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.

⇒In△BDC,⇒∠ADCis 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=12∠AOCand∠DCB=12∠DOE... (2)From (1) and (2), we have⇒12∠AOC=∠ABC+12∠DOE...[Since∠DBC=∠ABC]⇒∠ABC=12(∠AOC−∠DOE)Hence,∠ABCis equal to half the difference of angles subtended by the chordsACandDEat the centre.