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Question
Diagonals of rhombus ABCD intersect each other at point O.
Prove that: OA2 + OC2 = 2AD2 - `"BD"^2/2`
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Solution
Diagonals of the rhombus are perpendicular to each other.
In quadrilateral ABCD, ∠AOD = ∠COD = 90°.
So, ΔAOD and ΔCOD are right-angled triangles.
In ΔAOD using Pythagoras theorem,
AD2 = OA2 + OD2
⇒ OA2 = AD2 - OD2 ....(i)
In ΔCOD using Pythagoras theorem,
CD2 = OC2 + OD2
⇒ OC2 = CD2 - OD2 ....(ii)
LHS = OA2 + OC2
= AD2 - OD2 + CD2 - OD2 ...[ From(i) and (ii) ]
= AD2 + CD2 - 2OD2
= AD2 + AD2 - 2`("BD"/2)^2` ...[ AD = CD and OD = `"BD"/2`]
= 2AD2 - `("BD")^2/2`
= RHS.
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