Question

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.

Answer 1

Given – In the figure ABC is a triangle in which ∠A = 30°

To prove – BC is the radius of circumcircle of ∆ABC whose centre is O.

Construction – Join OB and OC.

Proof – ∠BOC = 2∠BAC = 2 × 30° = 60°

Now in ∆OBC,

OB = OC    ...[Radii of the same circle]

∠OBC = ∠OCB

But, in ΔBOC,

∠OBC + ∠OCB + ∠BOC = 180°   ...[Angles of a triangle]

`=>` ∠OBC + ∠OBC + 60° = 180°

`=>` 2∠OBC + 60° = 180°

`=>` 2∠OBC = 180° – 60°

`=>` 2∠OBC = 120°

`=> ∠OBC = 120^circ/2 = 60^circ`

`=>` ∠OBC = ∠OCB = ∠BOC = 60°

`=>` ΔBOC is an equilateral triangle

`=>` BC = OB = OC

But, OB and OC are the radii of the circumcircle

∴ BC is also the radius of the circumcircle.