the angle subtended by an arc at the center of a circle is double the angel it subtends at any point on the remaining part of the circle
Answers
Answer:
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
Answer:
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
Step-by-step explanation:
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