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

**Solution:**

We know that tangents drawn from a point outside of the circle, subtend equal angles at the centre.

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

AS = AP --- (The tangents drawn from an external point to a circle are equal.)

∠SOA = ∠POA = ∠1 = ∠2

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

Similarly,

∠3 = ∠4, ∠5 = ∠6, ∠7 = ∠8

Since complete angle is,

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

from above fig., ∠AOD + ∠BOC = 180 (or) ∠AOB + ∠COD = 180°

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

Hence Proved.

**Video Solution:**

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

### NCERT Solutions Class 10 Maths - Chapter 10 Exercise 10.2 Question 13:

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

It has been proved that the opposite sides of a quadrilateral ABCD circumscribing a circle with centre O subtend supplementary angles at the centre of the circle.