Answer :
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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