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.