Question

In \(\Delta DEF\), \(AB \parallel EF\). The value of \(x\) is :

• A. \(0, 2\)
• B. \(2\) only
• C. \(- 2\)
• D. \(1\)

Hint:

Always check if your algebraic solutions are geometrically possible.
Since lengths of a triangle must be strictly positive, any value of \(x\) that makes a side length zero or negative must be discarded.

Answer 1

The Correct Option is B

Step 1: Use the whole-side ratio form of the Basic Proportionality Theorem.
Since $AB \parallel EF$ in $\Delta DEF$, instead of comparing $DA:AE$ with $DB:BF$, we can compare the whole sides $DE$ and $DF$ with the parts $DA$ and $DB$:
\[ \frac{DA}{DE} = \frac{DB}{DF} \]
This is an equivalent corollary of the same theorem, since $DE = DA+AE$ and $DF = DB+BF$.

Step 2: Write $DE$ and $DF$ in terms of $x$.
\[ DE = DA + AE = 2x + (3x+1) = 5x+1 \]
\[ DF = DB + BF = x + \left(2x-\frac{1}{2}\right) = 3x - \frac{1}{2} \]

Step 3: Set up the equation and cross-multiply.
\[ \frac{2x}{5x+1} = \frac{x}{3x-\frac{1}{2}} \]
\[ 2x\left(3x-\frac{1}{2}\right) = x(5x+1) \]
\[ 6x^{2} - x = 5x^{2} + x \]

Step 4: Simplify and solve.
\[ 6x^{2} - 5x^{2} - x - x = 0 \]
\[ x^{2} - 2x = 0 \]
\[ x(x-2) = 0 \]
So $x=0$ or $x=2$. Since $x=0$ would make the lengths $DA$ and $DB$ both zero, which cannot happen in an actual triangle, we reject it.

Final Answer:
The only valid value is $x=2$, matching option (B).
\[ \boxed{x = 2} \]