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I found this geometry question in IMO-2015 (conducted by the Science Olympiad Foundation, India) paper. It goes like this:

ABCD is a parallelogram and L is a point on DB. The produced line AL meets BC at M. Given that DL = 3LB, find $\frac{AB}{CN}$ or AB:CN.

Image

The options are:
[A] 1:2
[B] 4:5
[C] 1:4
[D] 3:2

I tried this:
Join the diagonal AC.
Let DL = 3x, so LB = x
Let AC intersect BD at E.
ABDC is parallelogram
$\therefore$ DE = EB.
$\therefore$ DE = 2x, EL = LB = x

I have done till this and I have not idea what to do next. Can anyone give a hint or solution for this problem?

insipidintegrator
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Arya
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2 Answers2

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No need to introduce new points. Since $AB,CD$ are parallel, $\triangle ABL$ and $\triangle NDL$ are similar and we have $ND:AB=ND:CD=3:1$, so $NC:CD=\color{blue}{NC:AB=2:1}$.

Parcly Taxel
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Triangles LAB and DLN are similar. Therefore, DN/AB = DL/LB But DN = DC+CN So, 1 + CN/AB =3 AB/CN = 1/2 Option A is correct

Vish
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