Answer :

Let the equilateral triangle be Δ ABC, Circumcentre and circumradius be E and r respectively.

We know that in equilateral triangle median and internal angle bisector are same.

Here AE, BE and CE are part of medians.

⇒ EA acts as an internal angle bisector for ∠CAB.

⇒ ∠CAE = ∠EAB

In Δ ABC,

∠CAB = ∠CAE + ∠EAB = 60°

⇒ ∠CAB = 2(∠CAE ) = 2(∠EAB) = 60°

⇒ ∠CAE = ∠EAB = 30°

Extend CE to meet at AB at H.

CH is a median as well as altitude.

⇒ CH ⊥ AB and ∠CHA = ∠CHB = 90°

In Δ AEH,

∠EAH + ∠AHE + ∠HEA = 180°

⇒ 30° + 90° + ∠HEA = 180°

⇒ ∠HEA = 60°

We know that sides of any triangle of angles 30°, 60° and 90°

are in the ratio 1: √3: 2 .

⇒ EH :AH :AE = 1: √3: 2

⇒ EH :4 :r = 1: √3: 2

⇒ 4 :r = √3: 2

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