Points $A, B$ and $C$ lie on the same line so that $CA = AB$. Square $ABDE$ and the equilateral triangle $CFA$, are constructed  on the same side of line $CB$. Find the acute angle between straight lines $CE$ and $BF$.
1. **Identify the given conditions and construct the geometric shapes:**
   - Points \( A, B, \) and \( C \) lie on the same line such that \( CA = AB \).
   - Construct square \( ABDE \) and equilateral triangle \( CFA \) on the same side of line \( CB \).

2. **Analyze the square \( ABDE \):**
   - Since \( ABDE \) is a square, all sides are equal, and all angles are \( 90^\circ \).
   - Let \( AB = x \). Then, \( EA = AB = x \).

3. **Analyze the equilateral triangle \( CFA \):**
   - Since \( CFA \) is an equilateral triangle, all sides are equal, and all angles are \( 60^\circ \).
   - Given \( CA = AB = x \), the side \( CF \) is also \( x \).

4. **Determine the angle \( \angle ECA \):**
   - Since \( EA = AB = x \), triangle \( ECA \) is isosceles with \( EA = CA \).
   - The angle \( \angle ECA \) is the angle between the line segments \( CE \) and \( CA \).
   - Since \( \angle EAB = 90^\circ \) (as \( ABDE \) is a square), and \( \angle CAB = 0^\circ \) (as \( A, B, \) and \( C \) are collinear), the angle \( \angle ECA \) is \( 45^\circ \).

5. **Determine the angle \( \angle CBF \):**
   - Let \( FT \perp CB \) such that \( T \in CB \).
   - Since \( CFA \) is an equilateral triangle, the height \( FT \) from \( F \) to \( CB \) can be calculated as:
     \[
     FT = \frac{x \sqrt{3}}{2}
     \]
   - The distance \( TB \) from \( T \) to \( B \) is:
     \[
     TB = \frac{3x}{2}
     \]
   - The tangent of the angle \( \angle CBF \) is:
     \[
     \tan \angle CBF = \frac{FT}{TB} = \frac{\frac{x \sqrt{3}}{2}}{\frac{3x}{2}} = \frac{1}{\sqrt{3}}
     \]
   - Therefore, \( \angle CBF = 30^\circ \).

6. **Calculate the acute angle between the lines \( CE \) and \( BF \):**
   - The acute angle between the lines \( CE \) and \( BF \) is the difference between \( \angle ECA \) and \( \angle CBF \):
     \[
     \text{Acute angle} = 45^\circ + 30^\circ = 75^\circ
     \]

The final answer is \( \boxed{75^\circ} \).