Question

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

Hint:

When cross-multiplying in BPT problems, look for algebraic identities like \((x+2)(x-2) = x^2-4\) to simplify your work.

Answer 1

Given:
In ΔABC, DE ∥ BC
AD = x, DB = x − 2
AE = x + 2, EC = x − 1

Step 1: Apply Basic Proportionality Theorem (Thales’ theorem)
Since DE ∥ BC,
\[ \frac{AD}{DB} = \frac{AE}{EC} \]

Substitute the values:
\[ \frac{x}{x - 2} = \frac{x + 2}{x - 1} \]

Step 2: Cross-multiply
\[ x(x - 1) = (x + 2)(x - 2) \] Expand both sides:
Left side: \[ x^2 - x \] Right side: \[ x^2 - 4 \]

Step 3: Form the equation
\[ x^2 - x = x^2 - 4 \] Cancel \(x^2\) from both sides:
\[ -x = -4 \] \[ x = 4 \]

Final Answer:
\[ \boxed{x = 4} \]