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.
