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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 (1/2)
$\triangle$ OAP $\cong \triangle$ OAS
$\therefore \angle 1 = \angle 2$ (1)
Similarly, $\angle 3 = \angle 4$, $\angle 5 = \angle 6$, $\angle 7 = \angle 8$ (1/2)
Also, $\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^\circ$ (1/2)
$\Rightarrow 2 (\angle 1 + \angle 4 + \angle 5 + \angle 8) = 360^\circ$
$\Rightarrow \angle$ AOB + $\angle$ COD = $180^\circ$
Similarly, $\angle$ BOC + $\angle$ AOD = $180^\circ$ (1/2)_a_1.png)
$\triangle$ OAP $\cong \triangle$ OAS
$\therefore \angle 1 = \angle 2$ (1)
Similarly, $\angle 3 = \angle 4$, $\angle 5 = \angle 6$, $\angle 7 = \angle 8$ (1/2)
Also, $\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^\circ$ (1/2)
$\Rightarrow 2 (\angle 1 + \angle 4 + \angle 5 + \angle 8) = 360^\circ$
$\Rightarrow \angle$ AOB + $\angle$ COD = $180^\circ$
Similarly, $\angle$ BOC + $\angle$ AOD = $180^\circ$ (1/2)