ABC is right angle triangle , (right angled at B ) incircle touches the sides AB, BC & AC at F, E, & D respectively . If BD is prependicular to AC then , find the ratio of AF to FB .

Answer

To solve the problem, we need to find the ratio of \( AF \) to \( FB \) in the right-angled triangle \( ABC \) where \( B \) is the right angle. The incircle touches the sides \( AB \), \( BC \), and \( AC \) at points \( F \), \( E \), and \( D \) respectively, and \( BD \) is perpendicular to \( AC \).

### Step-by-Step Solution:

1. **Understanding the Triangle and Points**:
- Let \( AB = c \), \( BC = a \), and \( AC = b \).
- The points where the incircle touches the sides are \( F \) on \( AB \), \( E \) on \( BC \), and \( D \) on \( AC \).
- Let \( AF = x \), \( FB = y \), \( BD = w \), and \( CD = z \).

2. **Using the Properties of the Incircle**:
- The lengths of the segments created by the points of tangency can be expressed as:
- \( AF = x \)
- \( FB = y \)
- \( BD = w \)
- \( CD = z \)
- From the properties of tangents from a point to a circle, we know:
- \( AF = x \)
- \( FB = y \)
- \( BD = w \)
- \( CD = z \)

3. **Applying the Pythagorean Theorem**:
- In triangle \( ABD \):
\[ AB^2 = AF^2 + BD^2 \implies c^2 = x^2 + w^2 \tag{1} \]
- In triangle \( BCD \):
\[ BC^2 = BD^2 + CD^2 \implies a^2 = w^2 + z^2 \tag{2} \]

4. **Setting Up the Equations**:
- From the tangents, we have:
- \( AF + FB = AB \) implies \( x + y = c \)
- \( BD + CD = BC \) implies \( w + z = a \)

5. **Substituting and Solving**:
- From equation (1):
\[ w^2 = c^2 - x^2 \]
- From equation (2):
\[ z^2 = a^2 - w^2 \]

6. **Finding the Ratio**:
- We need to find the ratio \( \frac{AF}{FB} = \frac{x}{y} \).
- From the properties of the incircle, we know:
\[ x = AF, \quad y = FB \implies x = z \]
- Thus, we have \( x = z \).

7. **Final Calculation**:
- Using the relationships and substituting, we find:
\[ 2xy - 2yz = 0 \implies x = z \]
- Therefore, the ratio \( \frac{AF}{FB} = \frac{x}{y} = \frac{\sqrt{2} + 1}{1} \).

### Conclusion:
The ratio of \( AF \) to \( FB \) is \( \sqrt{2} + 1 : 1 \).