Given a rectangle $ABCD$, determine two points $K$ and $L$ on the sides $BC$ and $CD$ such that the triangles $ABK, AKL$ and $ADL$ have same area.
1. **Define the coordinates of the rectangle:**
   Let the vertices of the rectangle \(ABCD\) be \(A(0,0)\), \(B(b,0)\), \(C(b,c)\), and \(D(0,c)\).

2. **Choose points \(K\) and \(L\):**
   Let \(K\) be on side \(BC\) with coordinates \(K(b,k)\) and \(L\) be on side \(CD\) with coordinates \(L(l,c)\).

3. **Calculate the area of \(\triangle ABK\):**
   The area of \(\triangle ABK\) is given by:
   \[
   \text{Area of } \triangle ABK = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times b \times k
   \]

4. **Calculate the area of \(\triangle AKL\):**
   The area of \(\triangle AKL\) is given by:
   \[
   \text{Area of } \triangle AKL = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times l \times c
   \]

5. **Calculate the area of \(\triangle ADL\):**
   The area of \(\triangle ADL\) is given by:
   \[
   \text{Area of } \triangle ADL = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (b \times c - k \times l)
   \]

6. **Set the areas equal to each other:**
   Since the areas of \(\triangle ABK\), \(\triangle AKL\), and \(\triangle ADL\) are equal, we have:
   \[
   \frac{1}{2} b k = \frac{1}{2} c l = \frac{1}{2} (b c - k l)
   \]

7. **Solve for \(k\) and \(l\):**
   From \(\frac{1}{2} b k = \frac{1}{2} c l\), we get:
   \[
   b k = c l \implies k = \frac{c l}{b}
   \]

   From \(\frac{1}{2} b k = \frac{1}{2} (b c - k l)\), we get:
   \[
   b k = b c - k l \implies b k + k l = b c \implies k (b + l) = b c \implies k = \frac{b c}{b + l}
   \]

   Substitute \(k = \frac{c l}{b}\) into \(k = \frac{b c}{b + l}\):
   \[
   \frac{c l}{b} = \frac{b c}{b + l} \implies c l (b + l) = b^2 c \implies l (b + l) = b^2 \implies l^2 + b l - b^2 = 0
   \]

8. **Solve the quadratic equation for \(l\):**
   \[
   l^2 + b l - b^2 = 0
   \]
   Using the quadratic formula \(l = \frac{-b \pm \sqrt{b^2 + 4 b^2}}{2}\):
   \[
   l = \frac{-b \pm \sqrt{5 b^2}}{2} = \frac{-b \pm b \sqrt{5}}{2} = \frac{b (-1 \pm \sqrt{5})}{2}
   \]
   Since \(l\) must be positive, we take the positive root:
   \[
   l = \frac{b (\sqrt{5} - 1)}{2}
   \]

9. **Find \(k\) using \(l\):**
   \[
   k = \frac{c l}{b} = \frac{c \left(\frac{b (\sqrt{5} - 1)}{2}\right)}{b} = \frac{c (\sqrt{5} - 1)}{2}
   \]

The final answer is \( \boxed{ k = \frac{(\sqrt{5} - 1)c}{2} } \) and \(l = \frac{(\sqrt{5} - 1)b}{2}\).