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

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Step 1: Quadrilateral ABCD circumscribing a circle with centre O, tangent points P, Q, R, S and angles 1-8 marked
Step 2: $	riangle OAP \cong 	riangle OAS$ (RHS congruence, as $OA$ is common, $OP = OS$ radii, $\angle OPA = \angle OSA = 90°$). So $\angle 1 = \angle 2$.
Step 3: Similarly, $\angle 3 = \angle 4$, $\angle 5 = \angle 6$, $\angle 7 = \angle 8$. Also, $\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360°$
Step 4: $\Rightarrow 2(\angle 1 + \angle 4 + \angle 5 + \angle 8) = 360° \Rightarrow \angle AOB + \angle COD = 180°$. Similarly, $\angle BOC + \angle AOD = 180°$.

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