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We are given the following conditions about triangle \( \triangle ABC \):

- \( AB = 13 \), \( BC = 14 \), and \( AC = 15 \).
- Point \( D \) lies on \( BC \) such that \( AD \perp BC \), and \( E \) is the midpoint of \( AD \).
- Rectangle \( CDEF \) is constructed such that points \( C, D, E, F \) form its vertices.

We are tasked with finding the area of triangle \( \triangle AEF \).

### Step-by-Step Solution:

1. **Right Triangle Identification**:
   The triangle \( ABC \) is a right triangle with \( \angle B = 90^\circ \). This is because the sides 13, 14, and 15 follow the Pythagorean theorem, confirming it as a right triangle.

2. **Coordinate Setup**:
   Since \( \triangle ABC \) is a right triangle, we can set up coordinates for the vertices:
   - \( B(0, 0) \)
   - \( C(14, 0) \)
   - \( A(0, 13) \)
   
   Now, \( D \) lies on \( BC \) and is the foot of the altitude from \( A \). The altitude from \( A \) to \( BC \) implies that \( AD \perp BC \).

3. **Finding \( D \)**:
   Since \( BC \) is horizontal (on the x-axis), the line through \( A \) perpendicular to \( BC \) will have an undefined slope. Hence, the point \( D \) must be directly beneath \( A \) on the x-axis.
   - We can compute \( D \)'s coordinates by using similar triangles or by finding the length of the altitude using geometric properties of right triangles.

4. **Midpoint \( E \) and Rectangle**:
   \( E \) is the midpoint of \( AD \), and rectangle \( CDEF \) ensures that the lines \( EF \parallel BC \) and \( CF \parallel AD \), helping in setting up coordinates for all points.

5. **Area of Triangle \( \triangle AEF \)**:
   Once we calculate the coordinates of \( A \), \( E \), and \( F \), the area of triangle \( \triangle AEF \) can be computed using the formula for the area of a triangle with known vertices:
   \[
   \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
   \]
   where \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) are the coordinates of points \( A \), \( E \), and \( F \).

---

Let me go ahead and calculate the specific coordinates and the area for you!The \( x \)-coordinate of point \( D \) is approximately \( 12.96 \), and the length of the altitude \( AD \) is approximately \( 5 \) units (after rounding). Now, \( E \) is the midpoint of \( AD \), which we can calculate next.

The coordinates of \( D \) are approximately \( D(12.96, 0) \), and to find the coordinates of \( E \), we use the midpoint formula:

\[
E = \left( \frac{x_A + x_D}{2}, \frac{y_A + y_D}{2} \right)
\]

Finally, we compute the area of \( \triangle AEF \), using these coordinates.The coordinates of point \( E \) (the midpoint of \( AD \)) are approximately \( E(0.52, 6.5) \), and the coordinates of point \( F \) are \( F(14, 6.5) \), as it lies directly above \( C \) at the same height as \( E \).

The area of triangle \( \triangle AEF \) is approximately **43.81 square units**.

Would you like more details on any part of the solution, or do you have any further questions?

### Related Questions:
1. How would you compute the length of diagonal \( AF \) in this setup?
2. What are the properties of a rectangle that help simplify such problems?
3. Can the area of rectangle \( CDEF \) be easily computed from these coordinates?
4. How can you prove that \( AD \) is perpendicular to \( BC \) geometrically?
5. How do the coordinates of point \( F \) change if the rectangle were not aligned with the axes?

### Tip:
In problems involving geometry and coordinate planes, assigning specific coordinates to key points can simplify calculations, especially when dealing with perpendicular lines or shapes like rectangles.

In triangle ABC, AB = 13, BC = 14, and AC = 15. Let D be the point on BC such that AD ⊥ BC, and let E be the midpoint of AD. If F is a point such that CDEF is a rectangle, compute the area of triangle AEF.

Learn how to compute the area of triangle AEF in a right triangle setup where AB = 13, BC = 14, AC = 15, with rectangle CDEF. Explore key geometric concepts like midpoints, perpendiculars, and triangle area calculations using coordinate geometry. Suitable for students in Grades 10-12.

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