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)]