A line PQ is drawn through point O parallel to AB. From triangle BOQ, we get: OB² = BQ² + OQ² — (i) From triangle POD, we get: OD² = PD² + OP² — (ii) Since ABCD is a rectangle and PQ is parallel to AB, Therefore, AP = BQ and PD = QC Adding equations (i) and (ii): OB² + OD² = BQ² + OQ² + PD² + OP² ⇒ OB² + OD² = AP² + OQ² + QC² + OP² ⇒ OB² + OD² = AP² + OP² + OQ² + QC²