1. **State the problem:** We are given two similar triangles, $\triangle CDE \sim \triangle CAB$, and need to find the length $EB$. 2. **Identify known lengths:** - In $\triangle CAB$, $CA=24$, $AB=18$, $CB=30$. - In $\triangle CDE$, $DE=14$. 3. **Set up the proportion using similarity:** Corresponding sides of similar triangles are proportional: $$\frac{DE}{AB} = \frac{CE}{CB} = \frac{CD}{CA}$$ 4. **Find $EB$:** Since $E$ lies on $AB$, and $AB=18$, then $EB = AB - AE$. 5. **Find $AE$ using the proportion:** $$\frac{DE}{AB} = \frac{CE}{CB}$$ We know $DE=14$, $AB=18$, $CB=30$. 6. **Express $CE$ in terms of $AE$:** Since $E$ is on $AB$, $CE = CB - EB = CB - (AB - AE) = 30 - (18 - AE) = 12 + AE$. 7. **Set up the proportion:** $$\frac{14}{18} = \frac{12 + AE}{30}$$ 8. **Solve for $AE$:** Multiply both sides by 30: $$30 \times \frac{14}{18} = 12 + AE$$ Simplify left side: $$\frac{30 \times 14}{18} = 12 + AE$$ $$\frac{420}{18} = 12 + AE$$ Simplify fraction: $$\frac{\cancel{420}}{\cancel{18}} = 12 + AE$$ Since $420 \div 6 = 70$ and $18 \div 6 = 3$, $$\frac{70}{3} = 12 + AE$$ 9. **Isolate $AE$:** $$AE = \frac{70}{3} - 12 = \frac{70}{3} - \frac{36}{3} = \frac{34}{3}$$ 10. **Find $EB$:** $$EB = AB - AE = 18 - \frac{34}{3} = \frac{54}{3} - \frac{34}{3} = \frac{20}{3}$$ **Final answer:** $$EB = \frac{20}{3} \approx 6.67$$