Question

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

Answer 1

Given: In parallelogram $ABCD$, $AC=BD$

To prove : Parallelogram $ABCD$ is rectangle.

Proof : in $△ACB$ and $△BDA$

$AC=BD$ $\mid$ Given

$AB=BA$ $\mid$ Common

$BC=AD$ $\mid$ Opposite sides of the parallelogram ABCD

$△ACB$ $\cong △BDA\mid SSS$ Rule

$\therefore \angle ABC=\angle BAD...\left(1\right)$ CPCT

Again $AD$ $\parallel$ $\mid$ Opposite sides of parallelogram $ABCD$

$AD$ $\parallel BC$ and the traversal $AB$ intersects them.

$\therefore \angle BAD+\angle ABC={180}^{\circ }$ ...$\left(2\right)$ _ Sum of consecutive interior angles on the same side of the transversal is

${180}^{\circ }$

From (1) and (2) ,
$\angle BAD=\angle ABC={90}^{\circ }$

$\therefore \angle A={90}^{\circ }$ and $\angle C={90}^{\circ }$

Parallelogram $ABCD$ is a rectangle.