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

## Solution :

We know that AD is a median of triangle ABC, then

$${AB}^2$$ + $${AC}^2$$ = 2$${AD}^2$$ + $${1\over 2}{BC}^2$$

Since the diagonals of a parallelogram bisect each other, therefore BO and DO are medians of triangles ABC and ADC respectively.

$$\therefore$$  $${AB}^2$$ + $${BC}^2$$ = 2$${BO}^2$$ + $${1\over 2}{AC}^2$$                 ………..(1)

and      $${AD}^2$$ + $${CD}^2$$ = 2$${DO}^2$$ + $${1\over 2}{AC}^2$$              …………(2)

Adding (1) and (2), we get

$${AB}^2$$ + $${BC}^2$$ + $${AD}^2$$ + $${CD}^2$$ = 2($${BO}^2$$ + $${DO}^2$$) + $${AC}^2$$

$$\implies$$  $${AB}^2$$ + $${BC}^2$$ + $${AD}^2$$ + $${CD}^2$$ = 2($${1\over 4}{BD}^2$$ + $${1\over 4}{BD}^2$$) + $${AC}^2$$

$$\implies$$   $${AB}^2$$ + $${BC}^2$$ + $${AD}^2$$ + $${CD}^2$$ = $${AC}^2$$ + $${BD}^2$$