Question:

Let $ABCD$ be a quadrilateral which circumscribes a circle with centre $O$.\
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Denote the points of contact of the circle with the sides $AB, BC, CD,$ and $DA$ by $P, Q, R,$ and $S$ respectively.\
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1. Angle between two tangents\
At vertex $A$, the sides $AB$ and $AD$ are tangents to the circle. By the theorem on two tangents,\
$$\angle DAB = 180^{\circ} - \angle AOB,$$\
where $\angle AOB$ is the central angle subtended by the chord $PS$ (the chord joining the points of contact of the two tangents). Since $PS$ lies on side $AB$, $\angle AOB$ is the angle subtended at the centre by side $AB$.\
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2. Similarly, at vertex $C$, the sides $BC$ and $CD$ are tangents, and we have\
$$\angle BCD = 180^{\circ} - \angle COD,$$\
where $\angle COD$ is the central angle subtended by side $CD$.\
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3. Opposite interior angles of a tangential quadrilateral\
For any quadrilateral that circumscribes a circle, the sum of a pair of opposite interior angles is $180^{\circ}$ (this follows from the equality of the two tangents drawn from each vertex). Hence\
$$\angle DAB + \angle BCD = 180^{\circ}.$$\
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4. Substituting the expressions from steps 1 and 2\
\[ (180^{\circ} - \angle AOB) + (180^{\circ} - \angle COD) = 180^{\circ} \]\
Simplifying,\
$$\angle AOB + \angle COD = 180^{\circ}.$$\
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Thus the angles subtended at the centre by the opposite sides $AB$ and $CD$ are supplementary. By the same reasoning, the opposite sides $BC$ and $DA$ also subtend supplementary angles at the centre.