In the given figure, AB = BC = CD and ∠ABC = 132°. Calculate :

(i) ∠AEB,

(ii) ∠AED,

(iii) ∠COD.

(i) Join EB and EC.

In cyclic quadrilateral ABCE,

⇒ ∠ABC + ∠AEC = 180° [Sum of opposite angles in cyclic quadrilateral = 180°]

⇒ 132° + ∠AEC = 180°

⇒ ∠AEC = 180° - 132° = 48°.

Since, AB = BC.

We know that,

Angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ ∠AEB = $\dfrac{1}{2}$∠AEC

= $\dfrac{1}{2} \times 48°$ = 24°.

Hence, ∠AEB = 24°.

(ii) We know that,

Equal chords subtend equal angles at the circumference of the circle.

∠AEB = ∠BEC = ∠CED = 24°

∠AED = ∠AEB + ∠BEC + ∠CED = 24° + 24° + 24° = 72°.

Hence, ∠AED = 72°.

(iii) We know that,

Angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ ∠COD = 2∠CED = 2 × 24° = 48°.

Hence, ∠COD = 48°.