In the given figure, ABC is a triangle in which ∠BAC = 30°. Show that BC is equal to the radius of the circumcircle of the triangle ABC, whose centre is O.

Join OB and OC.

We know that,

Angle at the centre is twice the angle at remaining circumference.

∴ ∠BOC = 2∠BAC = 2 × 30° = 60°

In △OBC,

⇒ OB = OC [Radii of same circle]

⇒ ∠OBC = ∠OCB = x (let) [As angles opposite to equal sides are equal]

⇒ ∠BOC + ∠OBC + ∠OCB = 180°

⇒ 60° + x + x = 180°

⇒ 2x + 60° = 180°

⇒ 2x = 180° - 60°

⇒ 2x = 120°

⇒ x = $\dfrac{120°}{2}$ = 60°.

∴ ∠OBC = ∠OCB = ∠BOC = 60°.

Hence, △OBC is an equilateral triangle.

∴ OB = OC = BC.

Hence, proved that BC is equal to the radius of the circumcircle of the triangle ABC.