The diagonals of convex quadrilateral $PQRS$ intersect at right angles. Prove that $PQ^{2} + RS^{2} = PS^{2} + QR^{2}$.

(A) Let the diagonals $PR$ and $QS$ of the quadrilateral $PQRS$ intersect at point $O$ at right angles.
Thus,$\angle POQ = \angle QOR = \angle ROS = \angle SOP = 90^{\circ}$.
In right-angled $\triangle POQ$,by Pythagoras theorem: $PQ^{2} = PO^{2} + OQ^{2}$.
In right-angled $\triangle ROS$,by Pythagoras theorem: $RS^{2} = RO^{2} + OS^{2}$.
Adding these two equations: $PQ^{2} + RS^{2} = PO^{2} + OQ^{2} + RO^{2} + OS^{2}$ --- (Equation $1$).
In right-angled $\triangle SOP$,by Pythagoras theorem: $PS^{2} = SO^{2} + OP^{2}$.
In right-angled $\triangle QOR$,by Pythagoras theorem: $QR^{2} = OQ^{2} + OR^{2}$.
Adding these two equations: $PS^{2} + QR^{2} = SO^{2} + OP^{2} + OQ^{2} + OR^{2}$ --- (Equation $2$).
Comparing Equation $1$ and Equation $2$,we see that the right-hand sides are identical.
Therefore,$PQ^{2} + RS^{2} = PS^{2} + QR^{2}$.