In the given figure, ABCD is a || gm and E is the mid-point of BC.


In the given figure, ABCD is a || gm and E is the mid-point of BC. Also, DE and AB when produced meet at F. Then,

(a) $A F=\frac{3}{2} A B$

(b) $A F=2 A B$

(c) $A F=3 A B$

(d) $A F^{2}=2 A B^{2}$



(b) AF = 2 AB

I​n parallelogram ABCD, we have:
AB || DC
DCE = ∠​ EBF            (Alternate interior angles)
In ∆ DCE and ​ ∆ BFE, we have:
DCE = ∠ EBF              (Proved above)

DEC = ∠ BEF              (Vertically opposite angles)
BE = CE           ( Given)
i.e., ∆ DCE ≅​ ∆ BFE     (By ASA congruence rule)
∴  DC = BF         (CPCT)

But DC= AB, as ABCD is a parallelogram.
∴ DC = AB =  BF                 ...(i)

Now, AF = AB + BF             ...(ii)
   From (i), we get:
∴ AF = AB + AB = 2AB

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