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Question
Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.
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Solution
Diagonals of a parallelogram bisect each other, i.e., O is the midpoint of AC and BD.
In ∆ABD, point O is the midpoint of side BD.
BO = OD = `1/2`BD
∴ by Apollonius theorem,
AB2 + AD2 = 2AO2 + 2BO2 ...(1)
In ∆CBD, point O is the midpoint of side BD.
BO = OD = `1/2`BD
∴ by Apollonius theorem,
CB2 + CD2 = 2CO2 + 2BO2 ...(2)
Adding (1) and (2), we get,
AB2 + AD2 + CB2 + CD2 = 2AO2 + 2BO2 + 2CO2 + 2BO2
⇒ AB2 + AD2 + CB2 + CD2 = 2AO2 + 4BO2 + 2AO2 ...(∵ OC = OA)
⇒ AB2 + AD2 + CB2 + CD2 = 4AO2 + 4BO2
⇒ AB2 + AD2 + CB2 + CD2 = (2AO)2 + (2BO)2
⇒ AB2 + AD2 + CB2 + CD2 = (AC)2 + (BD)2
⇒ AB2 + AD2 + CB2 + CD2 = AC2 + BD2
∴ the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.
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