Let us prove that

Given: ABCD is a rhombus.

To prove: AB² + BC² + CD² + DA² = AC² + BD²

Proof:

We know that diagonals of a rhombus are perpendicular bisector to each other.

So,

AO = OC = 1/2AC

BO = OD = 1/2BD

∠AOB = ∠BOC = ∠COD = ∠AOD = 90°

Also all sides of rhombus are equal,

AB = BC = CD = DA ………………… (1)

By applying Pythagoras Theorem to ΔAOB, we get,

AB² = AO² + BO² [∵ H² = P² + B²]

⇒ AB² = (1/2AC)² + (1/2BD)²

⇒ AB² = 1/4AC² + 1/4BD²

⇒ AB² = 1/4(AC² + BD²)

⇒ 4AB² = AC² + BD²

⇒ AB² + BC² + CD² + DA² = AC² + BD² [By using eqn. (1)]

Hence proved.