The diagonals of $\square XYZW$ intersect at right angles. Prove that $XY^{2} + ZW^{2} = YZ^{2} + XW^{2}$.

(N/A) Let the diagonals $XZ$ and $YW$ of the quadrilateral $\square XYZW$ intersect at point $O$ at right angles $(90^{\circ})$.
In $\triangle XOY$,$\triangle YOZ$,$\triangle ZOW$,and $\triangle WOX$,by the Pythagorean theorem:
$XY^{2} = XO^{2} + YO^{2}$ $(1)$
$YZ^{2} = YO^{2} + ZO^{2}$ $(2)$
$ZW^{2} = ZO^{2} + WO^{2}$ $(3)$
$XW^{2} = WO^{2} + XO^{2}$ $(4)$
Adding equations $(1)$ and $(3)$:
$XY^{2} + ZW^{2} = (XO^{2} + YO^{2}) + (ZO^{2} + WO^{2})$
Rearranging the terms:
$XY^{2} + ZW^{2} = (YO^{2} + ZO^{2}) + (WO^{2} + XO^{2})$
Substituting equations $(2)$ and $(4)$ into the expression:
$XY^{2} + ZW^{2} = YZ^{2} + XW^{2}$
Hence,it is proved.