$O$ is any point inside a rectangle $ABCD$ (see Figure). Prove that $OB^{2} + OD^{2} = OA^{2} + OC^{2}$.

(N/A) Through $O$,draw $PQ \parallel BC$ such that $P$ lies on $AB$ and $Q$ lies on $DC$.
Now,$PQ \parallel BC$.
Therefore,$PQ \perp AB$ and $PQ \perp DC$ (since $\angle B = 90^{\circ}$ and $\angle C = 90^{\circ}$).
So,$\angle BPQ = 90^{\circ}$ and $\angle CQP = 90^{\circ}$.
Therefore,$BPQC$ and $APQD$ are both rectangles.
Now,from $\Delta OPB$,by Pythagoras theorem:
$OB^{2} = BP^{2} + OP^{2}$ $...(1)$
Similarly,from $\Delta OQD$:
$OD^{2} = OQ^{2} + DQ^{2}$ $...(2)$
From $\Delta OQC$,we have:
$OC^{2} = OQ^{2} + CQ^{2}$ $...(3)$
And from $\Delta OAP$,we have:
$OA^{2} = AP^{2} + OP^{2}$ $...(4)$
Adding $(1)$ and $(2)$:
$OB^{2} + OD^{2} = BP^{2} + OP^{2} + OQ^{2} + DQ^{2}$
Since $BP = CQ$ and $DQ = AP$ (opposite sides of rectangles $BPQC$ and $APQD$):
$OB^{2} + OD^{2} = CQ^{2} + OP^{2} + OQ^{2} + AP^{2}$
Rearranging the terms:
$OB^{2} + OD^{2} = (CQ^{2} + OQ^{2}) + (AP^{2} + OP^{2})$
Using equations $(3)$ and $(4)$:
$OB^{2} + OD^{2} = OC^{2} + OA^{2}$.