Question

which value of x would make \\(\overline{fg} \parallel \overline{bc}\\)?
3
1
9
6

Explanation:

Step1: Apply Basic Proportionality Theorem

If \( \overline{FG} \parallel \overline{BC} \), by the Basic Proportionality Theorem (Thales' theorem), we have \( \frac{AF}{FB} = \frac{AG}{GC} \).
From the diagram, \( AF = x + 1 \), \( FB = x + 6 \), \( AG = x - 1 \), \( GC = x + 3 \). So the proportion is \( \frac{x + 1}{x + 6} = \frac{x - 1}{x + 3} \).

Step2: Cross - Multiply the Proportion

Cross - multiplying gives \( (x + 1)(x + 3)=(x - 1)(x + 6) \).
Expanding both sides:
Left side: \( x(x + 3)+1(x + 3)=x^{2}+3x+x + 3=x^{2}+4x + 3 \)
Right side: \( x(x + 6)-1(x + 6)=x^{2}+6x-x - 6=x^{2}+5x - 6 \)

Step3: Solve the Equation

Set the expanded forms equal: \( x^{2}+4x + 3=x^{2}+5x - 6 \)
Subtract \( x^{2} \) from both sides: \( 4x+3 = 5x - 6 \)
Subtract \( 4x \) from both sides: \( 3=x - 6 \)
Add 6 to both sides: \( x=9 \)

Answer:

9