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Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
To do:
We have to prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution:
Let a quadrilateral $ABCD$ is a quadrilateral circumscribing a circle with centre $O$.
$AB$ touches the circle at $P$.
Similarly,
$BC, CD$ and $DA$ touch the circle at $Q, R$ and $S$.
Join $OA, OB, OC, OD$ and $OP, OQ, OR, OS$.
$OA$ bisects $\angle \mathrm{POS}$
$\angle 1=\angle 2$
$\angle 3=\angle 4$
$\angle 5=\angle 6$
$\angle 7=\angle 8$
Therefore,
$\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8=360^{\circ}$
$\angle 1+\angle 1+\angle 4+\angle 4+\angle 5+\angle 5+\angle 8+\angle 8=360^{\circ}$
$2[\angle 1+\angle 4+\angle 5+\angle 8]=360^{\circ}$
$(\angle 1+\angle 8+\angle 4+\angle 5)=180^{\circ}$
$\angle \mathrm{AOD}+\angle \mathrm{BOC}=180^{\circ}$
Similarly,
$\angle \mathrm{AOB}+\angle \mathrm{COD}=180^{\circ}$
Hence, the opposite sides of quadrilateral circumscribing a circle subtend supplementary angles at the centre of a circle.
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