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Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
To do:
We have to prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
Solution:
Let $PA$ and $PB$ be two tangents, $A$ and $B$ the points of contact of the tangents.
$\mathrm{OA} \perp \mathrm{AP}$
$\mathrm{OB} \perp \mathrm{BP}$
$\angle \mathrm{OAP}=\angle \mathrm{OBP}=90^{\circ}$
In quadrilateral $\mathrm{OAPB}$
$\angle \mathrm{OAP}+\angle \mathrm{OBP}+\angle \mathrm{APB}+\angle \mathrm{AOB}=360^{\circ}$
$90^{\circ}+90^{\circ}+\angle \mathrm{APB}+\angle \mathrm{AOB}=360^{\circ}$
$\angle \mathrm{APB}+\angle \mathrm{AOB}=360^{\circ}-180^{\circ}$
$=180^{\circ}$
Therefore,
$\angle \mathrm{APB}$ and $\angle \mathrm{AOB}$ are supplementary angles.
Hence proved.
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