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Question
In three line segments OA, OB, and OC, points L, M, N respectively are so chosen that LM || AB and MN || BC but neither of L, M, N nor of A, B, C are collinear. Show that LN ||AC.
Solution
We have,
LM || AB and MN || BC
Therefore, by basic proportionality theorem,
We have,
`"QL"/"AL"="OM"/"MB"` ..........(i)
and, `"ON"/"NC"="OM"/"MB"` ..........(ii)
Comparing equation (i) and equation (ii), we get,
`"ON"/"AL"="ON"/"NC"`
Thus, LN divides sides OA and OC of ΔOAC in the same ratio. Therefore, by the converse of basic proportionality theorem,
we have, LN || AC
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