In the figure (i) given below, O is the centre of the circle. Prove that ∠AOC = 2(∠ACB + ∠BAC).

We know that sum of angles in a triangle = 180°.

In △ABC,

∠ACB + ∠BAC + ∠ABC = 180°
∠ABC = 180° - (∠ACB + ∠BAC) …..(i)

In the circle arc AC subtends Reflex ∠AOC at centre and ∠ABC at remaining part of the circle.

∴ Reflex ∠AOC = 2∠ABC (∵ angle subtended on centre is double the angle subtended at remaining part of the circle.)

From (i)

Reflex ∠AOC = 2(180° - (∠ACB + ∠BAC))

We know

Reflex ∠AOC = 360° - ∠AOC.

or,

360° - ∠AOC = 2(180° - (∠ACB + ∠BAC))
360° -∠AOC = 360° - 2(∠ACB + ∠BAC)
∠AOC = 360° - (360° - 2(∠ACB + ∠BAC))
∠AOC = 2(∠ACB + ∠BAC).

Hence, proved that ∠AOC = 2(∠ACB + ∠BAC).