Let s denote the semiperimeter of a triangle ABC in which BC = a, CA = b, AB = c. If a circle touches the sides BC, CA, AB at D, E and F respectively, find BD.

Options

A. s - b
B. 2s + b
C. b + s
D. 3b - s

Answer

To find the length of segment BD in triangle ABC where a circle touches the sides BC, CA, and AB at points D, E, and F respectively, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Semi-Perimeter (s)**:
The semi-perimeter \( s \) of triangle ABC is given by the formula:
\[ s = \frac{a + b + c}{2} \]
where \( a = BC \), \( b = CA \), and \( c = AB \).

2. **Identify the Tangent Segments**:
When a circle is inscribed in triangle ABC, the lengths of the tangent segments from each vertex to the points of tangency are equal. Therefore, we can denote:
- \( BD = BF = x \) (tangent from B to the circle)
- \( CE = CD = y \) (tangent from C to the circle)
- \( AF = AE = z \) (tangent from A to the circle)

3. **Express the Sides in Terms of Tangents**:
From the definitions of the tangent segments, we can express the sides of the triangle as:
- \( a = BC = BD + DC = x + y \)
- \( b = CA = AF + EC = z + y \)
- \( c = AB = AE + FB = z + x \)

4. **Set Up the Equation for Semi-Perimeter**:
Using the semi-perimeter formula:
\[ s = \frac{a + b + c}{2} \]
Substitute the expressions for \( a \), \( b \), and \( c \):
\[ s = \frac{(x + y) + (z + y) + (z + x)}{2} \]
Simplifying this, we get:
\[ s = \frac{2x + 2y + 2z}{2} = x + y + z \]

5. **Find BD**:
We know that:
\[ BD = x \]
From the semi-perimeter equation, we can express \( BD \) in terms of \( s \):
\[ BD = s - b \]
where \( b = z + y \). Thus, we can substitute:
\[ BD = s - (z + y) \]

6. **Final Expression**:
Since \( s = x + y + z \), we can substitute this into the equation:
\[ BD = (x + y + z) - (z + y) = x \]

### Conclusion:
Thus, the length of segment \( BD \) is given by:
\[ BD = s - b \]