ABCD is a parallelogram and E is the mid-point of BC. DE and AB when produced meet at F. Then, AF =
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