BCD and EFG are equilateral triangles, and B, C, D, F, and G lie on the same circle. Let R_1 be the circumradius of BCD, and let R_2 be the cirucmradius of EFG. Find R_2/R_1.
For simplicity, let the circumradius of BCD = 2 = R1
Then the distance from the center of this triangle to its base = 1
Then the circumradius of EFG must be [2 - 1]/2 =1/2 = R2
R2 /R1 = (1/2) / 2 = 1/4