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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.
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