Question

In \(\Delta ABC\), \(DE \parallel BC\). If \(AD = x\), \(DB = x - 2\), \(AE = x + 2\) and \(EC = x - 1\), then find the value of \(x\).

Hint:

When solving ratios involving \(x\), check the final answer. Lengths like \(x-2\) and \(x-1\) must be positive, which is true here as \(4-2=2\) and \(4-1=3\).

Answer 1

Step 1: Understanding the Concept:
According to the Basic Proportionality Theorem (Thales Theorem), if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Step 2: Key Formula or Approach:
\[ \frac{AD}{DB} = \frac{AE}{EC} \]
Step 3: Detailed Explanation:
Given \(DE \parallel BC\), we have:
\[ \frac{x}{x - 2} = \frac{x + 2}{x - 1} \]
Cross-multiplying the terms:
\[ x(x - 1) = (x + 2)(x - 2) \]
Expanding both sides:
\[ x^2 - x = x^2 - 4 \]
Subtracting \(x^2\) from both sides:
\[ -x = -4 \implies x = 4 \]
Step 4: Final Answer:
The value of \(x\) is 4.