Let AO intersect BC at D
ΔAOB = ΔBOC = ΔAOC (SAS)
AB = BC = AC
ΔABC is equilateral
O is the circumcenter of equilateral ΔABC
Therefore, O is also the centroid of ΔABC
AO / AD = 2/3
2[tex]\sqrt{3}[/tex] / AD = 2/3
AD = 3[tex]\sqrt{3}[/tex]
Right triangle ADB has ∠ABD = 60°
AB / AD = 2/[tex]\sqrt{3}[/tex]
AB / 3[tex]\sqrt{3}[/tex] = 2/[tex]\sqrt{3}[/tex]
AB = 6
Perimeter of ΔABC = 6 + 6 + 6 =18
