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In a $Δ\ ABC$, $D$ and $E$ are points on the sides $AB$ and $AC$ respectively such that $DE\ ||\ BC$.
If $AD\ =\ 8x\ –\ 7\ cm$, $DB\ =\ 5x\ –\ 3\ cm$, $AE\ =\ 4x\ –\ 3\ cm$, and $EC\ =\ (3x\ –\ 1)\ cm$, Find the value of $x$.
img src=/doubts_assets/images/158630-1605782187.png" style="width: 25%;">"
 Given:
In a $Δ\ ABC$, $D$ and $E$ are points on the sides $AB$ and $AC$ respectively such that $DE\ ||\ BC$.
$AD =8x-7 cm$, $AE =5x-3 cm$, $DB =4x-3 cm$ and $EC =3x-1 cm$.
To do:
We have to find the value of $x$.
Solution:
$DE\ ||\ BC$ (given)
Therefore,
By Basic proportionality theorem,
$\frac{AD}{DB}\ =\ \frac{AE}{EC}$
$ \begin{array}{l}
\frac{8x-7}{5x-3} =\frac{4x-3}{3x-1}\\
\\
( 8x-7)( 3x-1) =( 4x-3)( 5x-3)\\
\\
8x( 3x-1) -7( 3x-1) =4x( 5x-3) -3( 5x-3)\\
\\
24x^{2} -8x-21x+7=20x^{2} -12x-15x+9\\
\\
( 24-20) x^{2} +( -8-21+12+15) x+7-9=0\\
\\
4x^{2} -2x-2=0\\
\\
2x^{2} -x-1=0\\
\\
2x^{2} -2x+x-1=0\\
\\
2x( x-1) +1( x-1) =0\\
\\
( x-1)( 2x+1) =0\\
\\
x-1=0\ or\ 2x+1=0\\
\\
x=1\ or\ 2x=-1\\
\\
x=1\ or\ x=\frac{-1}{2}
\end{array}$
The side of a triangle cannot be negative. Therefore, the value of $x$ is $1 cm$.