Question

Prove that the angle subtended by an arc at the center is twice the angle subtended by it at

any other point on the remaining part of the circle.
​

Answer 1

Answer:

$Reflex \: \: angle, \: ∠POQ=2∠PAQ$

Step-by-step 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.

$To \: \: prove : ∠POQ=2∠PAQ$

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 :

∠BOQ=∠OAQ+∠AQO .....(1)

Also, in △ OAQ,

OA=OQ [Radii of a circle]

Therefore,

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

∠BOQ=2∠OAQ .......(2)

Similarly, BOP=2∠OAP ........(3)

Adding 2 & 3, we get,

∠BOP+∠BOQ=2(∠OAP+∠OAQ)

∠POQ=2∠PAQ .......(4)

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

$Reflex \: \: angle, \: ∠POQ=2∠PAQ$

☣Hope it helps....❤❤

Answer 2

Answer:

Reflexangle,∠POQ=2∠PAQ

Step-by-step explanation:

Given :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 : ∠POQ=2∠PAQToprove:∠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 :

∠BOQ=∠OAQ+∠AQO .....(1)

Also, in △ OAQ,

OA=OQ [Radii of a circle]

Therefore,

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

∠BOQ=2∠OAQ .......(2)

Similarly, BOP=2∠OAP ........(3)

Adding 2 & 3, we get,

∠BOP+∠BOQ=2(∠OAP+∠OAQ)

∠POQ=2∠PAQ .......(4)

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

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

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