Answer:

Given:- In the given fig., O is the centre of circle.

To prove:- ∠AOC=∠AFC+∠AEC

Proof:- In △BEC, Exterior angle at B,

∴∠ABC=∠AEC+∠BCD.....(1)(Exterior angle theorem)

2∠ABC=2∠AEC+∠BCD

∵2∠ABC=∠AOC(Angle subtended on circle is double the angle subtended at centre on same arc)

∴∠AOC=∠AEC+∠BCD+∠AEC+∠BCD

∠AOC=∠AEC+∠BCD+∠ABC(From (1))

∵∠ABC=∠ADC(∵Angle subtends on same arc are equal)

∴∠AOC=∠AEC+∠BCD+∠ADC.....(2)

Now, in △FDC

Exterior angle at F.

∴∠AFC=∠BCD+∠ADC.....(3)

Now, from equation (2)&(3), we have

∠AOC=∠AEC+∠AFC