In given figure, L, M and N are points on OA, OB and OC such that LM

1.	In given figure, L, M and N are points on OA, OB and OC such that LM \|\| AM and MN \|\| BC then show that LN \|\| AC.
Answer» Given : In ∆ABC, point L,M, N respectively lie OA, OB and OC such that LM \|\| AB and MN \|\| BC. To prove : LN \|\| AC Proof : In ∆ABO LM \|\| AB (given) OL/LA = OM/MB…..(ii) (By Basic prop. theorem) Again, In ∆BCO MN \|\| BC OM/MB = ON/NC ….(ii) (By Basc prop. theorem) From equation (i) and (ii), ⇒ OL/LA = OM/MB = ON/NC OL/LA = ON/NC (By Converse of B. P. theorem) ⇒ LN \|\| AC

In given figure, L, M and N are points on OA, OB and OC such that LM || AM and MN || BC then show that LN || AC.

Answer»

Given : In ∆ABC, point L,M, N respectively lie OA, OB and OC such that LM || AB and MN || BC.

To prove : LN || AC

Proof : In ∆ABO

LM || AB (given)

OL/LA = OM/MB…..(ii)

(By Basic prop. theorem)

Again, In ∆BCO

MN || BC

OM/MB = ON/NC ….(ii) (By Basc prop. theorem)

From equation (i) and (ii),

⇒ OL/LA = OM/MB = ON/NC

OL/LA = ON/NC (By Converse of B. P. theorem)

⇒ LN || AC

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In given figure, L, M and N are points on OA, OB and OC such that LM || AM and MN || BC then show that LN || AC.

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