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

## Solution :

Let the diagonals BD and AC of the rhombus ABCD intersect each other at O.

Since the diagonals of the rhombus bisect each other at right angles.

$$\therefore$$  $$\angle$$ AOB = $$\angle$$ BOC = $$\angle$$ COD = $$\angle$$ DOA = 90

and  AO = CO, BO = OD

Since AOB is a right triangle angled at O, therefore,

$${AB}^2$$ = $${OA}^2$$ + $${OB}^2$$

Since AO = CO, BO = OD,

$$\implies$$ $${AB}^2$$ = $$({1\over 2}{AB})^2$$ + $$({1\over 2}{BD})^2$$

$$\implies$$  $$4{AB}^2$$ = $${AC}^2$$ + $${BD}^2$$            ……….(1)

Similarly, we have :

$$4{BC}^2$$ = $${AC}^2$$ + $${BD}^2$$            ……..(2)

$$4{CD}^2$$ = $${AC}^2$$ + $${BD}^2$$            ……..(3)

$$4{AD}^2$$ = $${AC}^2$$ + $${BD}^2$$             ………(4)

Adding all the equations above, we get

$${AB}^2$$ + $${BC}^2$$ + $${CD}^2$$ + $${DA}^2$$ = $${AC}^2$$ + $${BD}^2$$