Answer :

Given: ADC = 130° , BC = BE


We know that,


(exterior AFC ) = (2×ADC)


(exterior AFC ) = (2×130)


(exterior AFC ) = 260


AFC = 360° – (exterior AFC) = 360°–260° = 100°


AFB = AFC + CFB = 180°


AFC + CFB = 180°


100° + CFB = 180°


CFB = 180° – 100° = 80°


In quadrilateral ABCD


ADC + ABC = 180° (opposite angles in cyclic quadrilateral are supplementary)


130° + ABC = 180°


ABC = 180° – 130° = 50°


In ΔBCF


By angle sum property


CBF + CFB + BCF = 180°


50° + 80° + BCF = 180°


BCF = 180° – 50° – 80° = 50°


Now,


CFE = CFB + BFE = 180°


CFB + BFE = 180°


80° + BFE = 180°


BFE = 180° – 80° = 100°


Here,


In ΔBCE


BC = BE (given)


BCE = BEC = 50° (angles opposite to equal sides are equal)


By angle sum property


BCE + BEC + CBE = 180°


50° + 50° + CBE = 180°


CBE = 180° – 50° – 50° = 100°



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