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.