Answer :
Let ABC be an equilateral triangle, its incircle and circumcircle are drawn with center O, let the radius of incircle be 'r' and circumcircle be 'R'
To prove: R = 2r
Construction: Draw OA ⊥ BC and OB ⊥ AC
Proof:
In ΔOAC and ΔOBC
∠OAC = ∠OBC [Both 90°]
OC = OC [Common]
OA = OB [Radii of incircle]
⇒ ΔOAC ≅ ΔOBC [By Right Angle - Hypotenuse - Side Criterion]
⇒ ∠OCA = ∠OCB [Corresponding parts of congruent triangles are equal]
But,
∠C = 60° [Angle of equilateral triangle]
⇒ ∠OCA + ∠OCB = 60°
⇒ ∠OCA + ∠OCA = 60°
⇒ 2∠OCA = 60°
⇒ ∠OCA = 30°
Now, In right-angled triangle OAC
[As, OA = radius of incircle = r and OC = radius of circumcircle = R]
⇒ R = 2r
Hence Proved.
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