Question

The diagonals AC and BD of a rhombus intersect each other at 0. Prove that:
AB2+ BC2 + CD2 + DA2 = 4(0A2 + OB2)​

Answer 1

Answer:

The diagonals AC and BD of a rhombus intersect each other at O.

Diagonal of rhombus bisect each other perpendicularly

=> OA = OC = AC/2

& OB = OD = BD/2

Applying Pythagorean theorem

AB² = OA² + OB²

BC² = OB² + OC²

CD² = OC² + OD²

DA² = OD² + OA²

Adding all

=> AB² + BC² + CD² + DA² = OA² + OB² + OB² + OC² + OC² + OD² + OD² + OA²

using OA = OC & OB = OD

=>AB² + BC² + CD² + DA² = OA² + OB² + OB² + OA² + OA² + OB² + OB² + OA²

=> AB² + BC² + CD² + DA² = 4OA² + 4OB²

=> AB² + BC² + CD² + DA² = 4(OA² + OB²)

QED

Proved