<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] This condition will always hold true in a rectangle. And since a square, by definition, is a rectangle, the condition will also hold true for the square.
Let us pick an arbitrary point P inside the quadrilateral, and let us create a line $\ell$ that passes through point P such that the line is perpendicular to side AB. Since AB and CD are parallel, it follows that $\ell$ is perpendicular to CD as well. Let's name the point of intersection of $\ell$ and AB point X, and $\ell$ and CD, point Y.
Instead of showing $ PA^2 + PC^2 = PB^2 + PD^2 $, I will show 
$$ PA^2 - PB^2 + PC^2 - PD^2 = 0 \\$$
Proof:
\begin{align*}
  PA^2 &= PX^2 + XA^2 \\
  PB^2 &= PX^2 + XB^2 \\
  PC^2 &= PY^2 + YC^2 \\
  PD^2 &= PY^2 + YD^2 \\
  PA^2 - PB^2 + PC^2 - PD^2 &= XA^2 - XB^2 + YC^2 - YD^2 \\
  PA^2 - PB^2 + PC^2 - PD^2 &= (XA^2 - YD^2) + (YC^2 - XB^2) \\
  PA^2 - PB^2 + PC^2 - PD^2 &= 0 + 0 \\
  PA^2 - PB^2 + PC^2 - PD^2 &= 0
\end{align*}
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