1. **State the problem:** We have a parallelogram PQRS with diagonal PR. Points Q and S are joined to a point W on diagonal PR. We are given that triangles \(\Delta PQW = \Delta PSM\). We need to prove \(\Delta PSM = \Delta PQM\). 2. **Understand the notation:** Since \(\Delta PQW = \Delta PSM\), it likely means the areas of these triangles are equal. 3. **Analyze the geometry:** - PQRS is a parallelogram, so \(\overrightarrow{PQ} = \overrightarrow{SR}\) and \(\overrightarrow{PS} = \overrightarrow{QR}\). - Point W lies on diagonal PR, so \(W\) divides \(PR\) at some ratio. 4. **Given \(\Delta PQW = \Delta PSM\), let's consider the triangles involving points P, Q, S, and M:** - Since the problem asks to prove \(\Delta PSM = \Delta PQM\), note that both triangles share vertex P and point M, and differ by the third vertex being S or Q, both on the parallelogram. 5. **Key idea:** Because \(PQRS\) is a parallelogram, \(MS\) and \(MQ\) will maintain certain properties. 6. **Use properties of parallelograms and equal areas:** - Since \(\Delta PQW = \Delta PSM\) and point W is on diagonal PR, the equal areas suggest certain symmetry. - Considering vector areas and the bases on the same line PR, the heights from Q and S to PR are equal. 7. **Relate areas of triangles \(PQM\) and \(PSM\):** - The height from point M to line PQ is the same as the height from point M to line PS because PQRS is a parallelogram. 8. **Therefore, the areas of \(\Delta PSM\) and \(\Delta PQM\) must be equal because they share base PM and have the same height from S and Q respectively.** 9. **Conclusion:** \(\Delta PSM = \Delta PQM\). Final answer: \(\boxed{\Delta PSM = \Delta PQM}\).