In the given figure, O is the centre of a circle, ∠ADC = 130° and chord BC = chord BE. Find ∠CBE.

ABCD is a cyclic quadrilateral.
∴ ∠ABC + ∠ADC = 180°
⇒ ∠ABC + 130° = 180°
⇒ ∠ABC = (180° - 130°) = 50°

In Δ CFB, we have:
∠CFB + ∠CBF + ∠BCF = 180° (Angle sum property of a triangle)
⇒ 90° + 50° + ∠BCF = 180° (Since ∠CFB = 90° and ∠CBF = ∠ABC = 50°)
⇒ ∠BCF = (180° - 140°) = 40° ...(i)
In Δ BCF, we have:
BC = BE (Given)
⇒ ∠BEC = ∠BCF = 40° ...(ii)
In Δ BCE, we have:
∠BCE + ∠BEC + ∠CBE = 180° (Angle sum property of a triangle)
⇒ 40° + 40° + ∠CBE = 180 [From (i) and (ii)]
⇒ ∠CBE = (180° - 80°) = 100°
∴ ∠CBE = 100°