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

Pythagoras theorem, we obtain

AB² = AO² + OB²     (i)

BC² = BO² + OC²     (ii)

CD² = CO² + OD²     (iii)

AD² = AO² + OD²     (iv)

Now after adding all equations, we get,

AB² + BC² + CD² + AD² = 2 (AO² + OB² + OC² + OD²)

As we know that the Diagonals bisect each other,

= (AC)² + (BD)²