We are given a triangle $\triangle ABC$ where line segment $DE$ is parallel to $BC$. The problem provides the following measurements:

• $AD = 5 \, \text{cm}$,
• $DB = 2.5 \, \text{cm}$,
• $BC = 12 \, \text{cm}$.

We need to find the length of $DE$.

Since $DE \parallel BC$, triangles $\triangle ADE$ and $\triangle ABC$ are similar by the Basic Proportionality Theorem (or Thales' Theorem), which states:
$\frac{AD}{AB} = \frac{DE}{BC}.$

Step 1: Compute $AB$

$AB = AD + DB = 5 \, \text{cm} + 2.5 \, \text{cm} = 7.5 \, \text{cm}.$

Step 2: Apply the proportionality

$\frac{AD}{AB} = \frac{DE}{BC} \quad \Rightarrow \quad \frac{5}{7.5} = \frac{DE}{12}.$

Simplify the ratio:
$\frac{5}{7.5} = \frac{2}{3}.$

Thus,
$\frac{2}{3} = \frac{DE}{12}.$

Step 3: Solve for $DE$

$DE = \frac{2}{3} \times 12 = 8 \, \text{cm}.$

Final Answer

The length of $DE$ is 8 cm.
Answer: (C) 8 cm.

Would you like a detailed explanation of similarity properties or related concepts?
Here are some related questions to practice:

1. If $DE = 4 \, \text{cm}$, what is the new length of $BC$?
2. Find the ratio of areas of $\triangle ADE$ and $\triangle ABC$.
3. If $BC = 18 \, \text{cm}$, how long would $DE$ be?
4. What theorem justifies that $\triangle ADE \sim \triangle ABC$?
5. What is the scale factor of similarity between the triangles?

Tip: When two triangles are similar, the ratio of their corresponding sides is equal, and the ratio of their areas is the square of the side ratio.