1. Stating the problem: We have triangle ABC with points D and E on segment AD, where segment EB is parallel to segment DC. 2. Given: $AB = 12$, $AC = 16$, and $ED = 5$. 3. Goal: Find the length $AE$. 4. Because $EB$ is parallel to $DC$, triangle $AEB$ is similar to triangle $ADC$ by the parallel lines similarity criterion. 5. Similar triangles imply corresponding sides are proportional. Therefore, $$\frac{AE}{AD} = \frac{AB}{AC}$$ 6. We know $AD = AE + ED$. Let $AE = x$. Then $AD = x + 5$. 7. Substitute these into the proportion: $$\frac{x}{x + 5} = \frac{12}{16} = \frac{3}{4}$$ 8. Cross-multiply and solve for $x$: $$4x = 3(x + 5)$$ $$4x = 3x + 15$$ $$4x - 3x = 15$$ $$x = 15$$ 9. Therefore, the length $AE$ is $15$ units. Final answer: $$AE = 15$$