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°