1. **Problem statement:** Prove using vectors that the diagonals of a rhombus bisect each other. 2. **Setup:** Let the rhombus have vertices represented by vectors $ \vec{A}, \vec{B}, \vec{C}, \vec{D} $ in order, with $ \vec{A} = \vec{0} $ for convenience. 3. Since a rhombus has all sides equal, let $ \vec{AB} = \vec{u} $ and $ \vec{AD} = \vec{v} $ such that $ |\vec{u}| = |\vec{v}| $. 4. Then the vertices are: $ \vec{A} = \vec{0} $, $ \vec{B} = \vec{u} $, $ \vec{D} = \vec{v} $, $ \vec{C} = \vec{u} + \vec{v} $. 5. The diagonals are $ \vec{AC} = \vec{u} + \vec{v} $ and $ \vec{BD} = \vec{v} - \vec{u} $. 6. To prove the diagonals bisect each other, we need to show their midpoints are the same. 7. Midpoint of $ AC $ is: $$\frac{\vec{A} + \vec{C}}{2} = \frac{\vec{0} + (\vec{u} + \vec{v})}{2} = \frac{\vec{u} + \vec{v}}{2}.$$ 8. Midpoint of $ BD $ is: $$\frac{\vec{B} + \vec{D}}{2} = \frac{\vec{u} + \vec{v}}{2}.$$ 9. Since both midpoints are equal, the diagonals bisect each other. **Final answer:** The diagonals of a rhombus bisect each other because their midpoints coincide at $$\frac{\vec{u} + \vec{v}}{2}$$.