Let $ABCD$ and $A^{\prime }B^{\prime}C^{\prime }D^{\prime }$ be two arbitrary parallelograms in the space, and let $M,$ $N,$ $P,$ $Q$ be points dividing the segments $AA^{\prime },$ $BB^{\prime },$ $CC^{\prime },$ $DD^{\prime }$ in equal ratios.

[b]a.)[/b] Prove that the quadrilateral $MNPQ$ is a parallelogram.

[b]b.)[/b] What is the locus of the center of the parallelogram $MNPQ,$ when the point $M$ moves on the segment $AA^{\prime }$ ?

(Consecutive vertices of the parallelograms are labelled in alphabetical order.
### Part (a)

1. **Define the ratio \( u \):**
   Since the points \( M, N, P, Q \) divide the segments \( AA', BB', CC', DD' \) in equal ratios, we can find a real number \( u \) such that:
   \[
   u = \frac{AM}{AA'} = \frac{BN}{BB'} = \frac{CP}{CC'} = \frac{DQ}{DD'}
   \]

2. **Express the position vectors:**
   Let \( O \) be an arbitrary point in space. We can express the position vectors of \( M, N, P, Q \) as follows:
   \[
   \overrightarrow{OM} = \overrightarrow{OA} + u \cdot \overrightarrow{AA'} = \overrightarrow{OA} + u \cdot (\overrightarrow{OA'} - \overrightarrow{OA}) = (1-u) \cdot \overrightarrow{OA} + u \cdot \overrightarrow{OA'}
   \]
   Similarly,
   \[
   \overrightarrow{ON} = (1-u) \cdot \overrightarrow{OB} + u \cdot \overrightarrow{OB'}
   \]
   \[
   \overrightarrow{OP} = (1-u) \cdot \overrightarrow{OC} + u \cdot \overrightarrow{OC'}
   \]
   \[
   \overrightarrow{OQ} = (1-u) \cdot \overrightarrow{OD} + u \cdot \overrightarrow{OD'}
   \]

3. **Calculate \( \overrightarrow{MN} \) and \( \overrightarrow{QP} \):**
   Since \( \overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} \), we have:
   \[
   \overrightarrow{MN} = \left( (1-u) \cdot \overrightarrow{OB} + u \cdot \overrightarrow{OB'} \right) - \left( (1-u) \cdot \overrightarrow{OA} + u \cdot \overrightarrow{OA'} \right)
   \]
   \[
   = (1-u) \cdot (\overrightarrow{OB} - \overrightarrow{OA}) + u \cdot (\overrightarrow{OB'} - \overrightarrow{OA'})
   \]
   \[
   = (1-u) \cdot \overrightarrow{AB} + u \cdot \overrightarrow{A'B'}
   \]

   Similarly, for \( \overrightarrow{QP} \):
   \[
   \overrightarrow{QP} = \left( (1-u) \cdot \overrightarrow{OD} + u \cdot \overrightarrow{OD'} \right) - \left( (1-u) \cdot \overrightarrow{OC} + u \cdot \overrightarrow{OC'} \right)
   \]
   \[
   = (1-u) \cdot (\overrightarrow{OD} - \overrightarrow{OC}) + u \cdot (\overrightarrow{OD'} - \overrightarrow{OC'})
   \]
   \[
   = (1-u) \cdot \overrightarrow{DC} + u \cdot \overrightarrow{D'C'}
   \]

4. **Use the properties of parallelograms:**
   Since \( ABCD \) and \( A'B'C'D' \) are parallelograms, we have:
   \[
   \overrightarrow{AB} = \overrightarrow{DC} \quad \text{and} \quad \overrightarrow{A'B'} = \overrightarrow{D'C'}
   \]

5. **Conclude that \( \overrightarrow{MN} = \overrightarrow{QP} \):**
   \[
   \overrightarrow{MN} = (1-u) \cdot \overrightarrow{AB} + u \cdot \overrightarrow{A'B'}
   \]
   \[
   \overrightarrow{QP} = (1-u) \cdot \overrightarrow{DC} + u \cdot \overrightarrow{D'C'}
   \]
   Since \( \overrightarrow{AB} = \overrightarrow{DC} \) and \( \overrightarrow{A'B'} = \overrightarrow{D'C'} \), it follows that:
   \[
   \overrightarrow{MN} = \overrightarrow{QP}
   \]
   Therefore, \( MNPQ \) is a parallelogram.

### Part (b)

1. **Define the centers \( U, V, W \):**
   Let \( U, V, W \) be the centers of the parallelograms \( ABCD, A'B'C'D' \), and \( MNPQ \), respectively. These points are the midpoints of the diagonals \( AC, A'C', MP \).

2. **Express the position vectors of the centers:**
   \[
   \overrightarrow{OU} = \frac{1}{2} \cdot (\overrightarrow{OA} + \overrightarrow{OC})
   \]
   \[
   \overrightarrow{OV} = \frac{1}{2} \cdot (\overrightarrow{OA'} + \overrightarrow{OC'})
   \]
   \[
   \overrightarrow{OW} = \frac{1}{2} \cdot (\overrightarrow{OM} + \overrightarrow{OP})
   \]

3. **Substitute the expressions for \( \overrightarrow{OM} \) and \( \overrightarrow{OP} \):**
   \[
   \overrightarrow{OW} = \frac{1}{2} \cdot \left( (1-u) \cdot \overrightarrow{OA} + u \cdot \overrightarrow{OA'} + (1-u) \cdot \overrightarrow{OC} + u \cdot \overrightarrow{OC'} \right)
   \]
   \[
   = \frac{1}{2} \cdot \left( (1-u) \cdot (\overrightarrow{OA} + \overrightarrow{OC}) + u \cdot (\overrightarrow{OA'} + \overrightarrow{OC'}) \right)
   \]
   \[
   = (1-u) \cdot \frac{1}{2} \cdot (\overrightarrow{OA} + \overrightarrow{OC}) + u \cdot \frac{1}{2} \cdot (\overrightarrow{OA'} + \overrightarrow{OC'})
   \]
   \[
   = (1-u) \cdot \overrightarrow{OU} + u \cdot \overrightarrow{OV}
   \]

4. **Determine the locus of \( W \):**
   The point \( W \) lies on the segment \( UV \) and the ratio \( \frac{UW}{UV} = u \). As \( M \) moves on the segment \( AA' \), the ratio \( u \) varies from 0 to 1. Therefore, the locus of \( W \) is the segment \( UV \).

The final answer is the segment \( \boxed{ UV } \).