Question

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals

Answer 1

In ΔAOB, ΔBOC, ΔCOD, ΔAOD,

Applying Pythagoras theorem, we obtain

AB² = AO² + OB² ....(1)

BC² = BO² + OC² ....(2)

CD² = CO² + OD² ....(3)

AD² = AO² + OD² ....(4)

Adding all  these equation we obtain

AB² + BC² + CD² + AD² = 2(AO² + OB² + OC² + OD²)

`= 2(((AC)/2)^2 + ((BD)/2)^2 + ((AC)/2)^2 + ((BC)/2)^2)`

(Diagonals bisect each other)

`= 2((AC)^2/2 + (BD)^2/2)`

`= (AC)^2 + (BD)^2`