A circle touches the sides of a quadrilateral ABCD at P, Q, R and S respectively. Show that the angles subtended at the centre by a pair of opposite sides are supplementary.

Ans:    To prove :- ∠AOB + ∠DOC = 180^o

∠BOC + ∠AOD = 180^o

Proof : - Since the two tangents drawn from an external point to a circle subtend equal angles at centre.

∴∠1 = ∠2, ∠3 = ∠4, ∠5 = ∠6, ∠7 = ∠8

but ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360o

2(∠2 + ∠3 + ∠6 + ∠7) = 360^o

∠2 + ∠3 + ∠6 + ∠7 = 360^o

∴∠AOB + ∠DOC = 180^o

Similarly

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360^o

2(∠1 + ∠8 + ∠4 + ∠5) = 360^o

∠1 + ∠8 + ∠5 = 180^o

∴∠BOC + ∠AOD = 180^o

Hence proved