The point $D$ at the side $AB$ of triangle $ABC$ is given. Construct points $E,F$ at sides $BC, AC$ respectively such that the midpoints of $DE$ and $DF$ are collinear with $B$ and the midpoints of $DE$ and $EF$ are collinear with $C.$
1. **Construct the reflection \( X \) of the point \( D \) with respect to the point \( C \).**
   - To find the reflection \( X \), we use the fact that \( C \) is the midpoint of \( DX \). Therefore, \( X \) is such that \( \overrightarrow{CX} = \overrightarrow{CD} \). This gives us the coordinates of \( X \) as \( X = 2C - D \).

2. **Construct the point \( Y \in AB \) such that \( DY \parallel AC \).**
   - To ensure \( DY \parallel AC \), we need to find \( Y \) such that the slope of \( DY \) is equal to the slope of \( AC \). If \( A = (x_1, y_1) \), \( B = (x_2, y_2) \), \( C = (x_3, y_3) \), and \( D = (x_4, y_4) \), then the slope condition gives:
     \[
     \frac{y_4 - y_Y}{x_4 - x_Y} = \frac{y_3 - y_1}{x_3 - x_1}
     \]
     Solving this equation will give the coordinates of \( Y \).

3. **Construct the point \( E \in CA \) such that \( YE \parallel BC \).**
   - To ensure \( YE \parallel BC \), we need to find \( E \) such that the slope of \( YE \) is equal to the slope of \( BC \). This gives:
     \[
     \frac{y_Y - y_E}{x_Y - x_E} = \frac{y_2 - y_3}{x_2 - x_3}
     \]
     Solving this equation will give the coordinates of \( E \).

4. **Construct the point \( F \in XE \cap AB \).**
   - The point \( F \) is the intersection of the line \( XE \) with the line \( AB \). The equation of the line \( XE \) can be found using the coordinates of \( X \) and \( E \), and the equation of the line \( AB \) can be found using the coordinates of \( A \) and \( B \). Solving these two equations simultaneously will give the coordinates of \( F \).

5. **Verify the relations:**
   - The positions of the points \( E \in CA \) and \( F \in AB \) are defined by the following relations:
     \[
     \frac{EA}{EC} = \frac{DC}{DB}
     \]
     and
     \[
     \frac{AF}{AB} = \left( \frac{CD}{CB} \right)^2
     \]
   - These relations ensure that the midpoints of \( DE \) and \( DF \) are collinear with \( B \) and the midpoints of \( DE \) and \( EF \) are collinear with \( C \).

6. **Construct the reflection \( Z \) of the point \( Y \) with respect to the point \( A \).**
   - To find the reflection \( Z \), we use the fact that \( A \) is the midpoint of \( YZ \). Therefore, \( Z \) is such that \( \overrightarrow{AZ} = \overrightarrow{AY} \). This gives us the coordinates of \( Z \) as \( Z = 2A - Y \).

7. **Verify the collinearity conditions:**
   - The point \( F \) is the conjugate of the point \( B \) with respect to the pair \( \{Y, Z\} \), i.e., \( FI \parallel AC \), where \( I \in XY \cap DZ \).

The final answer is the construction of points \( \boxed{ E } \) and \( F \) as described above.