$ and $∆CBA$,

$DB=CA$ [Given]

$DC=BA$ [Opposite side of parallelogram are equal]

$CB=BC$ [Common side]

Using $SSS$,

$∆DCB\cong ∆CBA$,

So, $\angle DCB=\angle CBA$……………………….$\left(1\right)$

Step $2$: Prove that $\angle DCB=\angle CBA=90°$.

We know that, $ABCD$ is a parallelogram.

Therefore, $\angle DCB+\angle CBA=180°$ ………………………$\left(2\right)$ [Co-interior angles of parallelogram]

From equation$1$ and equation$2$ ,

$\angle DCB=\angle CBA=90°$

Thus parallelogram $ABCD$ having all angles equal to $90°$ because opposite sides are equal.

Hence $ABCD$ is a rectangle.