In three line segments $OA, OB$ and $OC$, points $L, M, N$ respectively are so chosen that $LM \parallel AB$ and $MN \parallel BC$ but neither of $L, M, N$ nor of $A, B, C$ are collinear. Show that $LN \parallel AC$.

Given:

In three line segments $OA, OB$ and $OC$, points $L, M, N$ respectively are so chosen that $LM \parallel AB$ and $MN \parallel BC$ but neither of $L, M, N$ nor of $A, B, C$ are collinear. 

To do:

We have to show that $LN \parallel AC$.
Solution:

From the figure,

$LM \parallel AB$ and $MN \parallel BC$
Therefore,

By basic proportionality theorem, 

In $\triangle OAB$,
$\frac{OL}{AL}=\frac{OM}{MB}$....(i)

$\frac{ON}{NC}=\frac{OM}{MB}$....(ii)

From equations (i) and (ii), we get,

$\frac{OL}{AL}=\frac{ON}{NC}$

Therefore, by converse of basic proportionality theorem,

$LN \parallel AC$.

Hence proved.