Question

. In triangle ABC , line parallel to BC intersect side AB and side AC in point M and N respectively. If AM = 8 ,MB= 12 ,AN=6 ;then find NC? *

Answer 1

Answer:

in ∆ABC

line MN || side BC,

AM/MB=AN/NC

(by basic proportionality theorem)

let the value of NC be x

12/8=6/x

by cross multiplication

12x=48

x=48units

is my answer helpful

Answer 2

Answer:

NC = 2

Step-by-step explanation:

Given : MN || BC , AM = 8 , MB = 12 , AN = 6

To find : NC = ?

In triangle ABC

MN || BC

.

. .

<AMN = <ABC ( corresponding angle)

<ANM = <ACB ( corresponding angle)

∆AMN~∆ABC ( AA criterion for similarity)

$⇒  \frac{AC}{AN} = \frac{AB}{MB} \\ \frac{AC}{6} = \frac{8}{12} \\ 12AC = 48 \\ AC = \frac{48}{12} \\ AC = 4$

.

. .

$NC = AN - AC \\ NC = 6 - 4 \\ NC = 2$

Hope it helps