1. **Problem statement:** Given triangle ABC with points M on BC, X on AB, Z on AC such that XZ is parallel to BC. Given lengths: AZ = 3 cm, ZC = 2 cm, BM = 3 cm, MC = 5 cm. Find the length of XY. 2. **Key concept:** Since XZ is parallel to BC, triangles AXZ and ABC are similar by the Basic Proportionality Theorem (Thales' theorem). 3. **Step 1: Find AC.** $$AC = AZ + ZC = 3 + 2 = 5 \text{ cm}$$ 4. **Step 2: Find BC.** $$BC = BM + MC = 3 + 5 = 8 \text{ cm}$$ 5. **Step 3: Use similarity ratio.** Since XZ is parallel to BC, the ratio of sides is: $$\frac{AX}{AB} = \frac{AZ}{AC}$$ 6. **Step 4: Find ratio of AZ to AC.** $$\frac{AZ}{AC} = \frac{3}{5}$$ 7. **Step 5: Since M lies on BC dividing it into BM=3 and MC=5, the ratio BM:MC = 3:5. Point Y lies on XZ such that Y corresponds to M on BC in the similar triangles. 8. **Step 6: Length of XZ corresponds to BC, so: $$XZ = \frac{AZ}{AC} \times BC = \frac{3}{5} \times 8 = 4.8 \text{ cm}$$ 9. **Step 7: Since Y divides XZ in the same ratio as M divides BC (3:5), length XY is: $$XY = \frac{BM}{BC} \times XZ = \frac{3}{8} \times 4.8 = 1.8 \text{ cm}$$ **Final answer:** $$\boxed{XY = 1.8 \text{ cm}}$$