Question

in the given figure point is the centre of the circle show that angle AOC is equal to angle afc + angle AeC

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