Solve this question but giving reason

In the figure, 
Δ DCE nd Δ BFE,
ang.DEC = ang. BEF ( vertically opp. ang.)
EC =BE ( E is the mid pnt)
ang. DCB =ang. EBF (alternate ang....... DC parallel ro AfF)
 so ΔDCE  congruent to Δ BFE
therefore DC = BF--------- (1)
now, CD = AB (ABCD is a parallelogram)
so AF = AB + BF
         = AB + DC  from (1)
         = AB + AB
         = 2 AB
hence proved............. hope dis helps....