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We are given that in triangle \( \triangle ABC \), line \( DE \parallel AB \), and several values are provided:

- \( AB = a \)
- \( BE = b \)
- \( EC = c \)
- \( DE = x \)

Since \( DE \parallel AB \), by using the property of similar triangles, we know that triangles \( \triangle CDE \) and \( \triangle CAB \) are similar. This allows us to set up a proportion between the sides of these triangles.

Using the sides:

\[
\frac{DE}{AB} = \frac{BE}{BC}
\]

Substitute the known values:

\[
\frac{x}{a} = \frac{b}{b + c}
\]

Now, solve for \( x \):

\[
x = \frac{a \cdot b}{b + c}
\]

Thus, the value of \( x \) in terms of \( a \), \( b \), and \( c \) is:

\[
x = \frac{ab}{b + c}
\]

This corresponds to option **(d)**.

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Would you like a detailed explanation of the steps or have any questions about this problem?

Here are 5 related questions to further explore this topic:
1. What are the conditions for two triangles to be similar?
2. How do parallel lines in triangles help with proportion calculations?
3. Can we solve this problem using the area ratio of similar triangles?
4. How would this problem change if \( DE \) were not parallel to \( AB \)?
5. What are the implications of the intercept theorem in solving similar problems?

**Tip**: In geometry, recognizing similar triangles early on can help simplify the use of proportions and side relationships.

In ∆ABC, DE ∥ AB. If AB = a, DE = x, BE = b, and EC = c, then x expressed in terms of a, b, and c is:

Learn how to solve a geometry problem involving similar triangles and parallel lines. Given a triangle with DE ∥ AB, calculate the value of x using proportions and the Basic Proportionality Theorem (Thales' Theorem). Suitable for students in Grades 9-10.

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