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In the figure, 
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
3 5 3
Limzy is really so cleaver.Thankyou for clarifying my doubts
u r welcme dear nd at d same tym fr calling me clever :)
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