ABC is a triangle. PQ is a line segment intersecting AB in P and AC in Q such that PQI/ BC and divides triangle ABC into two parts equal in area. Find the value of ratio BP: AB. 

Answer

To solve the problem, we need to find the ratio \( \frac{BP}{AB} \) given that line segment \( PQ \) is parallel to \( BC \) and divides triangle \( ABC \) into two parts of equal area.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- Triangle \( ABC \) is given.
- Line segment \( PQ \) intersects \( AB \) at \( P \) and \( AC \) at \( Q \).
- \( PQ \parallel BC \) and divides triangle \( ABC \) into two equal areas.

2. **Using the Area Property**:
- Since \( PQ \parallel BC \), triangles \( APQ \) and \( ABC \) are similar by the Basic Proportionality Theorem (also known as Thales' theorem).
- The area of triangle \( APQ \) is equal to half the area of triangle \( ABC \).

3. **Setting Up the Area Ratio**:
- The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
- Let \( AP = x \) and \( PB = y \). Therefore, \( AB = x + y \).
- The area of triangle \( APQ \) can be expressed as:
\[ \text{Area of } APQ = \frac{1}{2} \text{Area of } ABC \]
- Since the areas are in the ratio \( \frac{APQ}{ABC} = \left(\frac{AP}{AB}\right)^2 \), we have:
\[ \frac{\frac{1}{2} \text{Area of } ABC}{\text{Area of } ABC} = \left(\frac{AP}{AB}\right)^2 \]
- This simplifies to:
\[ \frac{1}{2} = \left(\frac{x}{x+y}\right)^2 \]

4. **Solving for the Ratio**:
- Taking the square root of both sides gives:
\[ \frac{1}{\sqrt{2}} = \frac{x}{x+y} \]
- Rearranging this gives:
\[ x = \frac{1}{\sqrt{2}}(x + y) \]
- Expanding and rearranging:
\[ x = \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} \]
\[ x - \frac{x}{\sqrt{2}} = \frac{y}{\sqrt{2}} \]
- Factoring out \( x \):
\[ x\left(1 - \frac{1}{\sqrt{2}}\right) = \frac{y}{\sqrt{2}} \]
- Thus, we have:
\[ \frac{y}{x} = \sqrt{2}\left(1 - \frac{1}{\sqrt{2}}\right) \]

5. **Finding \( \frac{BP}{AB} \)**:
- Since \( BP = y \) and \( AB = x + y \), we can express the ratio:
\[ \frac{BP}{AB} = \frac{y}{x+y} \]
- Substituting \( y = \sqrt{2}x - x \):
\[ \frac{BP}{AB} = \frac{\sqrt{2}x - x}{x + (\sqrt{2}x - x)} = \frac{(\sqrt{2} - 1)x}{\sqrt{2}x} \]
- This simplifies to:
\[ \frac{BP}{AB} = \frac{\sqrt{2} - 1}{\sqrt{2}} \]

### Final Answer:
The value of the ratio \( \frac{BP}{AB} \) is \( \frac{\sqrt{2} - 1}{\sqrt{2}} \).