(Q4)'O' is any point inside a rectangle ABCD.Prove that OB² + OD² = OA² + OC²

Solution :

Through 'O' draw PQ ∥ BC so that P lies on AB and Q lies on DC.

Now PQ ∥ BC

∴ PQ ⊥ & PQ ⊥ (∵ ∠B = ∠C = 90^o)

So, ∠BPQ = 90^o and ∠CQP = 90^o

∴ BPQC and APQD are both rectangles

Now from ∆OPB, OB² = ( )² + OP² .....(1)

Similarly from ∆OQD, we have OD² = ( )² + DQ² .....(2)

From ∆OQC, we have OC² = OQ² + ( )².....(3)

and from ∆OAP, OA² = AP² + ( )²

Adding (1) & (2)

OB² + OD² = ( )² + OP² + ( )² + DQ²

  = CQ² + OP² + OQ² + AP²   (∵ BP = CQ and DQ = AP)

  = CQ² + OQ² + OP² + AP²

  = OC² + OA² (from (3) & (4))