Question

O os the centre of circle ⭕ then prove that aoc = afc+ aec

Answer 1

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

Answer 2

Step-by-step explanation:

Given: In the given figure O is the centre of the circle.

To Prove: \angle AOC=\angle AFC+\angle AEC∠AOC=∠AFC+∠AEC

Proof: In ΔBEC using exterior angle theorem.

Exterior angle theorem property: Sum of two interior angle of triangle is equal to opposite exterior angle. So, we get

\angle ABC=\angle AEC+\angle BCD∠ABC=∠AEC+∠BCD

Double the above equation both sides

2\angle ABC=2\angle AEC+2\angle BCD2∠ABC=2∠AEC+2∠BCD

Angle subtended on circle is half angle subtended at centre.

2\angle ABC=\angle AOC2∠ABC=∠AOC

\angle AOC=\angle AEC+\angle BCD+\angle AEC+\angle BCD∠AOC=∠AEC+∠BCD+∠AEC+∠BCD

\angle AOC=\angle AEC+\angle BCD+\angle ABC∠AOC=∠AEC+∠BCD+∠ABC

\text{But } \angle ABC=\angle ADCBut ∠ABC=∠ADC (∴ Angles subtended on same arc are equal)

\angle AOC=\angle AEC+\angle BCD+\angle ADC∠AOC=∠AEC+∠BCD+∠ADC

In ΔFDC using exterior angle theorem.

\angle AFC=\angle BCD+\angle ADC∠AFC=∠BCD+∠ADC

\therefore \angle AOC=\angle AEC+\angle AFC∴∠AOC=∠AEC+∠AFC

\text{Hence proved, } \angle AOC=\angle AFC+\angle AECHence proved, ∠AOC=∠AFC+∠AEC