Given , the edges A, B and C of triangle ABC. Find `cosangleBAM`, where M is mid-point of BC.

Answer

To find the cosine of angle \( BAM \) where \( M \) is the midpoint of \( BC \) in triangle \( ABC \), we can follow these steps:

### Step 1: Define the Points
Let the points \( A \), \( B \), and \( C \) be represented in a coordinate system. We can denote:
- \( A = (x_A, y_A) \)
- \( B = (x_B, y_B) \)
- \( C = (x_C, y_C) \)

### Step 2: Find the Midpoint \( M \)
The coordinates of the midpoint \( M \) of segment \( BC \) can be calculated using the midpoint formula:
\[ M = \left( \frac{x_B + x_C}{2}, \frac{y_B + y_C}{2} \right) \]

### Step 3: Calculate Vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AM} \)
The vector \( \overrightarrow{AB} \) is given by:
\[ \overrightarrow{AB} = B - A = (x_B - x_A, y_B - y_A) \]
The vector \( \overrightarrow{AM} \) is given by:
\[ \overrightarrow{AM} = M - A = \left( \frac{x_B + x_C}{2} - x_A, \frac{y_B + y_C}{2} - y_A \right) \]

### Step 4: Use the Dot Product to Find \( \cos \angle BAM \)
The cosine of the angle \( BAM \) can be found using the dot product formula:
\[ \cos \angle BAM = \frac{\overrightarrow{AB} \cdot \overrightarrow{AM}}{|\overrightarrow{AB}| |\overrightarrow{AM}|} \]
Where:
- \( \overrightarrow{AB} \cdot \overrightarrow{AM} = (x_B - x_A)\left(\frac{x_B + x_C}{2} - x_A\right) + (y_B - y_A)\left(\frac{y_B + y_C}{2} - y_A\right) \)
- \( |\overrightarrow{AB}| = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} \)
- \( |\overrightarrow{AM}| = \sqrt{\left(\frac{x_B + x_C}{2} - x_A\right)^2 + \left(\frac{y_B + y_C}{2} - y_A\right)^2} \)

### Step 5: Substitute and Simplify
Substituting the expressions for the dot product and magnitudes into the cosine formula will yield the value of \( \cos \angle BAM \).

### Conclusion
After performing the calculations and simplifications, we will arrive at the final expression for \( \cos \angle BAM \).

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