Question

1. In the three line segment OA,OBand OC , points L,M,N respectively are so chosen that LM parallel AB and MN parallel BC but neither of L,M,N not of A,N,C are collinear. Show that LN parallel AC.

Answer 1

__________Hey Megha___________

______________________________

» ᴛʜɪs ᴀᴛᴛᴀᴄʜᴍᴇɴᴛ ᴡɪʟʟ ʜᴇʟᴘ ʏᴏᴜ ✌

______________________________

✔✔✔✔✔✔✔✔✔✔✔✔✔✔✔✔✔✔

______________________________

❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤

_____________________________

_____◦•●◉✿[Tʜᴀɴᴋ ʏᴏᴜ]✿◉●•◦_____

●▬▬▬▬▬ஜ۩۞۩ஜ▬▬▬▬▬▬●

_-_-_-_✌☆$BE \: \: \: \: BRAINLY$☆✌_-_-_-_

Answer 2

Jio HEY MATE HERE IS YOUR ANSWER

IN ⛛ OAB
=> LM II AB
=>/_OML=/_OAB(CROSS ANGLE)
=>
$\frac{ol}{angle \: a} = \frac{om}{mb}$
=>
$\frac{angle \: a}{ol} = \frac{om}{mb}$
=>
$\frac{la}{ol} + 1 = \frac{om}{mb} + 1$
$= > \frac{la + ol}{ol} = \frac{om + mb}{mb}$

$\frac{oa}{ol} = \frac{ob}{om} - - - - - - 1$
Similarly in ⛛ OBC,
=>MN II BC
=>
$\frac{oc}{on} = \frac{ob}{om} - - - - - 2$
$= > \frac{oa}{ol} = \frac{oc}{on} {from \: 1 \: and \: 2}$
IN ⛛ OAC

LN II AC
HENCE, PROVED