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Let A B C D be a rectangle and P be any ...

Let `A B C D` be a rectangle and `P` be any point in its plane. Show that `A P^2+P C^2=P B^2+P D^2dot`

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The correct Answer is:
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Let `A-=(0,0),B-=(a,0),C-=(a,b)and D-=(0,b)`
Let point P be (h,k).

`therefore PA^2+PC^2=(h-0)^2+(k-0)^2+(h-a)^2+(k-b)^2=2h^2+2k^2-2k^2-2ah-2bk+a^2+b^2`
Also, `PB^2+PD^2=(h-a)^2+(k-0)^2+(h-0)^2+(k-b)^2=2h^2+2k^2-2ah-2bk+a^2+b^2`
Thus, `PA^2+PC^2=PB^2+PD^2`
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