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प्रश्न
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals
उत्तर
In ΔAOB, ΔBOC, ΔCOD, ΔAOD,
Applying Pythagoras theorem, we obtain
AB2 = AO2 + OB2 ....(1)
BC2 = BO2 + OC2 ....(2)
CD2 = CO2 + OD2 ....(3)
AD2 = AO2 + OD2 ....(4)
Adding all these equation we obtain
AB2 + BC2 + CD2 + AD2 = 2(AO2 + OB2 + OC2 + OD2)
`= 2(((AC)/2)^2 + ((BD)/2)^2 + ((AC)/2)^2 + ((BC)/2)^2)`
(Diagonals bisect each other)
`= 2((AC)^2/2 + (BD)^2/2)`
`= (AC)^2 + (BD)^2`
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