In \(\triangle ABC\), if \(DE \parallel BC\), \(AD = x\), \(DB = x - 2\), \(AE = x + 2\), and \(EC = x - 1\), then the value of \(x\) is:
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Use the Basic Proportionality Theorem (Thales' Theorem) when dealing with parallel lines in triangles. It helps relate the segments on two sides divided by a parallel line.
Using the Basic Proportionality Theorem (Thales' Theorem), which states that if a line divides two sides of a triangle in the same ratio, the line is parallel to the third side, we have:
\[
\frac{AD}{DB} = \frac{AE}{EC}
\]
Substituting the given values:
\[
\frac{x}{x - 2} = \frac{x + 2}{x - 1}
\]
Now, cross-multiply to solve for \(x\):
\[
x(x - 1) = (x - 2)(x + 2)
\]
Expanding both sides:
\[
x^2 - x = x^2 - 4
\]
Simplifying:
\[
-x = -4 \quad \Rightarrow \quad x = 4
\]
Thus, the correct answer is option (4).
