1. **Problem statement:** In parallelogram ABCD, point M is the midpoint of side BC. Given vectors $\overrightarrow{AB} = \vec{a}$ and $\overrightarrow{AD} = \vec{b}$, find vectors $\overrightarrow{AM}$ and $\overrightarrow{MD}$ in terms of $\vec{a}$ and $\vec{b}$. 2. **Recall vector properties in parallelograms:** - $\overrightarrow{AB} = \vec{a}$ - $\overrightarrow{AD} = \vec{b}$ - $\overrightarrow{BC} = \overrightarrow{AD} = \vec{b}$ (opposite sides equal and parallel) - $\overrightarrow{AC} = \vec{a} + \vec{b}$ 3. **Find $\overrightarrow{AM}$:** - Since M is midpoint of BC, $\overrightarrow{BM} = \frac{1}{2} \overrightarrow{BC} = \frac{1}{2} \vec{b}$. - Vector $\overrightarrow{AM} = \overrightarrow{AB} + \overrightarrow{BM} = \vec{a} + \frac{1}{2} \vec{b}$. 4. **Find $\overrightarrow{MD}$:** - Vector $\overrightarrow{MD} = \overrightarrow{MB} + \overrightarrow{BD}$. - Note $\overrightarrow{MB} = - \overrightarrow{BM} = - \frac{1}{2} \vec{b}$. - Vector $\overrightarrow{BD} = \overrightarrow{BA} + \overrightarrow{AD} = -\vec{a} + \vec{b}$. - So, $\overrightarrow{MD} = - \frac{1}{2} \vec{b} + (-\vec{a} + \vec{b}) = -\vec{a} + \frac{1}{2} \vec{b}$. **Final answers:** $$\overrightarrow{AM} = \vec{a} + \frac{1}{2} \vec{b}$$ $$\overrightarrow{MD} = -\vec{a} + \frac{1}{2} \vec{b}$$