Q19 of 77 Page 16

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

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