Solution: △AFC is an isosceles triangle for the following reasons:

In △BAD and △BCE,

∠∠B =∠B，∠BAD=∠BCE，BD=BE，

∴△BAD≌△BCE，

∴BA=BC,∠BAC=∠BCA，

∴∠BAC-∠BAD=∠BCA-∠BCE, that is ∠ FAC = ∠ FCA.

∴AF=CF，

△ AFC is an isosceles triangle.