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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