Question
$O$ is any point inside a rectangle $ABCD$ see Fig. Prove that $OB^2 + OD^2 = OA^2 + OC^2 .$
✓
Answer
Through $O,$ draw $PQ \| BC$ so that $P$ lies on $AB$ and $Q$ lies on $DC.$
Now, $PQ \| BC$
Therefore, $PQ \perp AB$ and $PQ \perp DC ( \angle B = 90^{\circ}$ and $\angle C = 90^\circ )$
So, $\angle BPQ = 90^\circ$ and $\angle CQP = 90^\circ$
Therefore, $BPQC$ and $APQD$ are both rectangles.
Now, from $\triangle OPB$
$OB^2 = BP^2 + OP^2 ...(1)$
Similarly, from $\triangle OQD,$
$OD^2 = OQ^2 + DQ^2 ...(2)$
From $\triangle OQC,$ we have
$OC^2 = OQ^2 + CQ^2...(3)$
and from $\triangle OAP,$ we have
$OA^2 = AP^2 + OP^2 ...(4)$
Adding $(1)$ and $(2),$
$OB^2 + OD^2 = BP^2 + OP^2+ OQ^2 + DQ^2$
$= CQ^2 + OP^2 + OQ^2 + AP^2 ($As $BP = CQ$ and $DQ = AP)$
$= CQ^2 + OQ^2 + OP^2 + AP^2$
$OB^2 + OD^2= OC^2 + OA^2 [$From $(3)$ and $(4)]$
Hence proved.
Now, $PQ \| BC$
Therefore, $PQ \perp AB$ and $PQ \perp DC ( \angle B = 90^{\circ}$ and $\angle C = 90^\circ )$
So, $\angle BPQ = 90^\circ$ and $\angle CQP = 90^\circ$
Therefore, $BPQC$ and $APQD$ are both rectangles.
Now, from $\triangle OPB$
$OB^2 = BP^2 + OP^2 ...(1)$
Similarly, from $\triangle OQD,$
$OD^2 = OQ^2 + DQ^2 ...(2)$
From $\triangle OQC,$ we have
$OC^2 = OQ^2 + CQ^2...(3)$
and from $\triangle OAP,$ we have
$OA^2 = AP^2 + OP^2 ...(4)$
Adding $(1)$ and $(2),$
$OB^2 + OD^2 = BP^2 + OP^2+ OQ^2 + DQ^2$
$= CQ^2 + OP^2 + OQ^2 + AP^2 ($As $BP = CQ$ and $DQ = AP)$
$= CQ^2 + OQ^2 + OP^2 + AP^2$
$OB^2 + OD^2= OC^2 + OA^2 [$From $(3)$ and $(4)]$
Hence proved.
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