ABCDE is a cyclic pentagon with centre of its circumcircle at point O such that AB = BC = CD and angle ABC = 120°.

Calculate :

(i) ∠BEC

(ii) ∠BED

Cyclic pentagon ABCDE with its circumcircle with centre at point O is shown in the figure below:

(i) Given,

AB = BC = CD and ∠ABC = 120°

So, ∠BCD = ∠ABC = 120°

OB and OC are the bisectors of ∠ABC and ∠BCD, respectively.

So, ∠OBC = ∠BCO = 60°

In ∆BOC,

⇒ ∠OBC + ∠BCO + ∠BOC = 180° [By angle sum property of triangle]

⇒ 60° + 60° + ∠BOC = 180°

⇒ ∠BOC = 180° - 120° = 60°.

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

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.

∠BEC = $\dfrac{1}{2}$∠BOC = $\dfrac{1}{2}$ x 60° = 30°.

Hence, ∠BEC = 30°.

(ii) In cyclic quadrilateral BCDE, we have

⇒ ∠BED + ∠BCD = 180° [Sum of opposite angles in cyclic quadrilateral = 180°]

⇒ ∠BED + 120° = 180°

⇒ ∠BED = 180° - 120°

⇒ ∠BED = 60°.

Hence, ∠BED = 60°.