The lengths of the sides `CB` and `CA` of a triangle `ABC` are given by `a` and ` b` and the angle `C` is `(2pi)/(3)`. The line `CD` bisects the angle `C` and meets `AB` at `D`. Then the length of `CD` is : (a) `(1)/(a+b)` (b) `(a^(2)+b^(2))/(a+b)` (c) `(ab)/(2(a+b))` (d) `(ab)/(a+b)`

Options

A. `(1)/(a+b)`
B. `(a^(2)+b^(2))/(a+b)`
C. `(ab)/(2(a+b))`
D. `(ab)/(a+b)`

Answer

To find the length of the angle bisector \( CD \) in triangle \( ABC \), where \( CB = a \), \( CA = b \), and \( \angle C = \frac{2\pi}{3} \), we can use the angle bisector theorem and some trigonometric properties.

### Step-by-Step Solution:

1. **Draw the Triangle**:
- Sketch triangle \( ABC \) with \( CB = a \), \( CA = b \), and \( \angle C = \frac{2\pi}{3} \).
- Mark the angle bisector \( CD \) which divides \( \angle C \) into two equal angles of \( \frac{\pi}{3} \).

2. **Use Cosine Rule**:
- We can find the lengths of the segments \( AD \) and \( DB \) using the angle bisector theorem, which states that:
\[ \frac{AD}{DB} = \frac{CA}{CB} = \frac{b}{a} \]
- Let \( AD = k \) and \( DB = m \). Then, we have:
\[ \frac{k}{m} = \frac{b}{a} \implies k = \frac{b}{a} m \]

3. **Express Total Length \( AB \)**:
- The total length \( AB \) can be expressed as:
\[ AB = AD + DB = k + m = \frac{b}{a} m + m = m \left( \frac{b}{a} + 1 \right) = m \left( \frac{b + a}{a} \right) \]
- Thus, we can express \( m \) as:
\[ m = \frac{AB \cdot a}{b + a} \]

4. **Find Length of Angle Bisector \( CD \)**:
- The formula for the length of the angle bisector \( CD \) is given by:
\[ CD = \frac{2ab}{a + b} \cdot \cos\left(\frac{C}{2}\right) \]
- Since \( C = \frac{2\pi}{3} \), we have:
\[ \frac{C}{2} = \frac{\pi}{3} \]
- Therefore, \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \).

5. **Substitute Values**:
- Substituting the values into the angle bisector formula gives:
\[ CD = \frac{2ab}{a + b} \cdot \frac{1}{2} = \frac{ab}{a + b} \]

6. **Final Result**:
- Thus, the length of \( CD \) is:
\[ CD = \frac{ab}{a + b} \]

### Conclusion:
The correct option is (d) \( \frac{ab}{a + b} \).