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.

Given: □ABCD is a parallelogram, diagonals AC and BD intersect at point M. To prove: AC² + BD² = AB² + BC² + CD² + AD²

Answer 1

□ABCD is a parallelogram. 

∴ AB = CD and BC = AD (i) [Opposite sides of a parallelogram] 

AM = 1/2 AC and BM = 1/2 BD (ii) [Diagonals of a parallelogram bisect each other] 

∴ M is the midpoint of diagonals AC and BD. (iii) 

In ∆ABC. 

seg BM is the median. [From (iii)

AB² + BC² = 2AM² + 2BM² (iv) [Apollonius theorem]

∴ AB² + BC² = 2(1/2 AC)² + 2(1/2 BD)² [From (ii) and (iv)] 

∴ AB² + BC² = 2 × (BD²/4) + 2 X (AC²/4)

∴ AB² + BC² = (BD²/2) + (AC²/4)

∴ 2AB² + 2BC² = BD² + AC² [Multiplying both sides by 2]

∴ AB² + AB² + BC² + BC² = BD² + AC² 

∴ AB² + CD² + BC² + AD² = BD² + AC² [From(i)] 

i.e. AC² + BD² = AB² + BC² + CD² + AD²