In $\triangle A B C, D$ and $E$ are points on sides $A B$ and $A C$, such that $D E$ II $B C$. If $A D=x+3, D B$ $=2 x-3, A E=x+1$ and $E C=2 x-2$, then the value of $x$ is:

$ \frac{3}{5}$

Concept Used

When two triangles are similar, their corresponding angles are equal and their corresponding sides are proportional.

Calculation

As DE is parallel to BC and \(\angle\)A is common in both \(\Delta \) ABC and \(\Delta \)ADE

So, ABC is similar to ADE

So,

\(\frac{AB}{AD}\) = \(\frac{AC}{AE}\)

= \(\frac{x\;+\;3\;+\;2x\;-\;3}{x\;+\;3}\) = \(\frac{x\;+\;1\;2x\;-\;2}{x\;+\;1}\)

= \(\frac{3x}{x\;+\;3}\) = \(\frac{3x\;-\;1}{x\;+\;1}\)

= \( { 3x}^{2 } \) + 3x = \( { 3x}^{2 } \) - x + 9x - 3

= 3x = 8x - 3

= 5x = 3

= x = \(\frac{3}{5}\)

Therefore, x is \(\frac{3}{5}\)