1. **Problem statement:** Given a parallelogram ABCD and points M, N, P, Q defined by $$\overrightarrow{AM} = \frac{3}{2} \overrightarrow{AB}, \quad \overrightarrow{BN} = \frac{3}{2} \overrightarrow{BC}, \quad \overrightarrow{CP} = \frac{3}{2} \overrightarrow{CD}, \quad \overrightarrow{DQ} = \frac{3}{2} \overrightarrow{DA},$$ show that $$\overrightarrow{MN} = \overrightarrow{QP}.$$\n\n2. **Recall properties of parallelograms:** - Opposite sides are equal and parallel: $$\overrightarrow{AB} = \overrightarrow{DC}, \quad \overrightarrow{BC} = \overrightarrow{AD}.$$ - Vector addition along the parallelogram: $$\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}.$$\n\n3. **Express vectors for points M, N, P, Q relative to A:** - Since $$\overrightarrow{AM} = \frac{3}{2} \overrightarrow{AB},$$ point M is on the line from A towards B, extended beyond B. - For N, start from A: $$\overrightarrow{AN} = \overrightarrow{AB} + \overrightarrow{BN} = \overrightarrow{AB} + \frac{3}{2} \overrightarrow{BC}.$$ - For P, start from A: $$\overrightarrow{AP} = \overrightarrow{AC} + \overrightarrow{CP} = (\overrightarrow{AB} + \overrightarrow{BC}) + \frac{3}{2} \overrightarrow{CD}.$$ Using $$\overrightarrow{CD} = -\overrightarrow{AB},$$ $$\overrightarrow{AP} = \overrightarrow{AB} + \overrightarrow{BC} + \frac{3}{2}(-\overrightarrow{AB}) = \overrightarrow{BC} - \frac{1}{2} \overrightarrow{AB}.$$ - For Q, start from A: $$\overrightarrow{AQ} = \overrightarrow{AD} + \overrightarrow{DQ} = \overrightarrow{AD} + \frac{3}{2} \overrightarrow{DA} = \overrightarrow{AD} - \frac{3}{2} \overrightarrow{AD} = -\frac{1}{2} \overrightarrow{AD}.$$ Using $$\overrightarrow{AD} = \overrightarrow{BC},$$ $$\overrightarrow{AQ} = -\frac{1}{2} \overrightarrow{BC}.$$\n\n4. **Calculate $$\overrightarrow{MN}$$:** $$\overrightarrow{MN} = \overrightarrow{AN} - \overrightarrow{AM} = \left(\overrightarrow{AB} + \frac{3}{2} \overrightarrow{BC}\right) - \frac{3}{2} \overrightarrow{AB} = -\frac{1}{2} \overrightarrow{AB} + \frac{3}{2} \overrightarrow{BC}.$$\n\n5. **Calculate $$\overrightarrow{QP}$$:** $$\overrightarrow{QP} = \overrightarrow{AP} - \overrightarrow{AQ} = \left(\overrightarrow{BC} - \frac{1}{2} \overrightarrow{AB}\right) - \left(-\frac{1}{2} \overrightarrow{BC}\right) = \overrightarrow{BC} - \frac{1}{2} \overrightarrow{AB} + \frac{1}{2} \overrightarrow{BC} = -\frac{1}{2} \overrightarrow{AB} + \frac{3}{2} \overrightarrow{BC}.$$\n\n6. **Conclusion:** Since $$\overrightarrow{MN} = \overrightarrow{QP} = -\frac{1}{2} \overrightarrow{AB} + \frac{3}{2} \overrightarrow{BC},$$ the vectors are equal as required. This shows $$\overrightarrow{MN} = \overrightarrow{QP}.$$