In the figure (ii) given below, AC is a diameter of the circle with centre O. If CD || BE, ∠AOB = 130° and ∠ACE = 20°, find :

(i) ∠BEC

(ii) ∠ACB

(iii) ∠BCD

(iv) ∠CED.

∠AOB = 130°.

∠AOB + ∠BOC = 180° (∵ these angles form linear pair.)

130° + ∠BOC = 180°
∠BOC = 180° - 130° = 50°.

(i) Arc BC subtends ∠BOC at the centre and ∠BEC at the remaining part of the circle.

∠BOC = 2∠BEC (∵ angle subtended on centre is twice the angle subtended on the remaining part of the circle).

⇒ 50° = 2∠BEC
⇒ ∠BEC = $\dfrac{50°}{2} = 25°.$

Hence, the value of ∠BEC = 25°.

(ii) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.

⇒ ∠AOB = 2∠ACB (∵ angle subtended on centre is twice the angle subtended on the remaining part of the circle).

⇒ 130° = 2∠ACB

⇒ ∠ACB = $\dfrac{130°}{2} = 65°.$

Hence, the value of ∠ACB = 65°.

(iii) Given, CD || EB,

∠ECD = ∠CEB = 25°. (∵ alternate angles are equal.)

From figure,

∠BCD = ∠ACB + ∠ACE + ∠ECD = 65° + 20° + 25° = 110°.

Hence, the value of ∠BCD = 110°.

(iv) EBCD is a cyclic quadrilateral hence the sum of opposite interior angles = 180°.

∴ ∠BED + ∠BCD = 180°
⇒ ∠BEC + ∠CED + ∠BCD = 180°
⇒ 25° + ∠CED + 110° = 180°
⇒ ∠CED + 135° = 180°
⇒ ∠CED = 180° - 135° = 45°.

Hence, the value of ∠CED = 45°.