Question

solve...............

Answer 1

Step-by-step explanation:

We know that any sector projects an angle at the center twice the value of angle projected at any point in the circle.

=> ∠AOC = 2∠ABC = 2∠ADC

=> ∠ABC = ∠ADC  and

∠AOC = ∠ABC + ∠ADC....................eq1

In the triangle BEC , ∠ABC is an exterior angle, so

∠ABC = ∠AEC + ∠ECB.......................eq2

In triangle FDC, ∠AFC is an exterior angle

=> ∠AFC = ∠ADC + ∠ECB

=> ∠ECB = ∠AFC - ∠ADC,

putting this value in the above equation(2) we get,

∠ABC = ∠AEC + ∠AFC - ∠ADC

=> ∠ABC + ∠ADC = ∠AEC + ∠AFC

from eqn 1 we know that, ∠AOC = ∠ABC + ∠ADC

Hence,

∠AOC = ∠AEC + ∠AFC

Hence proved