1. **Problem statement:** Given lines AC and DE are parallel, and DC is perpendicular to both AC and DE. Segment BD has length $\frac{4}{3}$. We need to find the length of side DE. 2. **Given:** - $AC \parallel DE$ - $DC \perp AC$ and $DC \perp DE$ - $BD = \frac{4}{3}$ - $DC = 4$ - $AB = 5$ 3. **Understanding the figure:** - Since DC is perpendicular to both AC and DE, DC is vertical, and AC and DE are horizontal. - Triangle ABD is right-angled at D because DC is vertical and BD is diagonal. 4. **Use the Pythagorean theorem in triangle ABD:** $$AB^2 = AD^2 + BD^2$$ Given $AB=5$ and $BD=\frac{4}{3}$, solve for $AD$: $$AD^2 = AB^2 - BD^2 = 5^2 - \left(\frac{4}{3}\right)^2 = 25 - \frac{16}{9} = \frac{225}{9} - \frac{16}{9} = \frac{209}{9}$$ $$AD = \sqrt{\frac{209}{9}} = \frac{\sqrt{209}}{3}$$ 5. **Find length DE:** - Since AC and DE are parallel and DC is perpendicular, the horizontal distance between points D and E equals the horizontal distance between points A and B minus the horizontal distance between A and D. - The length DE equals the horizontal distance from D to E. - The horizontal distance from A to B is $AB = 5$. - The horizontal distance from A to D is $AD = \frac{\sqrt{209}}{3}$. 6. **Calculate DE:** $$DE = AB - AD = 5 - \frac{\sqrt{209}}{3}$$ 7. **Approximate $\sqrt{209}$:** $$\sqrt{209} \approx 14.4568$$ 8. **Calculate DE numerically:** $$DE \approx 5 - \frac{14.4568}{3} = 5 - 4.819 = 0.181$$ 9. **Compare with options:** - A. $\frac{1}{3} \approx 0.333$ - B. $1$ - C. $3$ - D. $6$ The closest value to our calculation is approximately $0.181$, which is closest to option A ($\frac{1}{3}$). **Final answer:** $\boxed{\frac{1}{3}}$