Question

In the given figure point O is the centre of circle . Show that angle AOC = angle AFC + angle AEC

Answer 1

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

To Prove:

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

   

Double the above equation both sides



Angle subtended on circle is half angle subtended at centre.  

 





 (∴ Angles subtended on same arc are equal)



In ΔFDC using exterior angle theorem.

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Answer 2

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

Hence proved.