Question

Prove the Theorem 10.8 Circles Class 9th NCERT?

Answer 1

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Answer 2

(10.8)--->> "The angle subtended by an arc at the centre 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 centre 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 :-

(i) arc PQ is minor .
(ii) arc PQ is a semicircle .
(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.