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Question

There are three line segment $$OA, OB$$ and $$OC$$. $$L, M, N$$ respectively are the points on them. These points 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 $$LN\parallel AC$$.


Solution


In $$\triangle OAB$$, $$LM\parallel AB$$

$$\therefore$$  $$\dfrac{OL}{LA}=\dfrac{OM}{MB}$$          [ Basic Proportionality Theorem]

$$\Rightarrow$$  $$\dfrac{LA}{OL}=\dfrac{MB}{OM}$$

$$\therefore$$   $$\dfrac{LA}{OL}+1=\dfrac{MB}{OM}+1$$      [Adding 1 to both sides ]

$$\therefore$$   $$\dfrac{LA+OL}{OL}=\dfrac{MB+OM}{OM}$$

$$\therefore$$   $$\dfrac{OA}{OL}=\dfrac{OB}{OM}$$       ---- ( 1 )

Similarly, in $$\triangle OBC$$, $$MN\parallel BC$$

$$\therefore$$   $$\dfrac{OC}{ON}=\dfrac{OB}{OM}$$       ----- ( 2 )
From ( 1 ) and ( 2 ), we get

$$\Rightarrow$$  $$\dfrac{OA}{OL}=\dfrac{OC}{ON}$$

$$\Rightarrow$$  In $$\triangle OAC$$, $$LN\parallel AC$$         [ Converse of BPT ]

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Mathematics

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