Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals
In ΔAOB, ΔBOC, ΔCOD, ΔAOD,
Applying Pythagoras theorem, we obtain
AB2 = AO2 + OB2 (i)
BC2 = BO2 + OC2 (ii)
CD2 = CO2 + OD2 (iii)
AD2 = AO2 + OD2 (iv)
Now after adding all equations, we get,
AB2 + BC2 + CD2 + AD2 = 2 (AO2 + OB2 + OC2 + OD2)
= 2[
2 + 
2 + 
2 + 
2]
As we know that the Diagonals bisect each other,
= 2 [
2 + 
2]
= (AC)2 + (BD)2
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