Question

In the given figure, $\angle BAC={30}^{\circ }$. Show that BC is equal to the radius of the circumcircle of $△$ABC whose centre is O.

Answer 1

ANSWER:
Join OB and OC.
∠ BOC = 2 ∠ BAC (As angle subtended by an arc of a circle at the centre is double the angle subtended by the
arc at any point on the circumference)
= 2 × 30 °
[∵ ∠ BAC = 30 ° ]
= 60 °
...(i)
Consider Δ BOC, we have:
OB = OC
[Radii of a circle]
⇒ ∠ OBC = ∠ OCB
...(ii)
In Δ BOC, we have:
∠ BOC + ∠ OBC + ∠ OCB = 180
(Angle sum property of a triangle)
⇒ 60 ° + ∠ OCB + ∠ OCB = 180 °
[From (i) and (ii)]
⇒ 2 ∠ OCB = (180 ° - 60 ° ) = 120 °
⇒ ∠ OCB = 60 °
...(ii)
Thus we have:
∠ OBC = ∠ OCB = ∠ BOC = 60 °
Hence, Δ BOC is an equilateral triangle.
i.e., OB = OC = BC
∴ BC is the radius of the circumcircle.