Question

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Answer 1

In rhombus ABCD, AB = BC = CD = DA
We know that diagonals of a rhombus bisect each other perpendicularly.
That is AC ⊥ BD, ∠AOB=∠BOC=∠COD=∠AOD=90° and

Consider right angled triangle AOB
AB² = OA² + OB²   [By Pythagoras theorem]

⇒  4AB² = AC²+ BD²
⇒  AB² + AB² + AB² + AB² = AC²+ BD²
∴ AB² + BC² + CD² + DA² = AC²+ BD²
Thus the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.