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°

∴