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

Let 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.

Then we have to prove that ∠AOB + ∠COD = 180° and ∠AOD + ∠BOC = 180°

Now, join OP, OQ, OR and OS.

Since the two tangents drawn from an external point to a circle subtend equal angles at the centre,

⇒ ∠1 = ∠2, ∠3 = ∠4, ∠5 = ∠6 and ∠7 = ∠8 … (1)

Now,

⇒ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360° … (2)

[Since sum of all the angles subtended at a point is 360°]

From (1) and (2),

⇒ 2 (∠2 + ∠3 + ∠6 + ∠7) = 360°

⇒ (∠2 + ∠3) + (∠6 + ∠7) = 180°

⇒ ∠AOB + ∠COD = 180°

Similarly,

⇒ 2 (∠1 + ∠8 + ∠4 + ∠5) = 360°

⇒ (∠1 + ∠8) + (∠4 + ∠5) = 180°

⇒ ∠AOD + ∠BOC = 180°

Hence proved.