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

(N/A) Let $ABCD$ be a quadrilateral circumscribing a circle centered at $O$ such that it touches the circle at points $P, Q, R, S$. Let us join the vertices of the quadrilateral $ABCD$ to the center of the circle.
Consider $\triangle OAP$ and $\triangle OAS$:
$AP = AS$ (Tangents from the same point)
$OP = OS$ (Radii of the same circle)
$OA = OA$ (Common side)
Therefore,$\triangle OAP \cong \triangle OAS$ ($SSS$ congruence criterion).
Thus,$\angle POA = \angle AOS$,which implies $\angle 1 = \angle 8$.
Similarly,we can show that:
$\angle 2 = \angle 3$
$\angle 4 = \angle 5$
$\angle 6 = \angle 7$
Since the sum of angles around the center is $360^{\circ}$:
$\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^{\circ}$
Substituting the equal angles:
$(\angle 1 + \angle 8) + (\angle 2 + \angle 3) + (\angle 4 + \angle 5) + (\angle 6 + \angle 7) = 360^{\circ}$
$2\angle 1 + 2\angle 2 + 2\angle 5 + 2\angle 6 = 360^{\circ}$
$2(\angle 1 + \angle 2) + 2(\angle 5 + \angle 6) = 360^{\circ}$
$(\angle 1 + \angle 2) + (\angle 5 + \angle 6) = 180^{\circ}$
Thus,$\angle AOB + \angle COD = 180^{\circ}$.
Similarly,we can prove that $\angle BOC + \angle DOA = 180^{\circ}$.
Hence,opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.