Question

Diagonals of a parallelogram intersect each other at point O. If AO = 5 cm, BO = 12 cm and AB = 13 cm then show that ABCD is a rhombus.

Answer 1

Given :$AO=5cm,BO=12cm,AB=13cm$
To prove : ABCD is a rhombus.
Proof:
$A{O}^2+B{O}^2=5^2+{12}^2=25+144=169c{m}^2$
$A{B}^2={13}^2=169c{m}^2$
$A{O}^2+B{O}^2=A{B}^2$
Hence these form a Pythagorean triplet.
$△AOB$ is a right angled triangle.
$\angle AOB={90}^{\circ }$
$\text{seg}AC\perp \text{seg}BD$
i.e, the diagonals are perpendicular to each other.
Rhombus is a parallelogram whose diagonals meet at right angles.
Hence ABCD is a rhombus.
Hence proved.