In a `DeltaABC`, if DE is parallel to BC, `(AD)/(DB)=(3)/(4)` and AC = 15 cm, then find the length AE.

Answer

To solve the problem, we will use the properties of similar triangles. Since DE is parallel to BC, triangles ADE and ABC are similar.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- \( \frac{AD}{DB} = \frac{3}{4} \)
- \( AC = 15 \, \text{cm} \)

2. **Determine the Ratio of Segments:**
- Let \( AD = 3x \) and \( DB = 4x \) for some value \( x \).
- Therefore, the total length \( AB = AD + DB = 3x + 4x = 7x \).

3. **Use the Property of Similar Triangles:**
- Since DE is parallel to BC, by the Basic Proportionality Theorem (or Thales' theorem), we have:
\[ \frac{AD}{AB} = \frac{AE}{AC} \]
- Substituting the values we have:
\[ \frac{3x}{7x} = \frac{AE}{15} \]

4. **Simplify the Ratio:**
- The \( x \) cancels out:
\[ \frac{3}{7} = \frac{AE}{15} \]

5. **Cross-Multiply to Solve for AE:**
- Cross-multiplying gives:
\[ 3 \cdot 15 = 7 \cdot AE \]
\[ 45 = 7 \cdot AE \]

6. **Isolate AE:**
- Divide both sides by 7:
\[ AE = \frac{45}{7} \approx 6.43 \, \text{cm} \]

### Final Answer:
The length of \( AE \) is approximately \( 6.43 \, \text{cm} \).