1. The problem asks to find all segments on line AF that are exactly two-fifths the length of AF. 2. Since points A, B, C, D, E, F lie on AF, the length AF is divided into segments by these points. 3. The construction with circles suggests equal spacing or segments related by compass and straightedge rules. 4. To find segments equal to $\frac{2}{5}$ of AF, we consider the length $AF$ and calculate $\frac{2}{5} AF$. 5. If the points divide AF into equal parts, then segments spanning two of these parts will be $\frac{2}{5} AF$. 6. From the labeling, segments AB, BC, CD, DE, EF are likely equal parts, each $\frac{1}{5} AF$. 7. Therefore, segments spanning two consecutive parts are: AC, BD, CE, DF. 8. Each of these segments is $AB + BC = \frac{1}{5}AF + \frac{1}{5}AF = \frac{2}{5}AF$. 9. Hence, the segments that are two-fifths the length of AF are AC, BD, CE, and DF. Final answer: AC, BD, CE, DF