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Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
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Correct figure. $\Delta OAP \cong \Delta OAS \implies \angle 1 = \angle 2$. Similarly, $\angle 3 = \angle 4, \angle 5 = \angle 6, \angle 7 = \angle 8$. Also, $\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^\circ \implies 2(\angle 1 + \angle 4 + \angle 5 + \angle 8) = 360^\circ \implies \angle AOB + \angle COD = 180^\circ$. Similarly, $\angle BOC + \angle AOD = 180^\circ$.