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
$\angle O A P=90^{\circ}....... (1)$ [Angle between tangent and radius through the point of contact is $90^{\circ}$ ]
$\angle O B P=90^{\circ} .......(2)$ [Angle between tangent and radius through the point of contact is $90^{\circ}$ ]
$\therefore OAPB$ is quadrilateral
$ \therefore \angle A P B+\angle A O B+\angle O A P+\angle O B P=360^{\circ}[\text { Angle sum property of a quadrilateral }] $
$ \Rightarrow \angle A P B+\angle A O B+90^{\circ}+90^{\circ}=360^{\circ}[\text { From (1) and (2) }] $
$ \Rightarrow \angle A P B+\angle A O B=180^{\circ} $
$\Rightarrow \angle A P B \text { and } \angle A O B \text { are supplementary }$
$\angle O B P=90^{\circ} .......(2)$ [Angle between tangent and radius through the point of contact is $90^{\circ}$ ]
$\therefore OAPB$ is quadrilateral
$ \therefore \angle A P B+\angle A O B+\angle O A P+\angle O B P=360^{\circ}[\text { Angle sum property of a quadrilateral }] $
$ \Rightarrow \angle A P B+\angle A O B+90^{\circ}+90^{\circ}=360^{\circ}[\text { From (1) and (2) }] $
$ \Rightarrow \angle A P B+\angle A O B=180^{\circ} $
$\Rightarrow \angle A P B \text { and } \angle A O B \text { are supplementary }$
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