Prove that any four vertices of a regular pentagon are concyclic.

Let the regular pentagon be ABCDE.

Since it's regular, and all interior angles are equal to 108°.

∴ ∠ABC = ∠BCD = ∠CDE = ∠DEA = ∠EAB

Interior angle of a regular polygon is given by :

$\dfrac{n - 2}{n} \times 180°$

So, each interior angle of a regular pentagon = $\dfrac{5 - 2}{5} \times 180° = \dfrac{3}{5} \times 180°$ - 108°.

In triangle AED,

AE = ED [Sides of a regular pentagon are equal]

∴ ∠EAD = ∠EDA [Angles opposite to equal sides are equal] ……(1)

⇒ ∠AED + ∠EAD + ∠EDA = 180°

⇒ 108° + ∠EAD + ∠EAD = 180°

⇒ 2∠EAD = 180° - 108°

⇒ 2∠EAD = 72°

⇒ ∠EAD = $\dfrac{72°}{2}$

⇒ ∠EAD = 36°

∴ ∠EDA = 36°.

From figure,

⇒ ∠BAD = ∠BAE - ∠EAD = 108° - 36° = 72°.

In quadrilateral ABCD,

∠BAD + ∠BCD = 72° + 108° = 180°.

Since, sum of opposite angles = 180°,

which is possible when the quadrilateral is cyclic quadrilateral.

Hence, proved that any four vertices of a regular pentagon are concylic.