An arc of a circle has an angle of subtendence at its center that is twice as wide as the angle it has anywhere else on the circumference.
Let's say that on the circumference, an arc PQ of a circle intersects POQ at the center O and PAQ at a point A.
We must demonstrate that the central angle of an arc, POQ=2 PAQ, is true. To do this, the arc AB will be taken into consideration in three distinct contexts: semi-circle AB, major arc AB, and minor arc AB.
What's more, in OA=OQ [Radii of a circle]
∠BOQ=∠OAQ+∠AQO .....(1)
Additionally, in △ OAQ,
OA=OQ [Radii of a circle]
In this way,
∠OAQ=∠OQA [Angles inverse to approach sides are equal]
∠BOQ=2∠OAQ .......(2)
Likewise, BOP=2∠OAP ........(3)
Adding 2 and 3, we get,
∠BOP+∠BOQ=2(∠OAP+∠OAQ)
∠POQ=2∠PAQ ....(4)
Condition 4 is on the grounds that the reflex point, ∠POQ=2∠PAQ, for situation 3, where PQ is the primary curve.
POQ=2 PAQ is the conclusion.
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