# 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 $4\ cm$.

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Updated on: 10-Oct-2022

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