Question

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

Answer 1

Given : A parallelogram $ABCD$ , in which $AC=BD$

TO Prove : $ABCD$ is a rectangle .

Proof : In $△ABC$ and $△ABD$

$AB=AB$ [common]

$AC=BD$ [given]

$BC=AD$ [opp. sides of a parallelogram]

$⟹△ABC\cong △BAD$ [ by SSS congruence axiom]

$⟹\angle ABC=△BAD$ [corresponding parts of congruent triangles]

Also, $\angle ABC+\angle BAD={180}^{\circ }$ [co - interior angles]

$⟹\angle ABC+\angle ABC={180}^{\circ }$ [Since, $\angle ABC=\angle BAD$]

$⟹2\angle ABC={180}^{\circ }$

$⟹\angle ABC=\frac{1}{2}\times {180}^{\circ }={90}^{\circ }$

Hence, parallelogram $ABCD$ is a rectangle.