Question : In $\triangle ABC$, $D$ and $E$ are points on sides $AB$ and $AC$, such that $DE$ II $BC$. If $AD=x+3$, $DB =2 x-3$, $A E=x+1$ and $EC=2 x-2$, then the value of $x$ is:
Option 1: $\frac{4}{5}$
Option 2: $\frac{1}{2}$
Option 3: $\frac{3}{5}$
Option 4: $\frac{1}{5}$
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Correct Answer: $\frac{3}{5}$
Solution : As $DE \parallel BC$, and $\angle A$ is common in both $\triangle ABC$ and $\triangle ADE$. ⇒ $\triangle ABC \sim \triangle ADE$ ⇒ $\frac{AB}{AD} = \frac{AC}{AE}$ ⇒ $\frac{AD+DB}{AD} = \frac{AE+EC}{AE}$ ⇒ $\frac{ (x + 3 + 2x - 3)}{(x + 3)} = \frac{(x + 1 + 2x-2)}{(x + 1)}$ ⇒ $\frac{(3x)}{(x+3)}=\frac{(3x-1)}{(x + 1)}$ ⇒ $3x^2+3x= 3x^2 - x + 9x - 3$ ⇒ $3x = 8x - 3$ ⇒ $5x = 3$ $\therefore x= \frac{3}{5}$ Hence, the correct answer is $\frac{3}{5}$.
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