1. **Problem statement:** Given triangles $\triangle ABC$ and $\triangle ADE$ are similar, find $x$. 2. **Known sides:** - $AB = 8$ - $AD = 20$ - $AC = 2x + 3$ - $DE = 15$ (perpendicular to $AB$) 3. **Similarity rule:** Corresponding sides of similar triangles are proportional: $$\frac{AB}{AD} = \frac{AC}{AE} = \frac{BC}{DE}$$ 4. Since $DE$ is perpendicular to $AB$, $DE$ corresponds to $BC$ in $\triangle ABC$. 5. Using the ratio of sides $AB$ and $AD$: $$\frac{AB}{AD} = \frac{8}{20} = \frac{2}{5}$$ 6. Using the ratio for $AC$ and $AE$: $$\frac{AC}{AE} = \frac{2x + 3}{AE} = \frac{2}{5}$$ 7. To find $AE$, note that $AE = AD + DE$ is not correct since $E$ lies on $DE$ perpendicular to $AB$. Instead, use the similarity ratio for $DE$ and $BC$: $$\frac{DE}{BC} = \frac{15}{BC} = \frac{2}{5}$$ 8. Solve for $BC$: $$15 \times 5 = 2 \times BC \Rightarrow 75 = 2 BC \Rightarrow BC = \frac{75}{2} = 37.5$$ 9. Now, use the similarity ratio for $AC$ and $AE$: $$\frac{2x + 3}{AE} = \frac{2}{5}$$ 10. Since $AE$ corresponds to $AC$ in the smaller triangle, and $AE$ is part of $AD$, but $AD$ is given as 20, so $AE$ must be 8 (corresponding to $AB$), so $AE = 8$. 11. Substitute $AE = 8$: $$\frac{2x + 3}{8} = \frac{2}{5}$$ 12. Cross multiply: $$5(2x + 3) = 2 \times 8$$ $$10x + 15 = 16$$ 13. Solve for $x$: $$10x = 16 - 15$$ $$10x = 1$$ $$x = \frac{1}{10} = 0.1$$ **Final answer:** $x = 0.1$