In the given figure, $A B C D$ is parallelogram and $E$ is the mid-point of $A D$ . A line through $D$ , drawn parallel to $E B$ , meets $A B$ produced at $F$ and $B C$ at $L$ . Then,

A F = 2 D C

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2 A F = 3 D C

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2 D C = 3 D L

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A F + D F = 3 D C + D L

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Solution

The correct option is A $A F = 2 D C$
Given, E is mid point of AD

Also $E B ∥ D F$

$\Rightarrow B$ is mid point of AF [mid--point theorem]

so, $A F = 2 A B$

$⟹ A F = 2 C D$ ( since $A B = C D$ )

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Similar questions

Q. In the given figure, ABCD is parallelogram and E is the mid-point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L. Prove that

(i) A F = 2 D C

(i i) D F = 2 D L

Q. 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) AF = 2AB
(c) AF = 3AB
(d) AF² = 2AB²

Q. In the given 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.

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

(a)

\frac{3}{2} A B

(b) 2 AB

(c) 3 AB

(d)

\frac{5}{4} A B

The given figure shows a parallelogram ABCD. E is a point in AD and CE produced meets BA produced at point F. If AE= 4cm, AF= 8 cm and AB = 12 cm, find the perimeter of the parallelogram ABCD.

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In the given figure, ABCD is parallelogram and E is the mid-point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L. Then,

The correct option is A AF=2DCGiven, E is mid point of ADAlso EB∥DF⇒B is mid point of AF [mid--point theorem]so, AF=2AB⟹AF=2CD ( since AB=CD)

In the given figure, $A B C D$ is parallelogram and $E$ is the mid-point of $A D$ . A line through $D$ , drawn parallel to $E B$ , meets $A B$ produced at $F$ and $B C$ at $L$ . Then,

The correct option is A $A F = 2 D C$
Given, E is mid point of AD

Also $E B ∥ D F$

$\Rightarrow B$ is mid point of AF [mid--point theorem]

so, $A F = 2 A B$

$⟹ A F = 2 C D$ ( since $A B = C D$ )