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The exercise provided is from vector geometry and involves a trapezoid \(ABCD\) where \(BC \parallel AD\). Points \(E\) and \(F\) are the midpoints of \(BC\) and \(AD\) respectively. The vector equation is given as:

\[
\overrightarrow{EF} = m \overrightarrow{DB} + n \overrightarrow{CA}
\]

We need to find the value of \(m^2 + n^2\).

### Solution:

1. **Midpoint Formula**:
   Since \(E\) and \(F\) are midpoints:
   \[
   \overrightarrow{E} = \frac{\overrightarrow{B} + \overrightarrow{C}}{2}, \quad \overrightarrow{F} = \frac{\overrightarrow{A} + \overrightarrow{D}}{2}
   \]
   
   Therefore:
   \[
   \overrightarrow{EF} = \overrightarrow{F} - \overrightarrow{E} = \frac{\overrightarrow{A} + \overrightarrow{D}}{2} - \frac{\overrightarrow{B} + \overrightarrow{C}}{2} = \frac{\overrightarrow{AD} - \overrightarrow{BC}}{2}
   \]
   
2. **Vectors in terms of \( \overrightarrow{DB} \) and \( \overrightarrow{CA} \)**:
   We express the vectors \( \overrightarrow{DB} \) and \( \overrightarrow{CA} \) in terms of the known vectors:
   \[
   \overrightarrow{DB} = \overrightarrow{B} - \overrightarrow{D}, \quad \overrightarrow{CA} = \overrightarrow{A} - \overrightarrow{C}
   \]
   
   Thus:
   \[
   \overrightarrow{EF} = m(\overrightarrow{B} - \overrightarrow{D}) + n(\overrightarrow{A} - \overrightarrow{C})
   \]
   
3. **Comparison of Coefficients**:
   Comparing this with the expression:
   \[
   \overrightarrow{EF} = \frac{\overrightarrow{AD} - \overrightarrow{BC}}{2}
   \]
   We know that:
   \[
   \overrightarrow{AD} = \overrightarrow{A} - \overrightarrow{D}, \quad \overrightarrow{BC} = \overrightarrow{B} - \overrightarrow{C}
   \]
   Therefore:
   \[
   \frac{\overrightarrow{AD} - \overrightarrow{BC}}{2} = \frac{(\overrightarrow{A} - \overrightarrow{D}) - (\overrightarrow{B} - \overrightarrow{C})}{2}
   \]
   Simplifying, we get:
   \[
   \overrightarrow{EF} = \frac{\overrightarrow{A} - \overrightarrow{B} + \overrightarrow{C} - \overrightarrow{D}}{2}
   \]
   
   Comparing with \( \overrightarrow{EF} = m(\overrightarrow{B} - \overrightarrow{D}) + n(\overrightarrow{A} - \overrightarrow{C}) \), we equate coefficients:
   \[
   m = -\frac{1}{2}, \quad n = \frac{1}{2}
   \]
   
4. **Calculation of \(m^2 + n^2\)**:
   \[
   m^2 + n^2 = \left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}
   \]
   
Therefore, the value of \(m^2 + n^2\) is \(\boxed{\frac{1}{2}}\).

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Do you want any more details or have any questions?

Here are some related questions:
1. What happens if \(BC\) and \(AD\) were not parallel?
2. How does the trapezoid's specific dimensions affect \(m\) and \(n\)?
3. What if \(E\) and \(F\) were not midpoints?
4. How would the solution change if \(EF\) was perpendicular to \(BC\)?
5. Can you generalize this problem for a non-trapezoidal quadrilateral?

**Tip:** Understanding vector operations and their geometric interpretations can significantly simplify problems involving vector equations.

Sobre un trapecio ABCD (BC // AD) se toman los E y F, puntos medios de BC y AD respectivamente. Se sabe que EF = mDB + nCA. Hallar m^2 + n^2.

This exercise involves vector geometry in a trapezoid where BC is parallel to AD. Midpoints E and F divide the sides BC and AD. The task is to find m^2 + n^2 using vector equations. Detailed steps are provided, making this problem suitable for students in Grades 10-12.

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