Given,

ABCD is a parallelogram

E is mid point of BC

DE & AB after producing meet at F

In ∆ECD & ∆BEF ,

∠BEF = ∠CED [vertically opposite angles]

BE = EC

∠EDC = ∠EFB [ alternate angles]

∴ ∆ECD ≅ ∆BEF

So, CD= BF

∵ AB=CD

Thus, AF= AB+ BF

= AF = AB + CD

= AF = AB+AB

= AF = 2AB