Maths geometry (triangles)

a)
∠ BFD = ∠ CEB ...(1) (corr. ∠s FD parallel EC)
BD = BC ....(2) (given)
∠ BDC = ∠ ABD + ∠ A (ext. ∠ of △)
∠ DCE = ∠ ADF (corr. ∠s FD parallel EC)
FA = FD (given)
∠ A = ∠ ADF (base ∠ s, isos. △)
∴∠ DCE = ∠ A
BC = BD (given)
∠ BCD = ∠ BDC (base ∠ s, isos. △)
∴∠ BCD = ∠ ABD + ∠ A
∠ DBF = ∠ BCE .....(3)
FROM (1)(2)(3) △BFD congruent △CEB (AAS)

b)
BF = CE (corr. sides, congruent △)
∠ DCE = ∠ A (proved in answer a)
EA = EC (sides oppo. equal ∠s)
∴EA = BF
EA - EF = BF - EF
AF = BE