1. **Problem statement:** In pentagon OABCD, given that OA is parallel to DC, AB is parallel to OD, with OD = 2AB and OA = 2DC, and vectors \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{AB} = \mathbf{b}\), find in terms of \(\mathbf{a}\) and \(\mathbf{b}\) the vectors \(\overrightarrow{AD}\) and \(\overrightarrow{BC}\). 2. **Recall vector addition and parallelism:** - Vector addition: \(\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BD}\). - Since OA is parallel to DC and OA = 2DC, then \(\overrightarrow{DC} = \frac{1}{2} \mathbf{a}\). - Since AB is parallel to OD and OD = 2AB, then \(\overrightarrow{OD} = 2 \mathbf{b}\). 3. **Find \(\overrightarrow{AD}\):** - Express \(\overrightarrow{AD} = \overrightarrow{AO} + \overrightarrow{OD}\). - Note \(\overrightarrow{AO} = -\mathbf{a}\) (opposite direction of \(\overrightarrow{OA}\)). - Given \(\overrightarrow{OD} = 2 \mathbf{b}\). - So, \(\overrightarrow{AD} = -\mathbf{a} + 2 \mathbf{b}\). 4. **Find \(\overrightarrow{BC}\):** - Since \(\overrightarrow{AB} = \mathbf{b}\) and \(\overrightarrow{BC} = \overrightarrow{DC} - \overrightarrow{DB}\). - But \(\overrightarrow{DC} = \frac{1}{2} \mathbf{a}\) and \(\overrightarrow{DB} = \overrightarrow{DO} + \overrightarrow{OB} = -\overrightarrow{OD} + \overrightarrow{OB}\). - \(\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = \mathbf{a} + \mathbf{b}\). - So, \(\overrightarrow{DB} = -2 \mathbf{b} + (\mathbf{a} + \mathbf{b}) = \mathbf{a} - \mathbf{b}\). - Therefore, \(\overrightarrow{BC} = \frac{1}{2} \mathbf{a} - (\mathbf{a} - \mathbf{b}) = \frac{1}{2} \mathbf{a} - \mathbf{a} + \mathbf{b} = -\frac{1}{2} \mathbf{a} + \mathbf{b}\). **Final answers:** (i) \(\overrightarrow{AD} = -\mathbf{a} + 2 \mathbf{b}\) (ii) \(\overrightarrow{BC} = -\frac{1}{2} \mathbf{a} + \mathbf{b}\)