∠BFD+∠B+∠FDB= 180

∠FDE+∠EDC+∠FDB= 180

Because < FDE = < b

So ∠EDC=∠BFD

BD=CE，BF=CD

It can also be concluded that △BDF is similar to △CDE.

So ∠DEC=∠BDF

In the internal angle sum relation of △BDF and △CDE

You can get ∠ b = ∠ c.