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Question
In the given Figure, ABC is a triangle in which ∠BAC = 30°. Show that BC is the radius of the circumcircle of A ABC, whose center is O.
Solution
Join OB and OC. Since the angle subtended by an arc of a circle at its center is twice the angle subtended by the same arc at a point on the circumference.
∴ ∠ BOC = 2 ∠ BAC
⇒ ∠ BOC = 2 x 30° = 60°
Now in Δ BOC, we have
OB = OC ...(Each equal to the radius of the circle)
⇒ ∠ OBC = ∠ OCB ...(∵ Angles opposite to equal sides of a triangle are equal)
But ∠ OBC + ∠ OCB + ∠ BOC = 180°
∴ 2 ∠ OBC + 60° = 180°
⇒ 2 ∠ OBC = 120°
⇒ ∠ OBC = 60°
Thus,
∠ OBC = ∠ OCB
∠ BOC = 60°
⇒ Triangle OBC is an equilateral
⇒ OB = BC ...(Showed)
⇒ BC is the radius of the circumcircle of Δ ABC.
Hence proved.
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