Given triangle $\triangle ABC$, with $DE \parallel BC$, we can use the Basic Proportionality Theorem (also known as Thales' theorem). This theorem states that if 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.

Step-by-step solution:

We are given:

• $AD = x$
• $DB = x - 2$
• $AE = x + 2$
• $EC = x - 1$

Since $DE \parallel BC$, the Basic Proportionality Theorem tells us:

$\frac{AD}{DB} = \frac{AE}{EC}$

Substitute the given values into this equation:

$\frac{x}{x - 2} = \frac{x + 2}{x - 1}$

Step 1: Cross-multiply

$x(x - 1) = (x - 2)(x + 2)$

Step 2: Expand both sides

Expanding both sides:

$x^2 - x = (x - 2)(x + 2)$ $x^2 - x = x^2 - 4$

Step 3: Simplify the equation

Subtract $x^2$ from both sides:

$-x = -4$

Step 4: Solve for $x$

Multiply both sides by -1:

$x = 4$

Conclusion:

The value of $x$ is $\boxed{4}$.

Would you like any further clarification or details on this?

Here are 5 related questions to explore further:

1. What is the Basic Proportionality Theorem (Thales' Theorem)?
2. How do parallel lines affect the properties of a triangle?
3. What happens when two sides of a triangle are divided in the same ratio?
4. Can the Basic Proportionality Theorem be applied to quadrilaterals?
5. How would the solution change if the triangle were isosceles?

Tip: Always check if lines are parallel in a triangle to apply the Basic Proportionality Theorem.