Let $ABCD$ be a square of side $1$ and $A'B'C'D'$ its image by a rotation $r$ of center $D$ . The intersection of these two squares is the quadrilateral $DALC'$. By $C'$, we draw the parallel to $AD$, which intersects $AB$ in $E$ .
(a) What is the area of the quadrilateral $DALC'$ if $E$ is the midpoint of $[AB]$?
(b) Find the position of $E$ in $AB$ so that the area of $DALC'$ is half the area of the initial square.
(c) Determine the position of $E$ in $AB$ for the area $DALC'$ to be $1/3 ,1/4, ...,1/m$, the area of the initial square ($m$ being a nonzero natural). 
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Given the square \(ABCD\) with vertices \(A(0,1)\), \(B(1,1)\), \(C(1,0)\), and \(D(0,0)\), we need to find the area of the quadrilateral \(DALC'\) under different conditions.

### Part (a)
We need to find the area of the quadrilateral \(DALC'\) if \(E\) is the midpoint of \([AB]\).

1. **Rotation and Coordinates:**
   - The square \(A'B'C'D'\) is obtained by rotating \(ABCD\) around \(D\) by an angle \(\alpha\).
   - The coordinates of \(C'\) after rotation are \((\cos \alpha, \sin \alpha)\).

2. **Intersection Point \(L\):**
   - The line through \(C'\) parallel to \(AD\) intersects \(AB\) at \(E\).
   - Since \(E\) is the midpoint of \([AB]\), \(E\) has coordinates \(\left(\frac{1}{2}, 1\right)\).

3. **Finding \(L\):**
   - The line through \(C'\) parallel to \(AD\) is vertical, so it has the equation \(x = \cos \alpha\).
   - This line intersects \(AB\) at \(L\), where \(L\) has coordinates \(\left(\cos \alpha, 1\right)\).

4. **Area Calculation:**
   - The area of quadrilateral \(DALC'\) can be found by subtracting the area of triangle \(DLC'\) from the area of triangle \(DAC'\).
   - The area of triangle \(DAC'\) is \(\frac{1}{2} \times 1 \times 1 = \frac{1}{2}\).
   - The area of triangle \(DLC'\) is \(\frac{1}{2} \times \cos \alpha \times \sin \alpha = \frac{1}{2} \sin \alpha \cos \alpha\).

5. **Final Area:**
   \[
   \text{Area of } DALC' = \frac{1}{2} - \frac{1}{2} \sin \alpha \cos \alpha = \frac{1}{2} (1 - \sin \alpha \cos \alpha)
   \]

### Part (b)
We need to find the position of \(E\) in \(AB\) so that the area of \(DALC'\) is half the area of the initial square.

1. **Area Condition:**
   - The area of the initial square is \(1\).
   - We want the area of \(DALC'\) to be \(\frac{1}{2}\).

2. **Equation Setup:**
   \[
   \frac{1}{2} (1 - \sin \alpha \cos \alpha) = \frac{1}{2}
   \]
   \[
   1 - \sin \alpha \cos \alpha = 1
   \]
   \[
   \sin \alpha \cos \alpha = 0
   \]

3. **Solving for \(\alpha\):**
   - \(\sin \alpha \cos \alpha = 0\) implies \(\alpha = 0\) or \(\alpha = \frac{\pi}{2}\).

4. **Position of \(E\):**
   - If \(\alpha = 0\), \(E\) is at \(A\).
   - If \(\alpha = \frac{\pi}{2}\), \(E\) is at \(B\).

### Part (c)
We need to determine the position of \(E\) in \(AB\) for the area \(DALC'\) to be \(\frac{1}{m}\) of the area of the initial square.

1. **Area Condition:**
   - The area of \(DALC'\) should be \(\frac{1}{m}\).

2. **Equation Setup:**
   \[
   \frac{1}{2} (1 - \sin \alpha \cos \alpha) = \frac{1}{m}
   \]
   \[
   1 - \sin \alpha \cos \alpha = \frac{2}{m}
   \]
   \[
   \sin \alpha \cos \alpha = 1 - \frac{2}{m}
   \]

3. **Solving for \(\alpha\):**
   - Using the double-angle identity \(\sin 2\alpha = 2 \sin \alpha \cos \alpha\):
   \[
   \sin 2\alpha = 2 \left(1 - \frac{2}{m}\right)
   \]
   \[
   \sin 2\alpha = 2 - \frac{4}{m}
   \]

4. **Position of \(E\):**
   - The position of \(E\) depends on the value of \(\alpha\) which satisfies the above equation.

The final answer is \(\boxed{\frac{1}{2} (1 - \sin \alpha \cos \alpha)}\)