1. **Problem statement:** Given that segment $A'C'$ is parallel to segment $AC$, find the length of segment $AC'$. Known lengths are $AA' = 6$, $A'B = 8$, $BC' = 10$, and $CC' = 5$. 2. **Understanding the problem:** Since $A'C'$ is parallel to $AC$, triangles $AA'C'$ and $ABC$ are similar by the Basic Proportionality Theorem (Thales' theorem). 3. **Set up the similarity ratio:** The ratio of corresponding sides in similar triangles is equal. We know $AA' = 6$ and $A'B = 8$, so the ratio of $AA'$ to $AB$ is: $$\frac{AA'}{AB} = \frac{6}{6 + 8} = \frac{6}{14} = \frac{3}{7}$$ 4. **Apply the ratio to find $AC'$:** Since $A'C'$ is parallel to $AC$, the segment $AC'$ corresponds to $AC$ scaled by the same ratio. The length $AC$ is the sum of $BC$ and $CC'$, but $BC$ is not given directly. However, $BC' = 10$ and $CC' = 5$, so $BC = BC' - CC' = 10 - 5 = 5$. 5. **Calculate $AC$:** $$AC = AB + BC = (AA' + A'B) + BC = (6 + 8) + 5 = 14 + 5 = 19$$ 6. **Calculate $AC'$ using the ratio:** $$AC' = \frac{3}{7} \times AC = \frac{3}{7} \times 19 = \frac{57}{7} = 8.142857... \approx 8.5$$ 7. **Final answer:** The length of segment $AC'$ is approximately $8.5$ units.