Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that `\ /_A B C` is equal to half the difference of the angles subtended by the chords AC and DE at

Explanation

Solution
`=>`In `/_\ BDC`,
`=>/_ADC` is the exterior angle

`:./_ADC=/_DBC+/_DCB` .... `(1)`
(The measure of an exterior angle is equal to the sum of the remote interior angles)
`=>`By inscribed angle theorem,

`=>/_ADC= 1/2 /_AOC` and `/_DCB= 1/2 ∠DOE` ....... `(2)`
From `(1)` and `(2)`, we have

`=>1/2 /_AOC=/_ABC+1/2 /_DOE` ...[Since `/_DBC=/_ABC`]
`=>/_ABC=1/2(/_AOC−/_DOE)`
Hence, `/_ABC` is equal to half the difference of angles subtended by the chords `AC` and `DE` at the centre.