ABCD is a paralle A. AB

B. 2 AB

C. 3 AB

D. AB

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