In a `Delta ABC`, line `DE|\|BC`. DE divides the area of `Delta` in ratio `1:2`. Find `(AD)/(DB)`.

Answer

To solve the problem, we need to find the ratio \( \frac{AD}{DB} \) in triangle \( ABC \) where line \( DE \) is parallel to \( BC \) and divides the area of triangle \( ABC \) in the ratio \( 1:2 \).

### Step-by-step Solution:

1. **Understanding the Problem:**
- We have triangle \( ABC \) and a line \( DE \) parallel to \( BC \).
- The area of triangle \( ADE \) is \( \frac{1}{3} \) of the area of triangle \( ABC \) since the areas are in the ratio \( 1:2 \).

2. **Using the Properties of Similar Triangles:**
- Since \( DE \parallel BC \), triangles \( ADE \) and \( ABC \) are similar by the AA (Angle-Angle) criterion.
- Therefore, the ratios of corresponding sides are equal:
\[ \frac{AD}{AB} = \frac{DE}{BC} = \frac{AE}{AC} \]

3. **Relating Areas to Side Ratios:**
- The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides:
\[ \frac{\text{Area of } ADE}{\text{Area of } ABC} = \left(\frac{AD}{AB}\right)^2 \]
- Given that the area ratio is \( \frac{1}{3} \):
\[ \frac{1}{3} = \left(\frac{AD}{AB}\right)^2 \]

4. **Finding the Side Ratio:**
- Taking the square root of both sides:
\[ \frac{AD}{AB} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} \]

5. **Expressing \( AB \) in Terms of \( AD \):**
- Let \( AD = k \). Then:
\[ AB = \sqrt{3}k \]

6. **Finding \( DB \):**
- \( DB \) can be expressed as:
\[ DB = AB - AD = \sqrt{3}k - k = k(\sqrt{3} - 1) \]

7. **Finding the Ratio \( \frac{AD}{DB} \):**
- Now substituting \( AD \) and \( DB \):
\[ \frac{AD}{DB} = \frac{k}{k(\sqrt{3} - 1)} = \frac{1}{\sqrt{3} - 1} \]

### Final Answer:
\[ \frac{AD}{DB} = \frac{1}{\sqrt{3} - 1} \]