Let C be incircle of `DeltaABC`. If the tangents of lengths `t_(1),t_(2) and t_(3)` are drawn inside the given triangle parallel to side a,b, and c, respectively, then `(t_(1))/(a) + (t_(2))/(b) + (t_(3))/(c)` is equal to

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the problem, we need to find the value of \(\frac{t_1}{a} + \frac{t_2}{b} + \frac{t_3}{c}\) where \(t_1\), \(t_2\), and \(t_3\) are the lengths of the tangents drawn parallel to the sides \(a\), \(b\), and \(c\) of triangle \(ABC\) respectively.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Let \(C\) be the in-circle of triangle \(ABC\).
- The sides of the triangle opposite to angles \(A\), \(B\), and \(C\) are denoted as \(a\), \(b\), and \(c\) respectively.
- The radius of the in-circle is denoted as \(r\).

2. **Identifying the Heights**:
- The height of triangle \(ABC\) from vertex \(A\) is denoted as \(h\).
- When we draw a tangent \(t_1\) parallel to side \(BC\), the height of triangle \(AP_1P_2\) (where \(P_1\) and \(P_2\) are points on the sides \(AB\) and \(AC\)) is \(h - 2r\).

3. **Using Similar Triangles**:
- Triangles \(AP_1P_2\) and \(ABC\) are similar.
- Therefore, we can set up the ratio:
\[ \frac{t_1}{a} = \frac{h - 2r}{h} \]
- Rearranging gives:
\[ t_1 = a \left(1 - \frac{2r}{h}\right) \]

4. **Substituting for \(r\)**:
- The in-radius \(r\) can be expressed as \(r = \frac{\Delta}{s}\), where \(\Delta\) is the area of the triangle and \(s\) is the semi-perimeter.
- The area \(\Delta\) can also be expressed as \(\Delta = \frac{1}{2}ah\).
- Therefore, substituting for \(r\):
\[ r = \frac{\frac{1}{2}ah}{s} \]

5. **Finding \(t_1\)**:
- Substituting \(r\) into the equation for \(t_1\):
\[ t_1 = a \left(1 - \frac{2 \cdot \frac{1}{2}ah/s}{h}\right) = a \left(1 - \frac{a}{s}\right) \]
- Thus, we have:
\[ \frac{t_1}{a} = 1 - \frac{a}{s} \]

6. **Repeating for \(t_2\) and \(t_3\)**:
- Similarly, we can derive:
\[ \frac{t_2}{b} = 1 - \frac{b}{s} \]
\[ \frac{t_3}{c} = 1 - \frac{c}{s} \]

7. **Combining the Results**:
- Now, we sum these results:
\[ \frac{t_1}{a} + \frac{t_2}{b} + \frac{t_3}{c} = \left(1 - \frac{a}{s}\right) + \left(1 - \frac{b}{s}\right) + \left(1 - \frac{c}{s}\right) \]
- This simplifies to:
\[ 3 - \left(\frac{a + b + c}{s}\right) \]

8. **Using the Semi-Perimeter**:
- The semi-perimeter \(s\) is defined as \(s = \frac{a + b + c}{2}\).
- Therefore:
\[ \frac{a + b + c}{s} = 2 \]
- Substituting this back gives:
\[ 3 - 2 = 1 \]

### Final Answer:
\[ \frac{t_1}{a} + \frac{t_2}{b} + \frac{t_3}{c} = 1 \]