Given: ABCD is a Rhombus in which AC and BD are diagonals which intersect each other at O.
To Prove : AB² + BC² + CD² + DA² = AC² + BD²
Proof : We know that diagonals of rhombus bisect each other at right angle.
Therefore,
∠AOB=∠BOC = ∠COD = ∠DOA = 90°
Now, in right angle ∆AOB
AB² = OA² + OB²    ...(i)
[Using Pythagoras theorem]
In, ∆BOC right triangle we have
BC² = OB² + OC²    ...(ii)
[Using Pythagoras theorem]
In, ∆COD right triangle we have
CD² = OC² + OD²    ...(iii)
[Using Pythagoras theorem]
In, ∆AOD right triangle we have
AD² = OD² + OA²    ...(iv)
[Using Pythagoras theorem]
Adding (i), (ii), (iii) and (iv), we get
      


           
        
                                 







Hence,