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Question
ABCDE is a cyclic pentagon with centre of its circumcircle at point O such that AB = BC = CD and angle ABC = 120°.
Calculate:
- ∠BEC
- ∠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 ∠BCD respectively.
∴ ∠OBC = ∠BCO = 60°
In ΔBOC,
∠BOC = 180° – (∠OBC + ∠BOC)
`=>` ∠BOC = 180° – (60° + 60°)
`=>` ∠BOC = 180° – 120° = 60°
Arc BC subtends ∠BOC at the centre and ∠BEC at the remaining part of the circle.
∴ `∠BEC = 1/2 ∠BOC = 1/2 xx 60^circ = 30^circ`
ii. In cyclic quadrilateral BCDE,
∠BED + ∠BCD = 180°
`=>` ∠BED + 120° = 180°
∴ ∠BED = 180° – 120° = 60°
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