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Let's break down the given problem. The exercise revolves around a parallelogram \( ABCD \) and two points \( M \) and \( N \) defined by specific vector relations. Here's the breakdown:

### Given Data:
- \( \overrightarrow{DM} = \frac{1}{3} \overrightarrow{DC} \)
- \( \overrightarrow{BN} = 3 \overrightarrow{BC} \)

### Questions:
1. **Construire une figure convenable (Construct an appropriate figure):**  
   You should draw the parallelogram \( ABCD \) and the points \( M \) and \( N \) following the given vector relations.  
   - For \( M \), place it on the line segment \( DC \) such that \( \overrightarrow{DM} = \frac{1}{3} \overrightarrow{DC} \). This means \( M \) divides \( DC \) in a 1:2 ratio.
   - For \( N \), we have \( \overrightarrow{BN} = 3 \overrightarrow{BC} \), so \( N \) lies on the line \( BC \) and extends beyond \( C \) (since the vector is 3 times longer).

2. **Montrer que

Let ABCD be a parallelogram. Points M and N are defined such that \( \overrightarrow{DM} = \frac{1}{3} \overrightarrow{DC} \) and \( \overrightarrow{BN} = 3 \overrightarrow{BC} \). The tasks are: 
1. Draw an appropriate figure. 
2. Show that \( \overrightarrow{AM} = \overrightarrow{BC} + \overrightarrow{DM} \). 
3. Prove that \( \overrightarrow{AN} = 3 \overrightarrow{BC} + 3 \overrightarrow{DM} \). 
4. Deduce that points A, M, and N are collinear.

Solve a vector geometry problem involving a parallelogram where you need to demonstrate vector relationships and prove collinearity. The tasks include vector addition and properties of parallelograms, suitable for students in Grades 10-12.

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