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.

Answer 1

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