According to the condition given in the question, we have

In triangle DCE and FBE

BE = EC (E is the mid-point of BC)

∠ CED = ∠ BEF (Vertically opposite angles)

∠ CDE = ∠ EFB (Alternate interior angles)

∴∆DCE∆FBE (By AAS congruence rule)

DC = BF (BY CPCT)

As AB is parallel to DC, then AB = DC

∴ AB = DC = BF

AF = AB + BF

AF = AB + AB

AF = 2AB

Hence, option (B) is correct