In the given figure, AC is the diameter of the circle with center O. CD and BE are parallel. Angle ∠AOB = 80° and ∠ACE = 10°. Calculate :

(i) Angle BEC,

(ii) Angle BCD,

(iii) Angle CED.

(i) From figure,

⇒ ∠BOC + ∠BOA = 180° [As AOC is a straight line.]

⇒ ∠BOC + 80° = 180°

⇒ ∠BOC = 180° - 80° = 100°.

We know that,

Angle at the centre is double the angle at the circumference subtended by the same chord.

⇒ ∠BOC = 2∠BEC

⇒ ∠BEC = $\dfrac{1}{2}$∠BOC = $\dfrac{100°}{2}$ = 50°.

Hence, ∠BEC = 50°.

(ii) Given,

DC || EB

∴ ∠DCE = ∠BEC = 50° [Alternate angles are equal]

We know that,

Angle at the centre is double the angle at the circumference subtended by the same chord.

⇒ ∠AOB = 2∠ACB

⇒ ∠ACB = $\dfrac{1}{2}$∠AOB = $\dfrac{80°}{2}$ = 40°.

From figure,

∠BCD = ∠ACB + ∠ACE + ∠DCE = 40° + 10° + 50° = 100°.

Hence, ∠BCD = 100°.

(iii) As sum of opposite angles of cyclic quadrilateral = 180°.

⇒ ∠BED + ∠BCD = 180°

⇒ ∠BED = 180° - ∠BCD = 180° - 100° = 80°.

From figure,

⇒ ∠BED = ∠BEC + ∠CED

⇒ 80° = 50° + ∠CED

⇒ ∠CED = 80° - 50° = 30°.

Hence, ∠CED = 30°.