<s>[INST] condition required for a quadrilateral $ABCD$ such that every point inside $ABCD$ satisfies $PA^2+PC^2 = PB^2+PD^2$ suppose there is a quadrilateral $ABCD$.  any point $P$ which lies inside the quadrilateral satisfies $PA^2+PC^2 = PB^2+PD^2$. Should such a condition exist always in a rectangle or a square?.can there be any other quadrilateral in which such a point exists.
 [/INST] If $P=A$, then $$AC^2 = AB^2+AD^2,$$
if $P=C$, then $$AC^2 = BC^2+BD^2,$$
so
$$AC^2 = AB^2+AD^2 = BC^2+CD^2.\tag{1}$$
Same way:
$$
AC^2 = AB^2+BC^2=CD^2+AD^2.\tag{2}
$$
Hence (subtracting eq. $(1),(2)$)
$$
AD^2-BC^2=-(AD^2-BC^2),\\
AB^2-CD^2=-(CD^2-AB^2),\\
$$
so $$AD=BC,\\AB=CD.\tag{3}$$
It is parallelogram (for now).
But $(1),(3) \Rightarrow$:
$$
AC^2=AB^2+BC^2,
$$
so $\angle B = 90^\circ$. So, it is rectangle.

Here is the image that shows that "to be rectangle" is sufficient condition for such quadrilateral. ($\color{blue}{PA^2+PC^2} = \color{red}{PB^2+PD^2}$). 

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