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