Vertex B of ABC is located outside the circle with centre O.Side AB intersects chord CE at point E and side BC intersects chord AD at point D with the circle.We have to prove thatABC = [AOC – DOE]Join OA, OC, OE and OD.Now AOC = 2AEC[Angle subtended by an arc at the centre of the circle is twice the angle subtended by the same arc at any point in the alternate segment of the circle]AOC = AEC …(i)Similarly DOE = DCE ….(ii)Subtracting eq. (ii) from eq. (i),[AOC – DOE] = AEC – DCE ….(iii)Now AEC = ADC[Angles in same segment in circle] ….(iv)Also DCE = DAE[Angles in same segment in circle] ….(v)Using eq. (iv) and (v) in eq. (iii),[AOC – DOE]= DAE + ABD – DAE[AOC – DOE] = ABDOr [AOC – DOE] = ABCHence proved.