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 collinera. Show that LN || AC.

Answer 1

$\Delta OLM$ similar to $\Delta OAB$

$\Rightarrow \angle O$ is common

$\angle OML=\angle OBA[LM\parallel AB]$, corresponding angles ----(i)
$\Rightarrow \angle OLM=\angle OAB(LM\parallel AB)$ ----(ii)

$\frac{OM}{OB}=\frac{OL}{OA}=\frac{LM}{AB}$ (by PRO. TH.) ----(iii)

similarly $\Delta OMN$ similar $\Delta OBC$

$\frac{OM}{OB}=\frac{ON}{OC}=\frac{NM}{CB}$ ----(iv)

From (iii) and (iv)

$\frac{NM}{CB}=\frac{LM}{AB}$ ----(v)

$\angle NML=\angle OMN+\angle OML=\angle OBC+\angle OBA$ ( Similar Triangles )

=$\angle CBA$ --------(vi)

Now, $\Delta NML$ and $\Delta CBA$

$\angle NML=\angle CBA$ --------(vi)

$\frac{NM}{CB}=\frac{LM}{AB}$ --------(v)

triangles are similar by S.A.S property for simiar triangle

$\Rightarrow \angle NLM=\angle CAB$ --------(vii)

Subtract (vii) from (ii)

$\angle OLM-\angle NLM=\angle OAB-\angle CAB$

$\angle OLN=\angle OAC$ (corresponding angles are equal)

$LN\parallel AC$