This is a question involving triangle similarity and the properties of parallel lines. Let's analyze the problem step-by-step:

We are given:

• $DE \parallel BC$
• $AE = a$ units
• $EC = b$ units
• $DE = x$ units
• $BC = y$ units

Since $DE \parallel BC$ and both $DE$ and $BC$ are corresponding segments in similar triangles $ADE$ and $ABC$, by the Basic Proportionality Theorem (Thales’ Theorem), we know that:

$\frac{AD}{AB} = \frac{AE}{AC}$

Let's find the expressions for these segments.

Step 1: Proportionality in Triangles

Using the similarity of triangles $ADE$ and $ABC$, the ratio of corresponding sides should be equal:

$\frac{DE}{BC} = \frac{AE}{AC}$

Step 2: Express AC in Terms of $a$ and $b$

From the diagram, $AC = AE + EC = a + b$.

Step 3: Apply the Proportionality

Using the relation we derived from similarity:

$\frac{x}{y} = \frac{a}{a + b}$

Rearranging this equation, we get:

$x = \frac{a}{a + b} \cdot y$

Thus, the correct answer is option (c):

$x = \frac{ay}{a + b}$

Would you like further explanation or clarification? Here are five related questions you might want to explore:

1. How does the Basic Proportionality Theorem apply to parallel lines in triangles?
2. Can this method be used for any triangle where a line parallel to one side is drawn?
3. What are some common applications of triangle similarity in geometry?
4. How would the solution change if $DE$ were not parallel to $BC$?
5. How can we use triangle similarity in real-world problems involving proportions?

Tip: Whenever you see a line parallel to one side of a triangle, it's a good idea to think about using the Basic Proportionality Theorem to find unknown lengths.