Question
O is any point inside a rectangle ABCD see Fig. Prove that $OB^2 + OD^2 = OA^2 + OC^2$
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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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