Question

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.

Answer 1

Draw a circle with center $O$ and take a external point P. PA and PB are the tangents.

As radius of the circle is perpendicular to the tangent.

$OA\perp PA$

Similarly $OB\perp PB$

$\angle OBP={90}^o$

$\angle OAP={90}^o$

In Quadrilateral $OAPB$, sum of all interior angles $={360}^o$

$\Rightarrow \angle OAP+\angle OBP+\angle BOA+\angle APB={360}^o$

$\Rightarrow {90}^o+{90}^o+\angle BOA+\angle APB={360}^o$

$\angle BOA+\angle APB={180}^o$

It proves the angle between the two tangents drawn from an external point to a circle supplementary to the angle subtented by the line segment