Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
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| ∠1 = ∠2, ∠3 = ∠4, ∠5 = ∠6, ∠7 = ∠8 |
| Since the 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° |