From Gogeometry

Using :

• Additional points
• Similar triangles
• Equilateral triangle

• Define 0 middle of BC. O is the center of semicircle OD diameter BC
• Define a=OB=OD=OC => AB=AC=BC=2a
• Arc BD=arc DE=arc EC => ∠BOD=∠DOE=∠EOC=Π/3

• OB=OD=a and ∠BOD=Π/3
• =>BD=a, ΔBOD is equilateral and BD//AC
• In the same way, DE=a, ΔDOE is equilateral and DE//BC
• and EC=a, ΔEOC is equilateral and EC//AB

• Define H the intersection of lines AB and DE
• Define I the intersection of lines AC and DE
• ∠ABC=Π/3 and ∠OBD=Π/3 => ∠HBD=Π/3
• ∠ABC=Π/3 and BC//DE => ∠BHD=Π/3
• ∠HBD=Π/3 and ∠BHD=Π/3 => ΔBHD is equilateral
• In the same way, ΔCIE is equilateral

• HD=DE=EI=a
• ΔAIE is similar to ΔACG => AI/IE=AC/CG =>3=2a/CG =>CG=2a/3
• ΔAHD is similar to ΔABF => AH/HD=AB/BF =>3=2a/BF =>BF=2a/3
• BC= 2a=BF+FG+GC=2a/3+FG+2a/3 => FG=2a/3
• Therefore BF=FG=GC