1. **Problem Statement:** We are given two similar triangles, $\triangle ABC$ and $\triangle BDE$. We need to determine if the areas of $\triangle ABL$ and $\triangle ALC$ are equal (assuming $L$ is a point on $AC$ or $AB$ as per the problem context, but since $L$ is not defined in the problem, we focus on the given parts a) and b)). 2. **Given:** - $\triangle ABC \sim \triangle BDE$ - $BD = 7$, $BE = 8$, $EC = 2$ - $DE = 10$ cm (for part b) 3. **Part a) Find length $DA$:** Since $D$ lies on $AB$, and $BD = 7$, let $AB = x$. Then $DA = AB - BD = x - 7$. 4. **Using similarity ratios:** From similarity $\triangle ABC \sim \triangle BDE$, corresponding sides are proportional: $$\frac{AB}{BD} = \frac{BC}{DE} = \frac{AC}{BE}$$ 5. **Calculate $AB$ using $BC$ and $DE$:** We know $BE = 8$, $EC = 2$, so $BC = BE + EC = 8 + 2 = 10$. Using the ratio: $$\frac{AB}{BD} = \frac{BC}{DE} \Rightarrow \frac{AB}{7} = \frac{10}{10} = 1$$ So, $$AB = 7$$ 6. **Find $DA$:** $$DA = AB - BD = 7 - 7 = 0$$ This implies $D$ coincides with $A$, which may be a special case or a misinterpretation. If $D$ lies between $A$ and $B$, and $BD=7$, $AB=7$, then $D$ is at $A$. 7. **Part b) Find length $AC$ given $DE=10$ cm:** Using similarity ratio: $$\frac{AC}{BE} = \frac{BC}{DE}$$ Substitute known values: $$\frac{AC}{8} = \frac{10}{10} = 1$$ So, $$AC = 8$$ **Final answers:** - a) $DA = 0$ (point $D$ coincides with $A$) - b) $AC = 8$ cm **Note:** The question about areas of $\triangle ABL$ and $\triangle ALC$ cannot be answered without definition or position of point $L$.