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,

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