Q5 of 27 Page 289

Let us prove that, the sum of squares drawn on the sides of a rhombus is equal to the sum of squares drawn on two diagonals.



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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