  Question

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

Solution

Let ABCD be a quadrilateral circumscribing a circle with centre O. Now join $$AO$$, $$BO$$, $$CO$$, $$DO$$. From the figure, $$\angle DAO = \angle BAO$$ [Since, AB and AD are tangents] Let $$\angle DAO = \angle BAO = 1$$ Also $$\angle ABO = \angle CBO$$ [Since, BA and BC are tangents] Let $$\angle ABO = \angle CBO = 2$$ Similarly we take the same way for vertices C and D Sum of the angles at the centre is $$360^o$$ Recall that sum of the angles in quadrilateral, ABCD = $$360^o$$  $$= 2 (1 + 2 + 3 + 4) = 360^o$$ $$= 1 + 2 + 3 + 4 = 180^o$$ In $$\Delta AOB, \angle BOA= 180 - (1 +2)$$ In $$\Delta COD, \angle COD = 180 - (3 +4)$$ $$\angle BOA + \angle COD = 360 - (1 + 2 + 3+ 4)$$ $$= 360^o – 180^o$$ $$= 180^o$$ 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. Maths

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