Option : (B)
ABCD is a parallelogram.
E is the midpoint of BC.
So,
BE = CE
DE produced meets the AB produced at F
Consider the triangles CDE and BFE
BE = CE (Given)
∠CED = ∠BEF (Vertically opposite angles)
∠DCE = ∠FBE (Alternate angles)
Therefore,
ΔCDE ≅ ΔBFE
So,
CD = BF (c.p.c.t)
But,
CD = AB
Therefore,
AB = BF
AF = AB + BF
AF = AB + AB
AF = 2 AB