Step 1: Convert the given information into a figure.

Step 2: Draw a line parallel to BC through the "O".

            EXPLANATION: Through ‘O’ draw PQ || BC so that P lies on AB and Q

              lies on DC.

          Now PQ || BC ∴ PQ ⊥ AB & PQ ⊥ DC (∵ ∠B = ∠C = 90°)  

Step 3: Observe the figure and note down the right angle triangles

            EXAMPLE: \triangle   OPB

                                 \triangle   OQD

                                 \triangle   OQC

                                 \triangle   OAP

Step 4: Apply the Pythagoras theorem to the right angle triangles.

            EXAMPLE:

            From ∆OPB. OB^2 = BP^2 + OP^2  ...............(1)

            Similarly from ∆OQD, OD^2 = OQ^2 + DQ^2 ...........(2)

              ∆OQC,   OC^2 = OQ^2 + CQ^2 ........................(3)

                  ∆OAP, OA^2 = AP^2 + OP^2 ......................(4)

Step 5: Add equation (1) and (2)

       OB^2 + OD^2 = BP^2 + OP^2 + OQ^2 + DQ^2

                                = CQ^2 + OP^2 + OQ^2 + AP^2   [∵ BP = CQ and DQ = AP (see the figure)]

                               = CQ^2 + OQ^2 + OP^2 + AP^2  

                               = OC^2 + OA^2  [from (3) & (4)]

        Hence

        = OB^2 + OD^2 = OC^2 + OA^2