So AF=CF (perpendicular bisector), so ∠A=∠ECF= ∠EOC and ∠OEC=∠CEF.
So △OEC∽△CEF
(2) Because △OEC∽△CEF, EC/EF=OC/CF Because △CEF is a right angle, EC/EF=tan∠OFC.
So CO/CF=tan∠OFC
(3) Because △AFC is an isosceles triangle and ∠ A = 60, △AFC is an equilateral triangle.
Because CD⊥AF, AD=DF=2= 1/2CF, namely AC=AF=CF=4 and EF⊥AC, so ∠ OFC = 1/2 ∠ AFC = 30 because Co/CF = Tan \.
So CO=tan∠OFC×CF=tan30×4=(4/3)√3 (three-fourths root number).