① When △ABC∽△B'FC: According to △ABC is an isosceles triangle, △B'FC is also an isosceles triangle.

Then ∠B'FC=∠C=∠B, let BF=x, then CF=6-x, B'F=B'C=x, according to △ ABC ∠△ b' fc,

Get: b' fab = cfbc，get X5 = 6？ X6，x = 30 1 1；

② When △ ABC ∽△ FB ′ c is FC = B'f = BF, then x=6-x and the solution is X = 3.

So BF=3 or 30 1 1.