Question

15.Prove that angle subtended by an arc at the centre of a circle is twice the angle subtended by it at any point on the remaining part of the circle​

Answer 1

$\huge{\fcolorbox{grey}{yellow}{Answer:↝}}$

☞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. 

☞Construction:

Join AO and extend it to a point B.

☞Proof :

There arises three cases: 

(A) are PQ is minor  (B) arc PQ s a semi - circle  (C) arc PQ is major. 

In all the cases, 

∠BOQ = ∠OAQ + ∠AQO ....(i)

• [∵An exterior angle of triangle is equal to the sum of the two interior opposite angles] 

In OAQ,  OA = OQ [Radii of a circle] 

∴ ∠OAQ = ∠OQA ...(ii)

• [Angles opposite equal of a triangle are equal] 

Now,

From (i) and (ii); we get,

  ∠BOQ = 2∠OAQ ....(iii) 

Similarly, 

∠BOP = 2∠OAP ....(iv) 

Adding (iii) and (iv), we get ;

∠BOP + ∠BOQ = 2(∠OAP + ∠OAQ) 

⇒ ∠POQ = 2∠PA. ....(v) 

✩NOTE ↝: For the case (C), where PQ is the major arc, (v) is replaced by reflex angles. 

Thus, ∠POQ = 2∠PAQ

$\large{\fcolorbox{cyan}{pink}{Hence \ Proved.}}$

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