Question

ABCD is a rhombus. AC and BD intersect at O, show that
AB2
+ BC2
+ CD2
+ DA2
= 4AO2
+ 4OD2​

Answer 1

Answer:

the answer to this question is

Step-by-step explanation:

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Answer 2

Answer and Step-by-step explanation:

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

Hence Proved