Comprehensive problems of mathematics and geometry in junior high school

(1)RT△ADC∽RT△OEC, then ∠A=∠EOC because E is the midpoint of AC and FE is perpendicular to AC.

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).