In $ \triangle \mathrm{ABC}, \mathrm{M} $ and $ \mathrm{N} $ are points of $ \mathrm{AB} $ and $ \mathrm{AC} $ respectively and $ \mathrm{MN} \| \mathrm{BC} $. If $ \mathrm{AM}=x $, $ \mathrm{MB}=x-2, \mathrm{AN}=x+2 $ and $ \mathrm{NC}=x-1 $, find the value of $ x $.

Given:

In \( \triangle \mathrm{ABC}, \mathrm{M} \) and \( \mathrm{N} \) are points of \( \mathrm{AB} \) and \( \mathrm{AC} \) respectively and \( \mathrm{MN} \| \mathrm{BC} \).

\( \mathrm{AM}=x \), \( \mathrm{MB}=x-2, \mathrm{AN}=x+2 \) and \( \mathrm{NC}=x-1 \).

To do:

We have to find the value of \( x \).

Solution:

We know that,

The line drawn parallel to one side of a triangle and cutting the other two sides divides the other two sides in equal proportion.

Therefore,

$\frac{AM}{BM}=\frac{AN}{NC}$

$\frac{x}{x-2}=\frac{x+2}{x-1}$

$x(x-1)=(x-2)(x+2)$

$x^2-x=x^2-2^2$

$x^2-x^2-x=-4$

$x=4$

Hence, the value of $x$ is $4\ cm$.