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Let's work through this question systematically.

### Given:
- \( \overrightarrow{OA} = 2a \)
- \( \overrightarrow{OB} = 2b \)
- \( \overrightarrow{OC} = 2c \)

Points \( P, Q, R, S \) are midpoints of sides:
- \( P \) is the midpoint of \( OA \)
- \( Q \) is the midpoint of \( AB \)
- \( R \) is the midpoint of \( BC \)
- \( S \) is the midpoint of \( CO \)

#### Part (a): Express in terms of \( a \), \( b \), and \( c \)

**i. \( \overrightarrow{AB} \)**:
\[
\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = 2b - 2a
\]

**ii. \( \overrightarrow{BC} \)**:
\[
\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = 2c - 2b
\]

**iii. \( \overrightarrow{PQ} \)**:
Since \( P \) is the midpoint of \( OA \) and \( Q \) is the midpoint of \( AB \), we have:
\[
\overrightarrow{PQ} = \frac{1}{2} \overrightarrow{AB} = \frac{1}{2}(2b - 2a) = b - a
\]

**iv. \( \overrightarrow{QR} \)**:
Similarly, \( Q \) is the midpoint of \( AB \) and \( R \) is the midpoint of \( BC \):
\[
\overrightarrow{QR} = \frac{1}{2} \overrightarrow{BC} = \frac{1}{2}(2c - 2b) = c - b
\]

**v. \( \overrightarrow{PS} \)**:
\( P \) is the midpoint of \( OA \) and \( S \) is the midpoint of \( CO \):
\[
\overrightarrow{PS} = \frac{1}{2} \overrightarrow{CO} = \frac{1}{2}(2c - 2a) = c - a
\]

#### Part (b): Relationship between \( QR \) and \( PS \)

We found that:
\[
\overrightarrow{QR} = c - b \quad \text{and} \quad \overrightarrow{PS} = c - a
\]
Both vectors share the component \( c \), but \( QR \) and \( PS \) differ in their directional components: \( \overrightarrow{QR} \) points from \( Q \) to \( R \), while \( \overrightarrow{PS} \) points from \( P \) to \( S \). These two vectors are **parallel** because they both involve displacements proportional to the sides of the triangle, differing only in the subtraction of \( b \) and \( a \).

#### Part (c): Type of Quadrilateral \( PQRS \)

Since \( \overrightarrow{QR} \parallel \overrightarrow{PS} \), quadrilateral \( PQRS \) is a **parallelogram**. This is confirmed because opposite sides are parallel and equal in length.

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Do you want more details on any part? Here are five related questions for further exploration:

1. How can we verify that the opposite sides of a parallelogram are equal using vector algebra?
2. What are the conditions for two vectors to be parallel?
3. How do midpoints help in simplifying vector expressions?
4. Can you explain how a parallelogram's diagonal properties relate to its vector representation?
5. How would the calculations change if the midpoints were not at the exact center of each segment?

**Tip:** Remember, vectors simplify geometry by letting you break down complex shapes into manageable components based on direction and magnitude.

In the quadrilateral, OA = 2a, OB = 2b, OC = 2c. Points P, Q, R, and S are midpoints of sides OA, AB, BC, and CO respectively. Express AB, BC, PQ, QR, and PS in terms of a, b, and c. Describe the relationship between QR and PS, and determine the type of quadrilateral PQRS.

Learn how to express vector relations for quadrilateral sides OA, OB, and OC in terms of vectors a, b, and c. This problem covers vector algebra, midpoint theorem, and the classification of a quadrilateral as a parallelogram. Suitable for students in Grades 10-12.

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