Math Problem Statement

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.

Solution

Let's break down the given problem. The exercise revolves around a parallelogram ABCDABCD and two points MM and NN defined by specific vector relations. Here's the breakdown:

Given Data:

  • DM=13DC\overrightarrow{DM} = \frac{1}{3} \overrightarrow{DC}
  • BN=3BC\overrightarrow{BN} = 3 \overrightarrow{BC}

Questions:

  1. Construire une figure convenable (Construct an appropriate figure):
    You should draw the parallelogram ABCDABCD and the points MM and NN following the given vector relations.

    • For MM, place it on the line segment DCDC such that DM=13DC\overrightarrow{DM} = \frac{1}{3} \overrightarrow{DC}. This means MM divides DCDC in a 1:2 ratio.
    • For NN, we have BN=3BC\overrightarrow{BN} = 3 \overrightarrow{BC}, so NN lies on the line BCBC and extends beyond CC (since the vector is 3 times longer).
  2. **Montrer que

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Math Problem Analysis

Mathematical Concepts

Vector Geometry
Parallelogram Properties
Vector Addition

Formulas

Vector addition: \( \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC} \)
Collinearity condition for vectors

Theorems

Parallelogram vector properties

Suitable Grade Level

Grades 10-12