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

Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches the circle at point  P, Q, R, S.

Let us join the vertices of the quadrilateral ABCD to the center of the circle.

$\mathrm{Consider} ∆\mathrm{OAP} \mathrm{and} ∆\mathrm{OAS},$ 

AP = AS  (Tangents from the same point)

OP = OS  (Radii of the same circle)

OA = OA  ( common side )

$$\begin{gathered} ∆\mathrm{OAP} \cong ∆\mathrm{OAS} ( \mathrm{SSS} \mathrm{congruence} \mathrm{criterion} ) \\ \mathrm{Therefore}, \mathrm{A}\leftrightarrow \mathrm{A}, \mathrm{P}\leftrightarrow \mathrm{S}, \mathrm{O}\leftrightarrow \mathrm{O} \\ \mathrm{And} \mathrm{thus}, \angle \mathrm{POA} = \angle \mathrm{AOS} \\ \angle 1 = \angle 8 \\ \mathrm{Similarly}, \\ \angle 2 = \angle 3 \\ \angle 4 = \angle 5 \\ \angle 6 = \angle 7 \\ \angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 =360° \\ \left(\angle 1 + \angle 8 \right) + \left(\angle 2 + \angle 3 \right) + \left(\angle 4 + \angle 5 \right) + \left(\angle 6 + \angle 7 \right) = 360° \\ 2\angle 1 + 2\angle 2 + 2\angle 5 + 2\angle 6 = 360° \\ 2\left(\angle 1 + \angle 2 \right) + 2\left(\angle 5 + \angle 6\right) = 360° \\ \left(\angle 1 + \angle 2\right) + \left(\angle 5 + \angle 6\right) = 180° \\ \angle \mathrm{AOB} + \angle \mathrm{COD} = 180° \\ \mathrm{Similarly}, \mathrm{we} \mathrm{can} \mathrm{prove} \mathrm{that} \angle \mathrm{BOC} + \angle \mathrm{DOA} = 180° \end{gathered}$$

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