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प्रश्न
ABCD is a parallelogram and E is the mid-point of BC. DE and AB when produced meet at F. Then, AF =
पर्याय
\[\frac{3}{2}AB\]
2 AB
3 AB
\[\frac{5}{4}AB\]
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उत्तर
Parallelogram ABCD is given with E as the mid-point of BC.

DE and AB when produced meet at F
We need to find AF.
Since ABCD is a parallelogram, then DC || AB
Therefore, DC || AF
Then, the alternate interior angles should be equal.
Thus, ∠DCE = ∠CBF …… (I)
In ΔDEC and ΔFEB:
∠DEC = ∠CBF (From(I))
CE = BE (E is the mid-point of BC)
∠CED = ∠BEF (Vertically opposite angles)
ΔDEC ≅ ΔFEB (by ASA Congruence property)
We know that the corresponding angles of congruent triangles should be equal.
Therefore,
DC = BF
But,
DC = AB (Opposite sides of a parallelogram are equal)
Therefore,
BF = AB …… (II)
Now,
AF = BF + AB
From (II),we get:
AF = AB +AB
AF = 2AB
Hence the correct choice is (b).
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