**Answer:**

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

AB^{2} = OA^{2} + OB^{2} [By Pythagoras theorem]

⇒ 4AB^{2} = AC^{2}+ BD^{2}

⇒ AB^{2} + AB^{2} + AB^{2} + AB^{2} = AC^{2}+ BD^{2}

**∴ AB**^{2} + BC^{2} + CD^{2} + DA^{2} = AC^{2}+ BD^{2}

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