ABCD is a square and P, Q, R and S are points on the sides AB, BC, CD DA respectively such that AP = BQ = CR = DS. What is angle SPQ equal to?

Given that \( ABCD \) is a square and \( P \), \( Q \), \( R \), and \( S \) are points on sides \( AB \), \( BC \), \( CD \), and \( DA \) respectively, such that \( AP = BQ = CR = DS \), we want to find the angle \( \angle SPQ \).

### Solution:
1. **Understanding the Setup**:
   - Let the side length of the square be \( a \).
   - Let the length \( AP = BQ = CR = DS = x \), where \( 0 < x < a \).
   - Points \( P \), \( Q \), \( R \), and \( S \) divide each side of the square into segments of length \( x \) and \( a-x \).

2. **Coordinates of Points**:
   - Assign coordinates to the vertices:
     - \( A(0, 0) \)
     - \( B(a, 0) \)
     - \( C(a, a) \)
     - \( D(0, a) \)
   - Coordinates of the points \( P \), \( Q \), \( R \), and \( S \) are:
     - \( P(x, 0) \) (on \( AB \))
     - \( Q(a, x) \) (on \( BC \))
     - \( R(a-x, a) \) (on \( CD \))
     - \( S(0, a-x) \) (on \( DA \))

3. **Slope Calculation**:
   - Slope of line \( SP \):
     \[
     \text{slope of } SP = \frac{(a-x) - 0}{0 - x} = \frac{a-x}{-x} = \frac{x-a}{x}
     \]
   - Slope of line \( PQ \):
     \[
     \text{slope of } PQ = \frac{x - 0}{a - x} = \frac{x}{a-x}
     \]

4. **Angle Between Two Lines**:
   The angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) is given by:
   \[
   \tan \theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right|
   \]
   Substituting \( m_1 = \frac{x-a}{x} \) and \( m_2 = \frac{x}{a-x} \):
   \[
   \tan \angle SPQ = \left|\frac{\frac{x-a}{x} - \frac{x}{a-x}}{1 + \frac{x-a}{x} \cdot \frac{x}{a-x}}\right|
   \]
   Simplifying the expression:
   \[
   \tan \angle SPQ = \left|\frac{\frac{(x-a)(a-x) - x^2}{x(a-x)}}{1 + \frac{(x-a)x}{x(a-x)}}\right| = \left|\frac{-(a^2 - 2ax + x^2) + x^2}{x(a-x)}\right|
   \]
   Simplifying further:
   \[
   \tan \angle SPQ = \left|\frac{-a^2 + 2ax}{a^2 - 2ax + x^2}\right| = 1
   \]
   Therefore, \( \angle SPQ = 45^\circ \).

### Final Answer:
The angle \( \angle SPQ = 45^\circ \).

Would you like to explore further details or have any questions?

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Here are five related questions:
1. What is the area of the square if the side length \( a \) is 10 units?
2. How would the angle \( \angle SPQ \) change if \( AP = BQ = CR = DS \) were different lengths?
3. Can the angle \( \angle SPQ \) be calculated using vector methods?
4. How would you find the distance between points \( S \) and \( Q \)?
5. What would happen if \( P \) and \( R \) were located at the midpoints of their respective sides?

**Tip:** Understanding the relationship between the slopes of lines is crucial for finding the angle between them.

Explore the calculation of angle SPQ in square ABCD where points P, Q, R, and S divide sides AB, BC, CD, and DA respectively with equal segments AP = BQ = CR = DS. Detailed solution with geometric and algebraic methods provided.

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