1. **Problem Statement:** We have two triangles, $\triangle ABC$ and $\triangle DEC$, with $AB \parallel CD$ and $DE \parallel CA$. Given side lengths $AB = 20$ m, $DC = 14$ m, $DE = 11.2$ m, and $AC = x$ m, we need to: a) Identify the similarity criterion used. b) Find the value of $x$. c) Given the perimeter of $\triangle ABC = 50$ m, find the length $CE$. 2. **Similarity Criterion:** Since $AB \parallel CD$ and $DE \parallel CA$, corresponding angles are equal by the Alternate Interior Angles Theorem. Thus, $\triangle ABC \sim \triangle DEC$ by the **AA (Angle-Angle) similarity criterion**. 3. **Using similarity, corresponding sides are proportional:** $$\frac{AB}{DC} = \frac{BC}{EC} = \frac{AC}{DE}$$ Given: $$\frac{20}{14} = \frac{x}{11.2}$$ 4. **Find $x$:** Cross-multiply: $$20 \times 11.2 = 14 \times x$$ $$224 = 14x$$ Divide both sides by 14: $$\frac{224}{\cancel{14}} = \frac{14x}{\cancel{14}}$$ $$16 = x$$ So, $x = 16$ m. 5. **Find $CE$ given perimeter of $\triangle ABC = 50$ m:** Perimeter $P = AB + BC + AC = 50$ We know $AB = 20$, $AC = 16$, so: $$20 + BC + 16 = 50$$ $$BC = 50 - 36 = 14$$ Using similarity ratio for $BC$ and $EC$: $$\frac{BC}{EC} = \frac{AB}{DC} = \frac{20}{14} = \frac{10}{7}$$ So, $$\frac{14}{EC} = \frac{10}{7}$$ Cross-multiply: $$14 \times 7 = 10 \times EC$$ $$98 = 10 EC$$ Divide both sides by 10: $$\frac{98}{\cancel{10}} = \frac{10 EC}{\cancel{10}}$$ $$9.8 = EC$$ So, $CE = 9.8$ m. --- **Final answers:** - a) Similarity criterion: AA (Angle-Angle) - b) $x = 16$ m - c) $CE = 9.8$ m