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Diagonals of rhombus ABCD intersect each other at point O. Prove that OA2+ OC2=2AD2-BD2/2

Question: Diagonals of rhombus ABCD intersect each other at point O. Prove that OA2 + OC2= 2AD2 + BD2/2

Solution: In ∆ AOD
OA2 = AD2 - OD2 -------------Eqn (i)
In ∆ OBC
OC2 = BC2 - OB2 -------------Eqn (ii)
Adding Eqn (i) & Eqn (ii) We get,
OA2 + OC2 = AD2 - OD2 + BC2 - OB2
        = AD2 - OD2 + AD2 - OB2
        = 2AD2 - OD 2 OB 2
        = 2AD2 - (BD/2) 2 -(BD/2)2
        = 2AD2 - BD2/4 -BD2/4
        = 2AD2 - BD2/2 Proved

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