In triangle $ABC$, side $AC$ is $8$ cm. Two segments are drawn parallel to $AC$ that have their ends on $AB$ and $BC$ and that divide the triangle into three parts of equal area. What is the length of the parallel segment closest to $AC$?
1. **Set up the problem with coordinates:**
   Let the coordinates of the vertices of triangle \(ABC\) be \(A(0,0)\), \(B(a,b)\), and \(C(8,0)\).

2. **Determine the area of triangle \(ABC\):**
   The area of triangle \(ABC\) can be calculated using the formula for the area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):
   \[
   \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
   \]
   Substituting the coordinates of \(A\), \(B\), and \(C\):
   \[
   \text{Area} = \frac{1}{2} \left| 0(b - 0) + a(0 - 0) + 8(0 - b) \right| = \frac{1}{2} \left| -8b \right| = 4b
   \]

3. **Divide the triangle into three parts of equal area:**
   Since the total area of triangle \(ABC\) is \(4b\), each of the three parts must have an area of \(\frac{4b}{3}\).

4. **Determine the positions of the parallel segments:**
   Let the parallel segments be \(DE\) and \(FG\), where \(D\) and \(F\) are on \(AB\), and \(E\) and \(G\) are on \(BC\). The segments \(DE\) and \(FG\) divide the triangle into three parts of equal area.

5. **Find the coordinates of points \(D\) and \(F\):**
   Points \(D\) and \(F\) on \(AB\) can be parameterized as \(D(k, \frac{bk}{a})\) and \(F(l, \frac{bl}{a})\) with \(k < l\).

6. **Use the area condition to find \(k\) and \(l\):**
   The area of the smaller triangle \(ADE\) should be \(\frac{4b}{3}\). The area of triangle \(ADE\) is:
   \[
   \text{Area of } ADE = \frac{1}{2} \times k \times \frac{bk}{a} = \frac{bk^2}{2a}
   \]
   Setting this equal to \(\frac{4b}{3}\):
   \[
   \frac{bk^2}{2a} = \frac{4b}{3} \implies k^2 = \frac{8a}{3} \implies k = a \sqrt{\frac{8}{3}}
   \]

7. **Find the length of the segment \(DE\):**
   The length of segment \(DE\) is the horizontal distance between the points on \(AB\) and \(BC\) at the same height. Since \(DE\) is parallel to \(AC\), its length is proportional to the height at which it is drawn:
   \[
   \text{Length of } DE = \frac{8 \sqrt{6}}{3}
   \]

8. **Find the length of the segment \(FG\):**
   Similarly, the length of segment \(FG\) is:
   \[
   \text{Length of } FG = \frac{8 \sqrt{3}}{3}
   \]

9. **Determine which segment is closest to \(AC\):**
   The segment closest to \(AC\) is the one with the smaller height, which corresponds to the segment \(FG\).

The final answer is \(\boxed{\frac{8 \sqrt{3}}{3}}\).