1. **Problem statement:** Given that $AD$ intersects $BE$ at $C$, and $AB \parallel DE$, with $CD=6.6$ cm, $DE=3.4$ cm, and $BC=5.25$ cm, find the length of $AC$ to the nearest hundredth of a centimeter. 2. **Key concept:** Since $AB \parallel DE$, triangles $ABC$ and $CDE$ are similar by the AA (Angle-Angle) similarity criterion. 3. **Similarity ratios:** Corresponding sides of similar triangles are proportional: $$\frac{AC}{CD} = \frac{BC}{DE}$$ 4. **Substitute known values:** $$\frac{AC}{6.6} = \frac{5.25}{3.4}$$ 5. **Solve for $AC$:** $$AC = 6.6 \times \frac{5.25}{3.4}$$ 6. **Calculate the fraction:** $$\frac{5.25}{3.4} \approx 1.5441$$ 7. **Calculate $AC$:** $$AC = 6.6 \times 1.5441 = 10.1909$$ 8. **Round to nearest hundredth:** $$AC \approx 10.19 \text{ cm}$$ **Final answer:** The length of $AC$ is approximately $10.19$ cm.