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Question
Prove that any four vertices of a regular pentagon are concylic (lie on the same circle).
Solution
ABCDE is a regular pentagon.
∴ `∠BAE = ∠ABC = ∠BCD = ∠CDE = ∠DEA = ((5 - 2)/5) xx 180^circ = 180^circ`
In ΔAED,
AE = ED ...(Sides of regular pentagon ABCDE)
∴ ∠EAD = ∠EDA
In ΔAED,
∠AED + ∠EAD + ∠EDA = 180°
`=>` 108° + ∠EAD + ∠EAD = 180°
`=>` 2∠EAD = 180° − 108° = 72°
`=>` ∠EAD = 36°
∴ ∠EDA = 36°
∠BAD = ∠BAE − ∠EAD = 108° − 36° = 72°
In quadrilateral ABCD,
∠BAD + ∠BCD = 108° + 72° = 180°
∴ ABCD is a cyclic quadrilateral
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