# Prove that sum of the squares on the diagonals of a rhombus is equal to four times the square on the side of the rhombus?

Jul 26, 2016

In a rhombus diagonals intersect each other at right angles. In the given rhombus $A B C D$ diagonal bisect each other at $O$ hence
$B {C}^{2} + A {D}^{2} = {\left(2 B O\right)}^{2} + {\left(2 A O\right)}^{2} = 4 B {O}^{2} + 4 A {O}^{2}$
= $4 \left(B {O}^{2} + A {O}^{2}\right) = 4 A {B}^{2}$