ABCD is a parallelogram and L is a point on DB. The produced line AL meets BC at M and DC produced at N. Given that DL $= 3$ LB, find $\frac{A B}{C N}$ .

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Solution

The correct option is A $1 / 2$
In $△ L A B$ and $△ D L N$ ,
$∠ A L B = ∠ D L N$ are vertically opposite and $∠ L A B = ∠ D N L$ (alternate angles) and third angle is also equal.
So both triangles are similar.
Therefore $\frac{D N}{A B} = \frac{D L}{L B}$
$D N = D C + C N$
So, $1 + \frac{C N}{A B} = 3$
$\frac{A B}{C N} = \frac{1}{2}$
Hence option B is correct.

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ABCD is a parallelogram and L is a point on DB. The produced line AL meets BC at M and DC produced at N. Given that DL=3LB, find ABCN.

The correct option is A 1/2In △LAB and △DLN , ∠ALB=∠DLN are vertically opposite and ∠LAB=∠DNL (alternate angles) and third angle is also equal. So both triangles are similar.Therefore DNAB=DLLBDN=DC+CN So, 1+CNAB=3 ABCN=12Hence option B is correct.

ABCD is a parallelogram and L is a point on DB. The produced line AL meets BC at M and DC produced at N. Given that DL $= 3$ LB, find $\frac{A B}{C N}$ .