D,E and F are the points on sides BC, CA and AB respectively, of `triangleABC` such that AD bisects `angleA`, BE bisects `angleB` and CF bisects `angleC`. If AB=5 cm, BC= 8cm and CA= 4 cm. find AF, CE and BD.

Answer

To solve the problem, we will use the Angle Bisector Theorem, which states that the ratio of the lengths of the two segments created by an angle bisector is equal to the ratio of the lengths of the other two sides of the triangle.

### Step-by-Step Solution:

1. **Identify the segments and sides**:
- Let \( AB = c = 5 \, \text{cm} \)
- Let \( BC = a = 8 \, \text{cm} \)
- Let \( CA = b = 4 \, \text{cm} \)
- We need to find \( AF, CE, \) and \( BD \).

2. **Finding \( BD \)**:
- By the Angle Bisector Theorem for \( \triangle ABC \) at vertex \( A \):
\[ \frac{BD}{DC} = \frac{AB}{AC} = \frac{c}{b} = \frac{5}{4} \]
- Let \( BD = x \) and \( DC = 8 - x \) (since \( BC = 8 \, \text{cm} \)).
- Therefore, we can set up the equation:
\[ \frac{x}{8 - x} = \frac{5}{4} \]
- Cross-multiplying gives:
\[ 4x = 5(8 - x) \]
- Expanding:
\[ 4x = 40 - 5x \]
- Adding \( 5x \) to both sides:
\[ 9x = 40 \]
- Solving for \( x \):
\[ x = \frac{40}{9} \, \text{cm} \]
- Thus, \( BD = \frac{40}{9} \, \text{cm} \).

3. **Finding \( CE \)**:
- By the Angle Bisector Theorem for \( \triangle ABC \) at vertex \( B \):
\[ \frac{CE}{EA} = \frac{BC}{AB} = \frac{8}{5} \]
- Let \( CE = y \) and \( EA = 4 - y \) (since \( CA = 4 \, \text{cm} \)).
- Therefore, we can set up the equation:
\[ \frac{y}{4 - y} = \frac{8}{5} \]
- Cross-multiplying gives:
\[ 5y = 8(4 - y) \]
- Expanding:
\[ 5y = 32 - 8y \]
- Adding \( 8y \) to both sides:
\[ 13y = 32 \]
- Solving for \( y \):
\[ y = \frac{32}{13} \, \text{cm} \]
- Thus, \( CE = \frac{32}{13} \, \text{cm} \).

4. **Finding \( AF \)**:
- By the Angle Bisector Theorem for \( \triangle ABC \) at vertex \( C \):
\[ \frac{AF}{FB} = \frac{CA}{BC} = \frac{4}{8} = \frac{1}{2} \]
- Let \( AF = z \) and \( FB = 5 - z \) (since \( AB = 5 \, \text{cm} \)).
- Therefore, we can set up the equation:
\[ \frac{z}{5 - z} = \frac{1}{2} \]
- Cross-multiplying gives:
\[ 2z = 5 - z \]
- Adding \( z \) to both sides:
\[ 3z = 5 \]
- Solving for \( z \):
\[ z = \frac{5}{3} \, \text{cm} \]
- Thus, \( AF = \frac{5}{3} \, \text{cm} \).

### Final Results:
- \( BD = \frac{40}{9} \, \text{cm} \)
- \( CE = \frac{32}{13} \, \text{cm} \)
- \( AF = \frac{5}{3} \, \text{cm} \)