Theorem :

A circle with centre O in which AB subtends 𢈊OB at centre and angle � at any point on the remaining part of the circle.

R.T.P. – 𢈊OB = 2 x �

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 𢈊OD = ∠OCA+ ∠OCA

or 𢈊OD = 2 . ∠ OCA – – – – Eqn. 1

4. Similary ot can be proved that 𢈋OD = 2. ∠OCB – – – – – – Eqn. 2

5. Adding Eqn 1 and Eqn. 2 we get

𢈊OD + 𢈋OD = 2 . ∠OCA + 2. ∠OCB

𢈊OB = 2 (∠OCA + ∠OCB)

∴ 𢈊OB = 2 ∠ACB Proved

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