1. **Stating the problem:** We have a parallelogram ABCD with vectors \(\vec{DA} = \vec{a}\) and \(\vec{DC} = \vec{c}\). Points M and N are midpoints of segments CB and AB respectively. We need to find the vector \(\vec{MN}\) in terms of \(\vec{a}\) and \(\vec{c}\). 2. **Recall properties and formulas:** - Since ABCD is a parallelogram, \(\vec{AB} = \vec{DC} = \vec{c}\) and \(\vec{BC} = \vec{DA} = \vec{a}\). - Midpoint of a segment between points P and Q is \(\frac{\vec{P} + \vec{Q}}{2}\). - Vector between two points X and Y is \(\vec{XY} = \vec{Y} - \vec{X}\). 3. **Express position vectors of points:** - Let \(\vec{D} = \vec{0}\) (origin for convenience). - Then \(\vec{A} = \vec{a}\). - \(\vec{C} = \vec{c}\). - \(\vec{B} = \vec{A} + \vec{AB} = \vec{a} + \vec{c}\). 4. **Find midpoints M and N:** - \(\vec{M} = \) midpoint of CB = \(\frac{\vec{C} + \vec{B}}{2} = \frac{\vec{c} + (\vec{a} + \vec{c})}{2} = \frac{\vec{a} + 2\vec{c}}{2} = \frac{\vec{a}}{2} + \vec{c}\). - \(\vec{N} = \) midpoint of AB = \(\frac{\vec{A} + \vec{B}}{2} = \frac{\vec{a} + (\vec{a} + \vec{c})}{2} = \frac{2\vec{a} + \vec{c}}{2} = \vec{a} + \frac{\vec{c}}{2}\). 5. **Calculate vector \(\vec{MN}\):** $$\vec{MN} = \vec{N} - \vec{M} = \left(\vec{a} + \frac{\vec{c}}{2}\right) - \left(\frac{\vec{a}}{2} + \vec{c}\right) = \vec{a} - \frac{\vec{a}}{2} + \frac{\vec{c}}{2} - \vec{c} = \frac{\vec{a}}{2} - \frac{\vec{c}}{2} = \frac{1}{2}(\vec{a} - \vec{c})$$ **Final answer:** $$\boxed{\vec{MN} = \frac{1}{2}(\vec{a} - \vec{c})}$$