1. **State the problem:** We are given a right triangle ABC with right angle at B. We know \( \cos C = \frac{4}{5} \), vertical side CB = 16, and a segment ED inside the triangle parallel to AB with length 6. We need to find the length of AE. 2. **Analyze the triangle:** Since \(\cos C = \frac{4}{5}\), triangle ABC is a right triangle with angle C adjacent side to CB and hypotenuse AC. Recall \(\cos C = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{CB}{AC} = \frac{4}{5} \). 3. **Find hypotenuse AC:** \[ AC = \frac{CB}{\cos C} = \frac{16}{\frac{4}{5}} = 16 \times \frac{5}{4} = 20. \] 4. **Find side AB:** Using Pythagoras, \( AB = \sqrt{AC^2 - CB^2} = \sqrt{20^2 - 16^2} = \sqrt{400 - 256} = \sqrt{144} = 12. \) 5. **Interpret the line segment ED:** ED is parallel to AB, which means triangles CED and CAB are similar. 6. **Use similarity:** Similarity ratio between smaller triangle CED and larger triangle CAB: \[ \frac{ED}{AB} = \frac{6}{12} = \frac{1}{2}. \] 7. **Find AE:** Because triangles are similar, the segments on AC satisfy: \[ \frac{AE}{AC} = \frac{1}{2} \implies AE = \frac{AC}{2} = \frac{20}{2} = 10. \] **Final answer:** The length of AE is \(\boxed{10}\).