In the adjoining figure, DE is a chord parallel to diameter AC of the circle with centre O

Angles in the same segment of a circle are equal.
i.e., ∠CAD = ∠CBD = 60°
We know that an angle in a semicircle is a right angle.
i.e., ∠ADC = 90°
In  ΔADC, we have:
∠ACD + ∠ADC + ∠CAD = 180°  (Angle sum property of a triangle)
⇒ ∠ACD + 90° + 60° = 180°
⇒∠ACD = 180° –  (90° + 60°) = (180° – 150°) = 30°
⇒∠CDE = ∠ACD = 30°  (Alternate angles as AC parallel to DE)
Hence, ∠CDE = 30°