1. **Stating the problem:** We have three parallel lines $l_1 \parallel l_2 \parallel l_3$ and three concurrent lines $l_4, l_5,$ and $l_6$ intersecting them. Points $A$ on $l_1$, $B$ and $C$ on $l_2$, and $D$ and $E$ on $l_3$ are arranged such that $A-B-D$ lie on one concurrent line and $A-C-E$ lie on another. We want to verify the given ratios and equalities involving segments on these lines. 2. **Formula and important rule:** When three parallel lines are cut by two or more transversals, the segments intercepted on the transversals are proportional. This is a direct consequence of the Basic Proportionality Theorem (Thales' theorem). 3. **Step-by-step verification:** - Given $l_1 \parallel l_2 \parallel l_3$ and concurrent lines $l_4, l_5, l_6$, the segments satisfy: $$\frac{AE}{BC} = \frac{AC}{DE}$$ - Similarly, for the other ratios, by applying the proportionality of segments intercepted by parallel lines on transversals, we get: $$\frac{AB}{AD} = \frac{AE}{AC}$$ $$\frac{AD}{AC} = \frac{BD}{CE}$$ $$\frac{AC}{CE} = \frac{BD}{BA}$$ $$\frac{BC}{CE} = \frac{DE}{BD}$$ $$\frac{AB}{AC} = \frac{AD}{AE}$$ 4. **Explanation:** Each equality comes from the fact that the ratios of segments on one transversal equal the ratios on the other transversal because the lines are parallel and the other lines are concurrent, creating similar triangles. 5. **Intermediate work example:** For the first equality, $$\frac{AE}{BC} = \frac{AC}{DE}$$ This can be seen by considering triangles $A E D$ and $B C D$ which are similar by AA similarity (corresponding angles equal due to parallel lines and concurrency). 6. **Final answer:** All the given equalities hold true by the properties of parallel lines and concurrency, confirming the proportionality of the segments as stated.