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

Solution: In &#x2206 AOD
OA2 = AD2 – OD2 ————-Eqn (i)
In &#x2206 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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