In three line segments $O A, O B$ and $O C$ points $L, M, N$ respectively are so chosen that $L M$ is parallel to $A B$ and $M N$ is parallel to $B C$ but neither of $L, M, N$ nor of $A, B, c$ are collinear. Show that $L N$ is parallel to $A C$ .

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Solution

In $Δ O L M$ and $Δ O A B$
$∠ O L M = ∠ O A B$ ( $∵ L M | | A B$ )
$∠ L O M = ∠ A O B$ (common)
$∴ Δ O L M \sim Δ O A B$
=> $\frac{O L}{O A} = \frac{O M}{O B} = \frac{M L}{A B}$ ....(1)
Similiarly, in $Δ O N M$ and $Δ O C B$
$∠ O N M = ∠ O C B$ ( $∵ N M | | C B$ )
$∠ N O M = ∠ C O B$ (common)
$∴ Δ O N M \sim Δ O C B$
$\frac{O N}{O C} = \frac{O M}{O B} = \frac{M N}{C B}$ ....(2)
From equation (1) and (2), we get
$\frac{O N}{O C} = \frac{O L}{O A}$
Now in $Δ L O N$ and $Δ A O C$
$∠ N O L = ∠ C O A$ (common)
$\frac{O N}{O C} = \frac{O L}{O A}$
$∴ Δ O L N \sim Δ O A C$
$∴ ∠ O L N = ∠ O A C$
But these are corresponding angles.
Therefore, $L N | | A C$

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