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 ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

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# 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 ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

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## Piyush Kumar Modi

This answer was edited.In △BDC,

∠ADC=∠DBC+∠DCB …. (1) ( exterior angle property)

⇒∠ADC= ½∠AOC and ∠DCB= ½∠DOE ……. (2)

[The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.]

From (1) and (2), we have

⇒ ½∠AOC=∠ABC+ ½ ∠DOE…[Since ∠DBC=∠ABC]

⇒∠ABC= ½ (∠AOC−∠DOE)

Hence, ∠ABC is equal to half the difference of angles subtended by the chords AC and DE at the centre.