Question

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) $AF=\frac{3}{2}AB$
(b) AF = 2AB
(c) AF = 3AB
(d) AF² = 2AB²

Answer 1

(b) AF = 2 AB

Explanation:
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