Question

In \( \triangle ABC \), \( DE \parallel BC \), if \( AD = x + 1 \), \( DB = 3x - 1 \), \( AE = x \), and \( EC = 4x - 3 \), then the value of \( x \) is:

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

For parallel lines in a triangle, use the basic proportionality theorem: \( \frac{AD}{DB} = \frac{AE}{EC} \).

Answer 1

The Correct Option is A

We are given that \( DE \parallel BC \) in \( \triangle ABC \), so by the basic proportionality theorem (also known as Thales' theorem), we know:
\[ \frac{AD}{DB} = \frac{AE}{EC} \] Substitute the given values into this proportion:
\[ \frac{x + 1}{3x - 1} = \frac{x}{4x - 3} \] Now, cross-multiply to solve for \( x \): \[ (x + 1)(4x - 3) = x(3x - 1) \] Expand both sides: \[ 4x^2 - 3x + 4x - 3 = 3x^2 - x \] Simplify the equation: \[ 4x^2 + x - 3 = 3x^2 - x \] Move all terms to one side: \[ 4x^2 + x - 3 - 3x^2 + x = 0 \] Simplify further: \[ x^2 + 2x - 3 = 0 \] Factor the quadratic equation: \[ (x + 3)(x - 1) = 0 \] Thus, the solutions are: \[ x = -3 \quad \text{or} \quad x = 1 \] Since \( x \) must be positive, the value of \( x \) is 1.