1. **State the problem:** We need to prove that if three parallel lines cut off equal segments on one transversal, then they cut off equal segments on any other transversal. 2. **Given:** Three parallel lines $l_1, l_2, l_3$ and two transversals $AC$ and $DF$. Segments $|AB| = |BC|$ on transversal $AC$. Lines $AZ \parallel DF$ and $XZ \parallel AC$ are drawn. 3. **Goal:** Prove $|DE| = |EF|$ on transversal $DF$. 4. **Step 1:** Given $|YZ| = |BC|$ and $|BC| = |AB|$, so by transitivity: $$|YZ| = |AB|$$ 5. **Step 2:** Angles $\angle BAW$ and $\angle WZY$ are equal because $AZ \parallel DF$ and $l_1, l_2, l_3$ are parallel lines cut by transversal $AWZ$ (alternate interior angles). 6. **Step 3:** Angles $\angle AWB$ and $\angle ZWY$ are equal because $XZ \parallel AC$ and $l_1, l_2, l_3$ are parallel lines cut by transversal $WBY$ (alternate interior angles). 7. **Step 4:** From the above, $\angle ABW = \angle ZYW$ by corresponding angles or angle subtraction. 8. **Step 5:** Triangles $\triangle ABW$ and $\triangle WZY$ have: - $|AB| = |YZ|$ (from step 4) - $\angle BAW = \angle WZY$ (step 5) - $\angle ABW = \angle ZYW$ (step 7) By ASA (Angle-Side-Angle) congruence criterion, $\triangle ABW \cong \triangle WZY$. 9. **Step 6:** From congruence, corresponding sides are equal: $$|AW| = |WZ|$$ 10. **Step 7:** Given $|AW| = |DE|$ and $|WZ| = |EF|$, so: $$|DE| = |EF|$$ **Final answer:** $|DE| = |EF|$ proving the segments cut by the parallel lines on transversal $DF$ are equal. This completes the proof.