Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Given : A circle with centre O touches the sides AB, BC, CD and DA of a quadrilateral ABCD at the points P, Q, R and S respectively.
To prove : ∠AOB + ∠COD = 180°
∠AOD + ∠BOC = 180°
Const. : Join OP, OQ, OR and OS.
Proof : Since, the two tangents drawn from an external point to a circle subtend equal angles at the centre.
∴ ∠1 = ∠2, ∠3 = ∠4, ∠5 = ∠6, ∠7 = ∠8
Since sum of all the angles subtended at a point is 360°.
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8
= 360°
⇒2 ∠2 + 2 ∠3 + 2 ∠6 + 2 ∠7) = 360°
⇒ 2 (∠2 + ∠3 + ∠6 + ∠7) = 360°
⇒ ∠2 + ∠3 + ∠6 + ∠7) = 180°
⇒ (∠6 + ∠7) + (∠2 + ∠3) = 180°
⇒ ∠AOB + ∠COD = 180°
Similarly, we can prove ∠AOD + ∠BOC = 180°
