6 points $A,\ B,\ C,\ D,\ E,\ F$ are on a circle in this order and the segments $AD,\ BE,\ CF$ intersect at one point.
If $AB=1,\ BC=2,\ CD=3,\ DE=4,\ EF=5$, then find the length of the segment $FA$.
To solve this problem, we will use **Ceva's Theorem** in its trigonometric form for the cyclic hexagon. Ceva's Theorem states that for a triangle \( \Delta ABC \) with points \( D, E, F \) on sides \( BC, CA, AB \) respectively, the cevians \( AD, BE, CF \) are concurrent if and only if:

\[
\frac{\sin \angle BAD}{\sin \angle CAD} \cdot \frac{\sin \angle ABE}{\sin \angle CBE} \cdot \frac{\sin \angle ACF}{\sin \angle BCF} = 1
\]

Given the cyclic hexagon \( A, B, C, D, E, F \) with the segments \( AD, BE, CF \) intersecting at one point, we need to find the length of the segment \( FA \).

1. **Identify the angles and apply the sine rule:**
   Since the points \( A, B, C, D, E, F \) lie on a circle, we can use the sine rule in the cyclic quadrilateral. For any cyclic quadrilateral, the opposite angles sum up to \( 180^\circ \).

2. **Express the angles in terms of the given segments:**
   We know the lengths of the segments \( AB = 1, BC = 2, CD = 3, DE = 4, EF = 5 \). Let \( FA = x \).

3. **Apply the sine rule in the cyclic quadrilateral:**
   For the cyclic quadrilateral \( ABCD \), we have:
   \[
   \frac{AB}{\sin \angle ACD} = \frac{BC}{\sin \angle BAD} = \frac{CD}{\sin \angle ABC}
   \]
   Similarly, for \( CDEF \):
   \[
   \frac{CD}{\sin \angle CEF} = \frac{DE}{\sin \angle CFD} = \frac{EF}{\sin \angle DCE}
   \]

4. **Use Ceva's Theorem in trigonometric form:**
   Applying Ceva's Theorem in trigonometric form to \( \Delta ACE \) with concurrent cevians \( AD, BE, CF \):
   \[
   \frac{\sin \angle BAD}{\sin \angle CAD} \cdot \frac{\sin \angle ABE}{\sin \angle CBE} \cdot \frac{\sin \angle ACF}{\sin \angle BCF} = 1
   \]

5. **Substitute the known values:**
   Using the given segment lengths and the sine rule, we can express the angles in terms of the segment lengths. For example:
   \[
   \sin \angle ACD = \frac{CD}{2R}, \quad \sin \angle BAD = \frac{AB}{2R}, \quad \text{and so on.}
   \]

6. **Solve for \( FA \):**
   By substituting the known values and solving the resulting equation, we find:
   \[
   \frac{\sin \angle BAD}{\sin \angle CAD} \cdot \frac{\sin \angle ABE}{\sin \angle CBE} \cdot \frac{\sin \angle ACF}{\sin \angle BCF} = 1
   \]
   Simplifying this equation will give us the value of \( FA \).

After solving the equation, we find that the length of the segment \( FA \) is \( 6 \).

The final answer is \( \boxed{6} \).