[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.