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

SOLUTION :  

In Δ OAB,

LM || AB

Then,  OL/LA = OM/MB  …………..(1)

[By using basic proportionality theorem]

In Δ OBC,

MN || BC

Then,  OM/MB=ON/NC   ………….(2)

[By using basic proportionality theorem]

From eq 1 & 2 ,

OL/LA=ON/NC

Thus, L & N are points on sides OA & OC of ∆OAC,

In a Δ OCA,

OL/LA=ON/NC

LN || AC

[By Converse of basic proportionality theorem

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Answer 2

Answer:

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