Join A and B, B and C, B and D.
In ∆ CDE,
∠1 + ∠2 + ∠CED = 180° ...(1)
∵ CE is a tangent to the circle CBA at point C,
∠CBA is an angle in the alternate segment.
∴ ∠1 = ∠3 ...(2)
∠2 = ∠4 ...(3)
From (1), (2) and (3) we have
∠3 + ∠4 + ∠CED = 180°
⇒ ∠CBD + ∠CED = 180°
⇒ C, B, D, E are concyclic