1. **State the problem:** Given triangle ABC, points x, y, z are midpoints of sides AB, BC, and CA respectively. M is the centroid (point of intersection of medians) of the triangle, and O is any point in the plane. Prove that: i) $$\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} = \overrightarrow{Ox} + \overrightarrow{Oy} + \overrightarrow{Oz}$$ ii) $$\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} = 3\overrightarrow{OM}$$ --- 2. **Prove part (i):** Since x, y, z are midpoints: $$\overrightarrow{Ox} = \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2}$$ $$\overrightarrow{Oy} = \frac{\overrightarrow{OB} + \overrightarrow{OC}}{2}$$ $$\overrightarrow{Oz} = \frac{\overrightarrow{OC} + \overrightarrow{OA}}{2}$$ Add these three vectors: $$\overrightarrow{Ox} + \overrightarrow{Oy} + \overrightarrow{Oz} = \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2} + \frac{\overrightarrow{OB} + \overrightarrow{OC}}{2} + \frac{\overrightarrow{OC} + \overrightarrow{OA}}{2}$$ Simplify the right side: $$= \frac{1}{2} (\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OC} + \overrightarrow{OA})$$ $$= \frac{1}{2} (2\overrightarrow{OA} + 2\overrightarrow{OB} + 2\overrightarrow{OC})$$ $$= \frac{1}{2} \times 2 (\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}) = \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}$$ This proves part (i). --- 3. **Prove part (ii):** The centroid M divides each median in a 2:1 ratio from the vertex, so $$\overrightarrow{OM} = \frac{\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}}{3}$$ Multiply both sides by 3: $$3\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}$$ This completes part (ii). --- **Final answers:** i) $$\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} = \overrightarrow{Ox} + \overrightarrow{Oy} + \overrightarrow{Oz}$$ ii) $$\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} = 3\overrightarrow{OM}$$