In the figure (ii) given below, O is the centre of a circle. Chord CD is parallel to the diameter AB. If ∠ABC = 25°, calculate ∠CED.

Join OC and OD as shown in the figure below:

AC subtends angle AOC at centre and ∠ABC at point B.

∴ ∠AOC = 2∠ABC = 2 × 25° = 50°.

From figure,

∠OCD = ∠AOC (Alternate angles)

Hence, ∠OCD = 50°.

In △OCD,

OC = OD (Both are radius of the circle)

so, ∠ODC = ∠OCD.

Since, sum of angles of a triangle is 180°.

⇒ ∠COD + ∠OCD + ∠ODC = 180°
⇒ ∠COD + 50° + 50° = 180°
⇒ ∠COD + 100° = 180°
⇒ ∠COD = 80°.

CD subtends ∠COD at center and ∠CED at point E of the circle.

∴ ∠COD = 2∠CED
⇒ 80° = 2∠CED
⇒ ∠CED = 40°.

Hence, ∠CED = 40°.