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

$$\begin{gathered} \mathrm{In}∆\mathrm{OAB},\mathrm{since}\mathrm{LM}\parallel \mathrm{AB},\mathrm{then} \\ \frac{\mathrm{OL}}{\mathrm{LA}}=\frac{\mathrm{OM}}{\mathrm{MB}}\left(\mathrm{By}\mathrm{BPT}\right)........\left(1\right) \\ \mathrm{In}∆\mathrm{OBC},\mathrm{since}\mathrm{MN}\parallel \mathrm{BC},\mathrm{then} \\ \frac{\mathrm{OM}}{\mathrm{MB}}=\frac{\mathrm{ON}}{\mathrm{NC}}\left(\mathrm{By}\mathrm{BPT}\right) \\ \Rightarrow \frac{\mathrm{ON}}{\mathrm{NC}}=\frac{\mathrm{OM}}{\mathrm{MB}}.........\left(2\right) \\ \mathrm{from}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ \frac{\mathrm{OL}}{\mathrm{LA}}=\frac{\mathrm{ON}}{\mathrm{NC}}.........\left(3\right) \\ \mathrm{In}∆\mathrm{OCA},\mathrm{we}\mathrm{have}, \\ \frac{\mathrm{OL}}{\mathrm{LA}}=\frac{\mathrm{ON}}{\mathrm{NC}} \\ \Rightarrow \mathrm{LN}\parallel \mathrm{AC}\left(\mathrm{By}\mathrm{converse}\mathrm{of}\mathrm{BPT}\right) \end{gathered}$$