# 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.

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