Question: Diagonals of rhombus ABCD intersect each other at point O. Prove that OA^{2} + OC^{2}= 2AD^{2} + BD^{2}/2

Solution: In ∆ AOD

OA^{2} = AD^{2} – OD^{2} ————-Eqn (i)

In ∆ OBC

OC^{2} = BC^{2} – OB^{2} ————-Eqn (ii)

Adding Eqn (i) & Eqn (ii) We get,

OA^{2} + OC^{2} = AD^{2} – OD^{2} + BC^{2} – OB^{2}

= AD^{2} – OD^{2} + AD^{2} – OB^{2}

= 2AD^{2} – OD ^{2} OB ^{2}

= 2AD^{2} – (BD/2) ^{2} -(BD/2)^{2}

= 2AD^{2} – BD^{2}/4 -BD^{2}/4

= 2AD^{2} – BD^{2}/2 __Proved__