If ABCD be a rectangle and P be any point in a plane of the rectangle, then prove that PA² + PC² = PB² + PD².

[Hint: Take A as the origin and AB and AD as x and y - axis respectively. Let AB = a, AD = b]

Let us assume A as the origin (0, 0) and AB and AD as x and y axis with length a and b units.

Then we get points B to be (a, 0), D to be (0, b) and C to be (a, b).

Let us assume P(x, y) be any point in a plane of the rectangle.

We need to prove PA² + PC² = PB² + PD².

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

Let us assume L.H.S,

⇒ PA² + PC² = ((x - 0)² + (y - 0)²) + ((x - a)² + (y - b)²)

⇒ PA² + PC² = x² + y² + x² - 2ax + a² + y² - 2by + b²

⇒ PA² + PC² = (x² - 2ax + a² + y²) + (x² + y² - 2by + b²)

⇒ PA² + PC² = ((x - a)² + (y - 0)²) + ((x - 0)² + (y - b)²)

⇒ PA² + PC² = PB² + PD²

⇒ L.H.S = R.H.S

∴ Thus proved.