Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Given, a quadrilateral ABCD circumscribes a circle with centre O.
To prove: $\angle A O B+\angle C O D=180^{\circ}$
and $\angle A O D+\angle B O C=180^{\circ}$
Join $O P, O Q, O R$ and $O S$.
We know that the tangents drawn from an external point of a circle
subtend equal angles at the centre.
$\therefore \angle 1=\angle 7, \angle 2=\angle 3, \angle 4=\angle 5$ and $\angle 6=\angle 8$
and $\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8=360^{\circ} \quad$ [angles at a point]
$\Rightarrow(\angle 1+\angle 7)+(\angle 3+\angle 2)+(\angle 4+\angle 5)+(\angle 6+\angle 8)=360^{0}$
$2 \angle 1+2 \angle 2+2 \angle 6+2 \angle 5=360^{\circ}$
$\Rightarrow \angle 1+\angle 2+\angle 5+\angle 6=180^{\circ}$
$\Rightarrow \angle A O B+\angle C O D=180^{\circ}$ and $\angle A O D+\angle B O C=180^{\circ}$
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