There are three line segment $O A, O B$ and $O C$ . $L, M, N$ respectively are the points on them. These points are so chosen that $L M ∥ A B$ and $M N ∥ B C$ but neither of $L, M, N$ nor of $A, B, C$ are collinear. Show that $L N ∥ A C$ .

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Solution

In $△ O A B$ , $L M ∥ A B$

$∴$ $\frac{O L}{L A} = \frac{O M}{M B}$ [ Basic Proportionality Theorem]

$\Rightarrow$ $\frac{L A}{O L} = \frac{M B}{O M}$

$∴$ $\frac{L A}{O L} + 1 = \frac{M B}{O M} + 1$ [Adding 1 to both sides ]

$∴$ $\frac{L A + O L}{O L} = \frac{M B + O M}{O M}$

$∴$ $\frac{O A}{O L} = \frac{O B}{O M}$ ---- ( 1 )

Similarly, in $△ O B C$ , $M N ∥ B C$

$∴$ $\frac{O C}{O N} = \frac{O B}{O M}$ ----- ( 2 )
From ( 1 ) and ( 2 ), we get

$\Rightarrow$ $\frac{O A}{O L} = \frac{O C}{O N}$

$\Rightarrow$ In $△ O A C$ , $L N ∥ A C$ [ Converse of BPT ]

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