1. **Problem Statement:** We need to find all segments that are two-thirds the length of $\overline{AG}$. 2. **Understanding the setup:** There are five congruent circles arranged vertically, intersecting at points $B, C, D, E, F, G$ on a vertical line. Points $B$ through $G$ are collinear. 3. **Key insight:** Since the circles are congruent and arranged vertically, the distances between consecutive intersection points are equal. Let the length of each segment between consecutive points be $x$. 4. **Expressing $\overline{AG}$:** The segment $\overline{AG}$ spans from $A$ (top) to $G$ (bottom). Since $B$ through $G$ are points on the vertical line, and assuming $A$ is above $B$, the total length $\overline{AG}$ can be expressed as a sum of segments: $$\overline{AG} = \overline{AB} + \overline{BC} + \overline{CD} + \overline{DE} + \overline{EF} + \overline{FG}$$ 5. **Using congruency:** Because the circles are congruent and arranged vertically, the segments $\overline{BC}, \overline{CD}, \overline{DE}, \overline{EF}, \overline{FG}$ are all equal to $x$. 6. **Length of $\overline{AG}$ in terms of $x$:** $$\overline{AG} = \overline{AB} + 5x$$ 7. **Finding segments equal to $\frac{2}{3} \overline{AG}$:** We want segments whose length is: $$\frac{2}{3} \overline{AG} = \frac{2}{3} (\overline{AB} + 5x)$$ 8. **Identifying such segments:** Since $\overline{AB}$ is unknown, but the problem likely implies $\overline{AB} = x$ (since $A$ is the top of the first circle), then: $$\overline{AG} = 6x$$ So, $$\frac{2}{3} \overline{AG} = \frac{2}{3} \times 6x = 4x$$ 9. **Segments of length $4x$:** These are segments spanning 4 consecutive points along the vertical line: - $\overline{AC}$ (from $A$ to $C$): length $2x$ (too short) - $\overline{AD}$ (from $A$ to $D$): length $3x$ (too short) - $\overline{AE}$ (from $A$ to $E$): length $4x$ (correct) - $\overline{BF}$ (from $B$ to $F$): length $4x$ (correct) - $\overline{CG}$ (from $C$ to $G$): length $4x$ (correct) 10. **Answer:** The segments that are two-thirds the length of $\overline{AG}$ are: $$\overline{AE}, \overline{BF}, \overline{CG}$$ --- **Final answer:** $\overline{AE}, \overline{BF}, \overline{CG}$