Question

.Prove that angle subtended by an Arc at the centre is double the angle subtended by it in the remaining part of the circle. ​

Answer 1

$\huge\bold\red{HOLA}$

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.

Answer 2

Answer:

Explanation:

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∠POQ=2∠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∠BOQ=∠OAQ+∠AQO       .....(1)

Also, in \triangle△ OAQ,

OA=OQOA=OQ                  [Radii of a circle]

Therefore,

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

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

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

Adding 2 & 3, we get,

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

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

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

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