Question

In \( \triangle ABC \), \( DE \parallel AB \), \( AD = x + 1 \), \( CD = x + 3 \), \( BE = x + 4 \) and \( CE = x + 7 \), then the value of \( x \) is _____.

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

After solving BPT problems, always verify by substituting the value back into both ratios. It quickly confirms correctness and avoids careless mistakes in exams.

Answer 1

The Correct Option is D

Concept: When a line is drawn parallel to one side of a triangle and intersects the other two sides, it divides those sides in the same ratio. This is known as the Basic Proportionality Theorem (Thales' Theorem). If \( DE \parallel AB \), then: \[ \frac{CD}{AD} = \frac{CE}{BE} \]

Step 1: Identify the given expressions. We are given: \[ AD = x + 1,\quad CD = x + 3,\quad BE = x + 4,\quad CE = x + 7 \]

Step 2: Apply Basic Proportionality Theorem. Since \( DE \parallel AB \), we write: \[ \frac{CD}{AD} = \frac{CE}{BE} \] Substituting values: \[ \frac{x + 3}{x + 1} = \frac{x + 7}{x + 4} \]

Step 3: Cross multiply carefully. \[ (x + 3)(x + 4) = (x + 7)(x + 1) \] Now expand both sides step by step. Left side: \[ (x + 3)(x + 4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12 \] Right side: \[ (x + 7)(x + 1) = x^2 + x + 7x + 7 = x^2 + 8x + 7 \]

Step 4: Simplify the equation. \[ x^2 + 7x + 12 = x^2 + 8x + 7 \] Cancel \(x^2\) from both sides: \[ 7x + 12 = 8x + 7 \]

Step 5: Solve for \(x\). Bring like terms together: \[ 12 - 7 = 8x - 7x \] \[ 5 = x \]

Step 6: Verification (important step). Substitute \(x = 5\): \[ AD = 6,\ CD = 8,\ BE = 9,\ CE = 12 \] Check ratio: \[ \frac{CD}{AD} = \frac{8}{6} = \frac{4}{3}, \quad \frac{CE}{BE} = \frac{12}{9} = \frac{4}{3} \] Both ratios match, so solution is correct. Final Answer: \[ \boxed{5} \]