Let ABCD be a quadrilateral circumscribing a circle with centre O. Now join AO, BO, CO, DO. From the figure, ∠DAO = ∠BAO [Since, AB and AD are tangents] Let ∠DAO = ∠BAO = 1 Also ∠ABO = ∠CBO [Since, BA and BC are tangents] Let ∠ABO = ∠CBO = 2 Similarly we take the same way for vertices C and D Sum of the angles at the centre is 360° Recall that sum of the angles in quadrilateral, ABCD = 360° = 2(1+2+3+4) = 360° = 1+2+3+4 = 180° In ΔAOB, ∠BOA = 180° − (1+2) In ΔCOD, ∠COD = 180° − (3+4) ∠BOA + ∠COD = 360° − (1+2+3+4) = 360° − 180° = 180° Since AB and CD subtend supplementary angles at O. Thus, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.