Theorem | the angle subtended by an arc of a circle | at the centre is double the angle subtended by it at | any point on the remaining part of the circle

Theorem :

A circle  with centre O in which AB subtends  ∠AOB at centre and angle ∠ACB at any point  on the remaining part of the circle.
R.T.P. - ∠AOB =  2 x ∠ACB

Solution :

Construction: Join CO and produce CO to point D .

1. In ∆AOC, OC = OA || Radii of same circle .
2. ∠OCA= ∠OAC || Base angles of isoscles ∆ in which OC = OA
3. Exterior ∠ AOD = ∠OCA+ ∠OAC || Exterior ∠ of a ∆ = sum of two opposite interior angles.
or ∠AOD = ∠OCA+ ∠OCA
or ∠AOD = 2 . ∠ OCA - - - - Eqn. 1
4. Similary ot can be proved that ∠BOD = 2. ∠OCB - - - - - - Eqn. 2
5. Adding Eqn 1 and Eqn. 2 we get
∠AOD + ∠BOD = 2 . ∠OCA + 2. ∠OCB
∠AOB = 2 (∠OCA + ∠OCB)
∴ ∠AOB = 2 ∠ACB Proved
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