Question

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

Answer 1

Diagonals of a parallelogram bisect each other, i.e., O is the midpoint of AC and BD.

In ∆ABD, point O is the midpoint of side BD.

BO = OD = ${\displaystyle \frac{1}{2}}$BD

∴ by Apollonius theorem,

AB² + AD² = 2AO² + 2BO²   ...(1)

In ∆CBD, point O is the midpoint of side BD.

BO = OD = ${\displaystyle \frac{1}{2}}$BD

∴ by Apollonius theorem,

CB² + CD² = 2CO² + 2BO²   ...(2)

Adding (1) and (2), we get,

AB² + AD² + CB² + CD² = 2AO² + 2BO² + 2CO² + 2BO²

⇒ AB² + AD² + CB² + CD² = 2AO² + 4BO² + 2AO²   ...(∵ OC = OA)

⇒ AB² + AD² + CB² + CD² = 4AO² + 4BO²

⇒ AB² + AD² + CB² + CD² = (2AO)² + (2BO)²

⇒ AB² + AD² + CB² + CD² = (AC)² + (BD)²

⇒ AB² + AD² + CB² + CD² = AC² + BD²

∴ the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.