Prove, using coordinates that the sum of squares of the diagonals of a rectangle is equal to the sum of squares of its sides.

Let us assume ABCD be a rectangle with A as the origin and AB and AD as x and y - axes having lengths a and b units.

We get the vertices of the rectangle as follows.

⇒ A = (0, 0)

⇒ B = (a, 0)

⇒ C = (a, b)

⇒ D = (0, b)

We need to prove that the sum of squares of the diagonals of a rectangle is equal to the sum of squares of its sides.

i.e., AC² + BD² = AB² + BC² + CD² + DA²

We know that distance between two points (x₁, y₁) and (x₂, y₂) is .

Assume L.H.S

⇒ AC² + BD² = ((0 - a)² + (0 - b)²) + ((a - 0)² + (0 - b)²)

⇒ AC² + BD² = a² + b² + a² + b²

⇒ AC² + BD² = 2(a² + b²) ..... - (1)

Assume R.H.S

⇒ AB² + BC² + CD² + DA² = ((0 - a)² + (0 - 0)²) + ((a - a)² + (0 - b)²) + ((a - 0)² + (b - b)²) + ((0 - 0)² + (b - 0)²)

⇒ AB² + BC² + CD² + DA² = a² + 0 + 0 + b² + a² + 0 + 0 + b²

⇒ AB² + BC² + CD² + DA² = 2(a² + b²) .... (2)

From (1) and (2), we can clearly say that,

⇒ AC² + BD² = AB² + BC² + CD² + DA²

∴ Thus proved.