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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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