In figure O is any point inside a rectangle ABCD. Prove that OB2 + OD2

1.	In figure O is any point inside a rectangle ABCD. Prove that OB2 + OD2 = OA2 + OC2.
Answer» Through O, draw PQ\|\|BC so that P lies on AB and Q lies on DC. Now, PQ\|\|BC PQ ⊥ AB and PQ ⊥ DC (∵ ∠B = 90° and ∠C = 90°) So, ∠BPQ = 90° and ∠CQP = 90° Therefore BPQC and APQD are both rectangles. Now from ΔOPB, OB² = BP² + OP² … (1) Similarly from ΔOQD, OD² = OQ² + DQ² … (2) From ΔOQC, we have OC² = OQ² + CQ² … (3) ΔOAP, we have OA² = AP² + OP² … (4) Adding (1) and (2) OB² + OD² = BP² + OP² + OQ² + DQ² (As BP = CQ and DQ = AP) = CQ² + OP² + OQ² + AP² = CQ² + OQ² + OP² + AP² = OC² + OA² [From (3) and (4)]

In figure O is any point inside a rectangle ABCD. Prove that OB2 + OD2 = OA2 + OC2.

Answer»

Through O, draw PQ||BC so that P lies on AB and Q lies on DC.

Now, PQ||BC

PQ ⊥ AB and PQ ⊥ DC (∵ ∠B = 90° and ∠C = 90°)

So, ∠BPQ = 90° and ∠CQP = 90°

Therefore BPQC and APQD are both rectangles.

Now from ΔOPB,

OB² = BP² + OP² … (1)

Similarly from ΔOQD,

OD² = OQ² + DQ² … (2)

From ΔOQC, we have

OC² = OQ² + CQ² … (3)

ΔOAP, we have

OA² = AP² + OP² … (4)

Adding (1) and (2)

OB² + OD² = BP² + OP² + OQ² + DQ² (As BP = CQ and DQ = AP)

= CQ² + OP² + OQ² + AP²

= CQ² + OQ² + OP² + AP²

= OC² + OA² [From (3) and (4)]