In the diagram above, \( A B C D \) is a parallelogram. The diagonals \( A C \) and \( B D \) intersect at \( E \) and the point \( F \) is the midpoint of \( A E . G \) is a point on \( A B \) such that \( 2 A G=G B . \overrightarrow{A G}=\boldsymbol{a} \) and \( \overrightarrow{A F}=\boldsymbol{b} \). (i) Express \( \overrightarrow{D F} \) in terms of \( \boldsymbol{a} \) and/or \( \boldsymbol{b} \).

To express \(\overrightarrow{D F}\) in terms of \(\boldsymbol{a}\) and \(\boldsymbol{b}\), we start by analyzing the given information about the parallelogram \(ABCD\) and the points \(E\), \(F\), and \(G\). 1. **Understanding the Points**: - Since \(ABCD\) is a parallelogram, we have: \[ \overrightarrow{C} = \overrightarrow{A} + \overrightarrow{B} - \overrightarrow{D} \] - The diagonals \(AC\) and \(BD\) intersect at point \(E\), which is the midpoint of both diagonals. Therefore: \[ \overrightarrow{E} = \frac{\overrightarrow{A} + \overrightarrow{C}}{2} = \frac{\overrightarrow{B} + \overrightarrow{D}}{2} \] 2. **Finding \(\overrightarrow{F}\)**: - Point \(F\) is the midpoint of segment \(AE\). Thus, we can express \(\overrightarrow{F}\) as: \[ \overrightarrow{F} = \frac{\overrightarrow{A} + \overrightarrow{E}}{2} \] - Substituting \(\overrightarrow{E}\): \[ \overrightarrow{F} = \frac{\overrightarrow{A} + \frac{\overrightarrow{A} + \overrightarrow{C}}{2}}{2} = \frac{2\overrightarrow{A} + \overrightarrow{A} + \overrightarrow{C}}{4} = \frac{3\overrightarrow{A} + \overrightarrow{C}}{4} \] 3. **Expressing \(\overrightarrow{G}\)**: - Given that \(2AG = GB\), we can express \(G\) as: \[ \overrightarrow{G} = \frac{2\overrightarrow{B} + \overrightarrow{A}}{3} \] - Since \(\overrightarrow{A G} = \boldsymbol{a}\), we have: \[ \overrightarrow{G} - \overrightarrow{A} = \boldsymbol{a} \implies \overrightarrow{G} = \overrightarrow{A} + \boldsymbol{a} \] 4. **Finding \(\overrightarrow{D}\)**: - From the properties of the parallelogram, we know: \[ \overrightarrow{D} = \overrightarrow{A} + \overrightarrow{B} - \overrightarrow{C} \] 5. **Finding \(\overrightarrow{D F}\)**: - Now, we can express \(\overrightarrow{D F}\) as: \[ \overrightarrow{D F} = \overrightarrow{F} - \overrightarrow{D} \] - Substituting the expressions for \(\overrightarrow{F}\) and \(\overrightarrow{D}\): \[ \overrightarrow{D F} = \left(\frac{3\overrightarrow{A} + \overrightarrow{C}}{4}\right) - \left(\overrightarrow{A} + \overrightarrow{B} - \overrightarrow{C}\right) \] - Simplifying this expression: \[ \overrightarrow{D F} = \frac{3\overrightarrow{A} + \overrightarrow{C}}{4} - \overrightarrow{A} - \overrightarrow{B} + \overrightarrow{C} \] \[ = \frac{3\overrightarrow{A} + \overrightarrow{C} - 4\overrightarrow{A} - 4\overrightarrow{B} + 4\overrightarrow{C}}{4} \] \[ = \frac{-\overrightarrow{A} - 4\overrightarrow{B} + 5\overrightarrow{C}}{4} \] 6. **Final Expression**: - We can express \(\overrightarrow{C}\) in terms of \(\overrightarrow{A}\) and \(\overrightarrow{B}\) using the properties of the parallelogram: \[ \overrightarrow{C} = \overrightarrow{D} + \overrightarrow{B} - \overrightarrow{A} \] - Thus, we can express \(\overrightarrow{D F}\) in terms of \(\boldsymbol