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

In the below figure, P, Q, R, S are points of contact

We know that,

The tangents drawn from an external point to a circle are equal.

AS = AP

We know that,

Tangents drawn from a point outside the circle, subtend equal angles at the centre.

∠1 = ∠2, ∠3 = ∠4, ∠5 = ∠6, ∠7 = ∠8.

Since complete angle is 360° at the centre,

We have,

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

⇒ ∠1 + ∠1 + ∠4 + ∠4 + ∠5 + ∠5 + ∠8 + ∠8 = 360° or ∠2 + ∠2 + ∠3 + ∠3 + ∠6 + ∠6 + ∠7 + ∠7 = 360°

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

⇒ ∠1 + ∠8 + ∠4 + ∠5 = 180° or ∠2 + ∠3 + ∠6 + ∠7 = 180° …………(1)

From above figure,

∠1 + ∠8 = ∠AOD, ∠4 + ∠5 = ∠BOC and ∠2 + ∠3 = ∠AOB, ∠6 + ∠7 = ∠COD

Substituting above values in equation (1), we get :

⇒ ∠AOD + ∠BOC = 180° or ∠AOB + ∠COD = 180°

∠AOD and ∠BOC are angles subtended by opposite sides of quadrilateral circumscribing a circle and the sum of the two is 180°.

Hence, proved that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.