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

Let a circle with centre O touch the sides AB, BC, CD and DA of a quadrilateral ABCD at the points P, Q, R and S respectively.

We have to prove that-

$$∠ \ AOB \ + \ ∠ \ COD = \ 180°$$

and, $$∠ \ AOD \ + \ ∠ \ BOC \ = \ 180°$$
Join OP, OQ, OR and OS.

$$\therefore$$ The two tangents drawn from an external point to a circle subtend equal angles at the centre. $$\therefore$$ $$∠ \ 1 \ = \ ∠ \ 2 \ , \ ∠ \ 3 \ = \ ∠ \ 4 \ , \ ∠ \ 5 \ = \ ∠ \ 6 \ and \ ∠ \ 7 \ = \ ∠ \ 8$$

Now, $$∠ \ 1 \ + \ ∠ \ 2 \ + \ ∠ \ 3 \ + \ ∠ \ 4 \ + \ ∠ \ 5 \ + \ ∠ \ 6 \ + \ ∠ \ 7 \ + \ ∠ \ 8 \ = \ 360°$$ [$$\because$$ Sum of all the ∠ s around a point is 360°]

$$\Rightarrow \ 2(∠ \ 2 \ + \ ∠ \ 3 \ + \ ∠ \ 6 \ + \ ∠ \ 7) \ = \ 360°$$

and, $$2(∠ \ 1 \ + \ ∠ \ 8 \ + \ ∠ \ 4 \ + \ ∠ \ 5) \ = \ 360°$$

$$\Rightarrow \ (∠ \ 2 \ + \ ∠ \ 3 \ + \ ∠ \ 6 \ + \ ∠ \ 7) \ = \ 180°$$

and $$(∠ \ 1 \ + \ ∠ \ 8) \ + \ (∠ \ 4 \ + \ ∠ \ 5) \ = \ 180°$$

Now since $$∠ 2 + ∠ 3 = ∠ AOB$$ ,
$$\ ∠ 6 + ∠ 7 = ∠ COD,$$
$$\ ∠ 1 + ∠ 8 = ∠ AOD \$$ and $$\ ∠ 4 + ∠ 5 = ∠ BOC,$$ we can say that

$$∠ \ AOB \ + \ ∠ \ COD \ = \ 180°$$
and, $$∠ \ AOD \ + \ ∠ \ BOC \ = \ 180°$$

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