Need Geometry Help, Having Trouble with Wording

The question is: Prove that the sum of the squares of the distances os any point P = (x,y) from two opposite vertices of any rectangle is equal to the sum of the squares of its distances from the other two vertices. You may choose the rectangle with vertices (0,0), (a,0), (0,b), (a,b).

I thought I had this, but I had some extra terms, when I was finished. I'm not sure what I did wrong. I tried putting P between (0,b) and (a,b). Any help is appreciated, please who your work. Thanks.

Let
A=(0,0)
B=(0,b)
C = (a,b)
D = (a,0)

You have to prove that AP^2 + CP^2 = BP^2 + DP^2

AP^2 = x^2 +y^2
CP^2 = (x-a)^2 +(y-b)^2
BP^2 = x^2 + (y-b)^2
DP^2 = (x-a)^2 +y^2

AP^2 + CP^2 = x^2+y^2 + (x-a)^2+(y-b)^2
BP^2 + DP^2 = x^2 + (y-b)^2 + (x-a)^2 + y^2
If you rearrange it, you have x^2+y^2 + (x-a)^2+(y-b)^2 which is the same as AP^2 + CP^2

I put P anywhere inside the rectangle
Using 2 opposite vertices
Square of sum of distances
=x^2+(y-b)^2 + (a-x)^2+y^2
Using the other 2 vertices
Square = (b-y)^2+(x-a)^2 + x^2 + y^2
As (b-y)^2 = (y-b)^2 and (a-x)^2 = (x-a)^2
The sums are equal.