Question

Prove that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle .

Answer 1

Given :

An arc PQ of a circle subtending angles POQ at the centre O and PAQ at a point A on the remaining part of the circle.

To prove : $\angle POQ=2\angle PAQ$

To prove this theorem we consider the arc AB in three different situations, minor arc AB, major arc AB and semi-circle AB.

Construction :

Join the line AO extended to B.

Proof :

$\angle BOQ=\angle OAQ+\angle AQO$ .....(1)

Also, in $△$ OAQ,

$OA=OQ$ [Radii of a circle]

Therefore,

$\angle OAQ=\angle OQA$ [Angles opposite to equal sides are equal]

$\angle BOQ=2\angle OAQ$ .......(2)

Similarly, $BOP=2\angle OAP$ ........(3)

Adding 2 & 3, we get,

$\angle BOP+\angle BOQ=2(\angle OAP+\angle OAQ)$

$\angle POQ=2\angle PAQ$ .......(4)

For the case 3, where PQ is the major arc, equation 4 is replaced by

Reflex angle, $\angle POQ=2\angle PAQ$