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:
Answer & explanation
Correct answer: option 3
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 \(\frac{3}{5}\)