Triangle Theorem

Triangle Theorem:
The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

Proof : Given an arc PQ of a circle subtending angles POQ at the center O and PAQ at a point A on the remaining part of the circle. We need to prove that ∠ POQ = 2 ∠ PAQ.

Consider the three different cases as given in Fig. (i), arc PQ is minor; in (ii),arc PQ is a semicircle and in (iii), arc PQ is major. Let us begin by joining AO and extending it to a point B.
In all the cases, ∠ BOQ = ∠ OAQ + ∠ AQO because an exterior angle of a triangle is equal to the sum of the two interior opposite angles.

Also in Δ OAQ, OA = OQ (Radii of a circle) Therefore, ∠ OAQ = ∠ OQA (Theorem 7.5)
This gives ∠ BOQ = 2 ∠ OAQ (1) Similarly, ∠ BOP = 2 ∠ OAP.... (2) From (1) and (2), ∠ BOP + ∠ BOQ = 2(∠ OAP + ∠ OAQ) This is the same as ∠ POQ = 2 ∠ PAQ ...(3) For the case (iii), where PQ is the major arc, (3) is replaced by reflex angle POQ = 2 ∠ PAQ

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