Question

In the given figure, Point O is the centre of the circle. Show that angleAOC=AngleAFC+AngleAEC

Answer 1

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