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Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

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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.

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