1. **Problem statement:** Prove that the diagonals of a parallelogram bisect each other using vectors, without assuming the origin. 2. **Setup:** Let the parallelogram have vertices represented by vectors $\vec{A}$, $\vec{B}$, $\vec{C}$, and $\vec{D}$ such that $\vec{A}$ and $\vec{C}$ are opposite vertices, and $\vec{B}$ and $\vec{D}$ are opposite vertices. 3. **Key property of parallelograms:** Opposite sides are equal and parallel, so $$\vec{B} - \vec{A} = \vec{D} - \vec{C}$$ 4. **Diagonals:** The diagonals are the vectors $\vec{AC} = \vec{C} - \vec{A}$ and $\vec{BD} = \vec{D} - \vec{B}$. 5. **Midpoints of diagonals:** We want to show the midpoint of $\vec{AC}$ equals the midpoint of $\vec{BD}$. 6. **Midpoint of diagonal $AC$:** $$\frac{\vec{A} + \vec{C}}{2}$$ 7. **Midpoint of diagonal $BD$:** $$\frac{\vec{B} + \vec{D}}{2}$$ 8. **Show these midpoints are equal:** Start with the midpoint of $BD$: $$\frac{\vec{B} + \vec{D}}{2} = \frac{\vec{B} + \vec{D}}{2}$$ Using the parallelogram property $\vec{B} - \vec{A} = \vec{D} - \vec{C}$, rearranged as $$\vec{D} = \vec{B} - \vec{A} + \vec{C}$$ Substitute into midpoint of $BD$: $$\frac{\vec{B} + (\vec{B} - \vec{A} + \vec{C})}{2} = \frac{2\vec{B} - \vec{A} + \vec{C}}{2}$$ 9. **Simplify:** $$= \vec{B} - \frac{\vec{A}}{2} + \frac{\vec{C}}{2}$$ 10. **Rewrite midpoint of $AC$:** $$\frac{\vec{A} + \vec{C}}{2} = \frac{\vec{A}}{2} + \frac{\vec{C}}{2}$$ 11. **Compare the two midpoints:** We want to prove $$\frac{\vec{A} + \vec{C}}{2} = \frac{\vec{B} + \vec{D}}{2}$$ From step 8 and 9, the midpoint of $BD$ is $$\vec{B} - \frac{\vec{A}}{2} + \frac{\vec{C}}{2}$$ But this is not immediately equal to midpoint of $AC$. Let's try a different approach. 12. **Alternative approach:** Express $\vec{D}$ in terms of $\vec{A}$, $\vec{B}$, and $\vec{C}$: Since $\vec{ABCD}$ is a parallelogram, $$\vec{D} = \vec{A} + (\vec{C} - \vec{B})$$ 13. **Midpoint of $BD$ becomes:** $$\frac{\vec{B} + \vec{D}}{2} = \frac{\vec{B} + \vec{A} + (\vec{C} - \vec{B})}{2} = \frac{\vec{A} + \vec{C}}{2}$$ 14. **Conclusion:** The midpoints of the diagonals $AC$ and $BD$ are equal, so the diagonals bisect each other. **Final answer:** The diagonals of a parallelogram bisect each other because their midpoints coincide: $$\boxed{\frac{\vec{A} + \vec{C}}{2} = \frac{\vec{B} + \vec{D}}{2}}$$