Answer :

Given: ABCD is a rhombus.

To prove: AB^{2} + BC^{2} + CD^{2} + DA^{2} = AC^{2} + BD^{2}

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,

AB^{2} = AO^{2} + BO^{2} [∵ H^{2} = P^{2} + B^{2}]

⇒ AB^{2} = (1/2AC)^{2} + (1/2BD)^{2}

⇒ AB^{2} = 1/4AC^{2} + 1/4BD^{2}

⇒ AB^{2} = 1/4(AC^{2} + BD^{2})

⇒ 4AB^{2} = AC^{2} + BD^{2}

⇒ AB^{2} + BC^{2} + CD^{2} + DA^{2} = AC^{2} + BD^{2} [By using eqn. (1)]

Hence proved.

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