ABCD is a parallleogram and E is the mid-point of BC. DE and AB when produced meet at F. Then , AF =

$\frac{3}{2} A B$

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2 AB

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3 AB

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$\frac{5}{4} A B$

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Solution

The correct option is B
2 AB

In || gm ABCD, E is the mid -point of BC. DE and AB are produced to meet at F.

$∵ E is mid point of BC ∴ B E = E C I n Δ B E F a n d Δ C E D B E = E C ∠ B E F = ∠ C E D (Vertically opposite angle) a n d ∠ E B F = ∠ E C D (Alternate angles) ∴ Δ B E F ≅ Δ C E D (A S A c r i t e r i a n) ∴ D C = B F B u t D C = A B ∴ A B = B F A F = A B + B F = A B + A B = 2 A B$

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Similar questions

\frac{3}{2} A B

\frac{5}{4} A B

A F = \frac{3}{2} A B

In the figure, ABCD is a parallelogram and E is the mid-point of side BC. If DE and AB when produced meet at F, prove that AF = 2AB.

In the given figure, ABCD is a ||gm and E is the midpoint of BC. Also, DE and All when produced meet at F. Then,

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

(b) AF = 2AB

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

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