Ques. 4 of exercise 10.6 maths …

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 that ABC = [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] = ABD Or [AOC – DOE] = ABC Hence proved.