In the adjoining figures, is a parallelogram and is the midpoint of side If and when produced meet at prove that

ABCD is parallelogram.

Also given that BE = CE

In ABCD, AB || DC

∠DCE = ∠EBF … Alternate angles with transversal DF

In ∆DCE and ∆BFE, we have,
∠DCE = ∠EBF …from above

∠DEC = ∠BEF … Vertically opposite angles
Also, BE = CE … given
Hence, by ASA congruence rule,

∆DCE ≅​ ∆BFE
∴ DC = BF … by cpct

But DC = AB, as ABCD is a parallelogram.
∴ DC = AB = BF
Now, AF = AB + BF
From above, we get,
AF = AB + AB = 2AB
Hence, proved.