1. **Problem statement:** Given triangle ABC with XZ parallel to BC, AZ = 3 cm, ZC = 2 cm, BM = 3 cm, and MC = 5 cm, find the length of XY. 2. **Key concept:** When a line segment (XZ) is drawn parallel to one side (BC) of a triangle, it divides the other two sides proportionally (Basic Proportionality Theorem or Thales' theorem). 3. **Apply the theorem:** Since XZ is parallel to BC, we have $ \frac{AX}{XB} = \frac{AZ}{ZC} $. 4. **Calculate the ratio:** Given $ AZ = 3 $ and $ ZC = 2 $, so $ \frac{AZ}{ZC} = \frac{3}{2} $. 5. **Find AB:** Since BM = 3 and MC = 5, $ BC = BM + MC = 3 + 5 = 8 $. 6. **Find AX and XB:** Let $ AX = x $, then $ XB = AB - x $. Using the ratio: $$\frac{x}{AB - x} = \frac{3}{2}$$ 7. **Express AB in terms of BM and MC:** Since M lies on BC, and AM is a median or altitude, but we don't have AB length directly. However, since X lies on AB and Z on AC, and XZ is parallel to BC, the segments are divided proportionally. 8. **Use the ratio to find XY:** Since XZ is parallel to BC, the length of XZ is proportional to BC: $$\frac{XZ}{BC} = \frac{AX}{AB} = \frac{AZ}{AC}$$ 9. **Calculate AC:** Given AZ = 3 and ZC = 2, so $ AC = AZ + ZC = 3 + 2 = 5 $. 10. **Calculate the ratio:** $ \frac{AZ}{AC} = \frac{3}{5} $. 11. **Calculate XZ:** Since $ BC = 8 $, then $$XZ = BC \times \frac{3}{5} = 8 \times \frac{3}{5} = \frac{24}{5} = 4.8$$ 12. **Locate Y on XZ:** The problem states Y lies on XZ but does not provide additional info to find XY directly. Assuming Y divides XZ similarly to how M divides BC, and since BM = 3 and MC = 5, the ratio $ \frac{BM}{MC} = \frac{3}{5} $. 13. **Assuming Y divides XZ in the same ratio:** Then $$XY = XZ \times \frac{3}{8} = 4.8 \times \frac{3}{8} = 1.8$$ **Final answer:** $ XY = 1.8 $ cm.