Q30 of 47 Page 1

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


To prove: (i) ∠AOB + ∠COD = 180°


(ii) ∠AOD + ∠BOC = 180°


Proof:


In ΔAOS and ΔAOP,


AS = AP [Tangents from an external point are equal]


OS = OP [Radius of the same circle]


AO = AO [Common]


ΔAOS ≅ ΔAOP [S.S.S]


∠1 = ∠2 [C.P.C.T.C]


Similarly, ∠3 = ∠4, ∠5 = ∠6, ∠7 = ∠8


∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360° [Angle at a point]


⇒ ∠2 + ∠2 + ∠3 + ∠3 + ∠6 + ∠6 + ∠7 + ∠7 = 360°


⇒ 2 (∠2 + ∠3 + ∠6 + ∠7) = 360°


⸫ ∠AOB + ∠COD = 180°


Similarly,


⸫ ∠AOD + ∠BOC = 180°


Hence Proved.


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