In Fig.10.13, ADC = 130° and chord BC = chord BE. Find CBE.


Given: ADC = 130°

chord BC = chord BE


Since, ABCD is a cyclic quadrilateral, therefore sum of its opposite angles is 180°.


ADC + OBC = 180°


130° + OBC = 180°


OBC = 50°


Now, in triangles BOC and BOE:


OB = OB (common)


BC = BE (given)


OC = OE (radius of the circle)


By SSS congruence rule, ΔBOC ΔBOE.


OBC = OBE = 50° (By CPCT)


Now, CBE = CBO + OBE


CBE = 50° + 50°


CBE = 100°


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