Question

In the given figure o is the centre of the circle Show that < AOC = <AFC +< AEC

Answer 1

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

To Prove: ∠AOC=∠AFC+∠AEC

Proof: In ΔBEC , Exterior angle at B

$\angle ABC=\angle AEC+\angle BCD$    (∴ Exterior angle theorem property)

Multiply both sides by 2

$2\angle ABC=2\angle AEC+2\angle BCD$

$2\angle ABC=\angle AOC$  (∴ angle subtended on circle is double angle subtended at centre on same arc)

$\angle AOC=\angle AEC+\angle BCD+\angle AEC+\angle BCD$

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

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

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

In ΔFDC , Exterior angle at F

$\angle AFC=\angle BCD+\angle ADC$

$\therefore \angle AOC=\angle AEC+\angle AFC$

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