#### Question

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

#### Solution

i) Join OC and OB.

AB = BC = CD and `∠`ABC =120°

∴ `∠`BCD = `∠`ABC = 120°

OB and OC are the bisectors of `∠`ABC and `∠`BCDrespectively.

∴ `∠`OBC = `∠`BCO = 60°

In ΔBOC,

`∠`BOC = 180° - (`∠`OBC + `∠`BOC)

⇒ `∠`BOC = 180° - ( 60° + 60°)

⇒ `∠`BOC = 180° - 120° = 60°

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

∴ `∠`BEC = `1/2` `∠`BOC = `1/2 xx 60° = 30°`

ii) In cyclic quadrilateral BCDE,

`∠`BED + `∠`BCD = 180°

⇒ `∠`BED + 120° = 180°

∴ `∠` BED = 60°

Is there an error in this question or solution?

Solution 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 Concept: Cyclic Properties.