Q. 13

# In the given figu

Given: ADC = 130° , BC = BE

We know that,

(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°

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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