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The problem involves a parallelogram \( OABC \), where the vectors \( \overrightarrow{OA} = \vec{a} \) and \( \overrightarrow{OB} = \vec{b} \) are given. We need to express the vectors \( \overrightarrow{MO}, \overrightarrow{MA}, \overrightarrow{MB}, \overrightarrow{MC} \) through the vectors \( \vec{a} \) and \( \vec{b} \), where \( M \) is the point of intersection of the diagonals of the parallelogram.

### Solution:

#### Step 1: Define the relationship between the diagonals.
In a parallelogram, the diagonals bisect each other. So, the point \( M \) is the midpoint of both diagonals.

The position vector of the point \( C \) (the fourth vertex of the parallelogram) can be found using the properties of a parallelogram:
\[
\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{OB} = \vec{a} + \vec{b}
\]

#### Step 2: Express \( M \) as the midpoint of the diagonal.
Since \( M \) is the midpoint of the diagonal \( AC \), we can write the position vector of \( M \) as:
\[
\overrightarrow{OM} = \frac{\overrightarrow{OA} + \overrightarrow{OC}}{2} = \frac{\vec{a} + (\vec{a} + \vec{b})}{2} = \frac{2\vec{a} + \vec{b}}{2}
\]
Thus, 
\[
\overrightarrow{OM} = \vec{a} + \frac{\vec{b}}{2}
\]

#### Step 3: Express the vectors \( \overrightarrow{MO}, \overrightarrow{MA}, \overrightarrow{MB}, \overrightarrow{MC} \).

- **Vector \( \overrightarrow{MO} \):**
\[
\overrightarrow{MO} = \overrightarrow{OM} - \overrightarrow{OO} = \overrightarrow{OM} = \vec{a} + \frac{\vec{b}}{2}
\]

- **Vector \( \overrightarrow{MA} \):**
\[
\overrightarrow{MA} = \overrightarrow{OA} - \overrightarrow{OM} = \vec{a} - \left( \vec{a} + \frac{\vec{b}}{2} \right) = - \frac{\vec{b}}{2}
\]

- **Vector \( \overrightarrow{MB} \):**
\[
\overrightarrow{MB} = \overrightarrow{OB} - \overrightarrow{OM} = \vec{b} - \left( \vec{a} + \frac{\vec{b}}{2} \right) = \vec{b} - \vec{a} - \frac{\vec{b}}{2} = \frac{\vec{b}}{2} - \vec{a}
\]

- **Vector \( \overrightarrow{MC} \):**
\[
\overrightarrow{MC} = \overrightarrow{OC} - \overrightarrow{OM} = (\vec{a} + \vec{b}) - \left( \vec{a} + \frac{\vec{b}}{2} \right) = \vec{a} + \vec{b} - \vec{a} - \frac{\vec{b}}{2} = \frac{\vec{b}}{2}
\]

### Final expressions:
- \( \overrightarrow{MO} = \vec{a} + \frac{\vec{b}}{2} \)
- \( \overrightarrow{MA} = -\frac{\vec{b}}{2} \)
- \( \overrightarrow{MB} = \frac{\vec{b}}{2} - \vec{a} \)
- \( \overrightarrow{MC} = \frac{\vec{b}}{2} \)

Let me know if you need further clarification or details! 

### Related Questions:
1. How do you find the midpoint of a vector in a different geometric shape, like a triangle?
2. How does the property of diagonals differ in other quadrilaterals like rhombuses or rectangles?
3. Can these expressions be generalized for any parallelogram with arbitrary diagonal lengths?
4. How would this problem change if the shape was a trapezoid rather than a parallelogram?
5. What would the vector \( \overrightarrow{MC} \) be if the point \( M \) wasn't the midpoint of the diagonal?

### Tip:
When solving vector problems in geometry, always rely on geometric properties like midpoints, bisectors, and parallelogram rules to simplify vector calculations.

У паралелограмі OABC задані вектори \(\overrightarrow{OA} = \vec{a}\) та \(\overrightarrow{OB} = \vec{b}\). Вирахувати вектори \(\overrightarrow{MO}, \overrightarrow{MA}, \overrightarrow{MB}, \overrightarrow{MC}\), де \( M \) - точка перетину діагоналей.

Solve a problem involving vector operations in a parallelogram where diagonals bisect each other. Express vectors \(\overrightarrow{MO}, \overrightarrow{MA}, \overrightarrow{MB}, \overrightarrow{MC}\) through given vectors \(\vec{a}\) and \(\vec{b}\), using the midpoint formula and properties of parallelograms. Suitable for high school students.

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