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, LN || AC

Given:

• In three line segments OA, OB, and OC,
• Points L, M, N respectively are so chosen that
• LM||AB
• MN||BC
• Neither of L, M, N nor of A, B, C are collinear.

Show that:

LN||AC

Explanation:

• I can attached figure from the given explanation you can see it

• We have,

         LM || AB and MN || BC

• Therefore, by basic proportionality theorem,

• We have,

       $\frac{OL}{AL}$ = $\frac{OM}{MB}$

      And ,  $\frac{ON}{NC}$ = $\frac{OM}{MB}$

• Comparing above equation

                $\frac{ON}{AL}$ = $\frac{ON}{NC}$

• 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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