Consider `triangleABC` and `triangleA_1B_1C_1` in such a way that `bar(AB)=bar(A_1B_1)` and `M, N, M_1 ,N_1` be the midpoints of `AB, BC, A_1B_1 and B_1C_1` respectively, then

Options

A. `vec(M M_(1))=vec(N N_(1))`
B. `vec(C C_(1))=vec(MM_(1))`
C. `vec(C C_(1))=vec(N N_(1))`
D. `vec(MM_(1))=vec(BB_(1))`

Explanation

To solve the problem involving triangles \(ABC\) and \(A_1B_1C_1\) with the given conditions, we can follow these steps:

### Step 1: Define the Points
Let the coordinates of the points be:
- \( A(x_1, y_1) \)
- \( B(x_2, y_2) \)
- \( C(x_3, y_3) \)
- \( A_1(x_1', y_1') \)
- \( B_1(x_2', y_2') \)
- \( C_1(x_3', y_3') \)

Given that \( \overline{AB} = \overline{A_1B_1} \), we have:
\[ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(x_2' - x_1')^2 + (y_1' - y_1)^2} \]

### Step 2: Find the Midpoints
The midpoints \(M\) and \(N\) of segments \(AB\) and \(BC\) respectively are given by:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
\[ N = \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right) \]

Similarly, the midpoints \(M_1\) and \(N_1\) of segments \(A_1B_1\) and \(B_1C_1\) are:
\[ M_1 = \left( \frac{x_1' + x_2'}{2}, \frac{y_1' + y_2'}{2} \right) \]
\[ N_1 = \left( \frac{x_2' + x_3'}{2}, \frac{y_2' + y_3'}{2} \right) \]

### Step 3: Establish Relationships
Since \( \overline{AB} = \overline{A_1B_1} \), we can establish relationships between the midpoints:
\[ \overline{MN} = \overline{M_1N_1} \]

### Step 4: Calculate Vectors
The vector from \(M\) to \(N\) is:
\[ \overline{MN} = N - M = \left( \frac{x_2 + x_3}{2} - \frac{x_1 + x_2}{2}, \frac{y_2 + y_3}{2} - \frac{y_1 + y_2}{2} \right) \]
This simplifies to:
\[ \overline{MN} = \left( \frac{x_3 - x_1}{2}, \frac{y_3 - y_1}{2} \right) \]

Similarly, for \(M_1\) and \(N_1\):
\[ \overline{M_1N_1} = N_1 - M_1 = \left( \frac{x_2' + x_3'}{2} - \frac{x_1' + x_2'}{2}, \frac{y_2' + y_3'}{2} - \frac{y_1' + y_2'}{2} \right) \]
This simplifies to:
\[ \overline{M_1N_1} = \left( \frac{x_3' - x_1'}{2}, \frac{y_3' - y_1'}{2} \right) \]

### Step 5: Show Equality of Vectors
Since \( \overline{AB} = \overline{A_1B_1} \), the lengths and directions of the vectors must be equal. Therefore:
\[ \overline{MN} = \overline{M_1N_1} \]
This implies:
\[ \frac{x_3 - x_1}{2} = \frac{x_3' - x_1'}{2} \quad \text{and} \quad \frac{y_3 - y_1}{2} = \frac{y_3' - y_1'} \]

### Conclusion
Thus, we conclude that the midpoints \(M\) and \(M_1\) are related as follows:
\[ M = M_1 \quad \text{and} \quad N = N_1 \]
This leads us to the conclusion that option D is correct:
\[ M = M_1 \quad \text{and} \quad N = N_1 \]