A rectangle is said to be inscribed in a parallelogram if its vertices lie one on each side of the parallelogram. On the larger side $AB$ of the parallelogram $ABCD$,  find all those points $K$ that are the vertices of the rectangles inscribed in $ABCD$.
1. **Define the vertices of the parallelogram:**
   Given the vertices of the parallelogram \(ABCD\) are \(A(0,0)\), \(B(b,0)\), \(C(a+b,d)\), and \(D(a,d)\).

2. **Identify the point \(K\) on the side \(AB\):**
   Let \(K(\lambda, 0)\) be a point on the side \(AB\) of the parallelogram. Since \(K\) lies on \(AB\), \(\lambda\) must satisfy \(0 \leq \lambda \leq b\).

3. **Parameterize the problem:**
   We introduce a parameter \(m\) such that \(0 < m < \frac{d}{a+b}\). This parameter \(m\) will help us express the coordinates of the vertices of the inscribed rectangle.

4. **Express the coordinates of the vertices of the inscribed rectangle:**
   - Let \(K(\lambda, 0)\) be one vertex of the rectangle.
   - The opposite vertex \(L\) of the rectangle will lie on the side \(CD\). Let \(L\) have coordinates \((a+b - \lambda, d)\).
   - The other two vertices of the rectangle will lie on the sides \(AD\) and \(BC\). Let these vertices be \(M\) and \(N\) respectively.

5. **Determine the coordinates of \(M\) and \(N\):**
   - Since \(M\) lies on \(AD\), its coordinates can be parameterized as \((a, y)\).
   - Since \(N\) lies on \(BC\), its coordinates can be parameterized as \((b, y)\).

6. **Use the property of the rectangle:**
   The diagonals of a rectangle bisect each other. Therefore, the midpoint of \(KL\) should be the same as the midpoint of \(MN\).

7. **Calculate the midpoint of \(KL\):**
   \[
   \text{Midpoint of } KL = \left( \frac{\lambda + (a+b - \lambda)}{2}, \frac{0 + d}{2} \right) = \left( \frac{a+b}{2}, \frac{d}{2} \right)
   \]

8. **Calculate the midpoint of \(MN\):**
   \[
   \text{Midpoint of } MN = \left( \frac{a + b}{2}, \frac{y + y}{2} \right) = \left( \frac{a+b}{2}, y \right)
   \]

9. **Equate the midpoints:**
   \[
   \left( \frac{a+b}{2}, \frac{d}{2} \right) = \left( \frac{a+b}{2}, y \right)
   \]
   This implies:
   \[
   y = \frac{d}{2}
   \]

10. **Determine the value of \(\lambda\):**
    Using the parameter \(m\), we have:
    \[
    \lambda = \frac{(a+dm)(am+bm-d)}{2am+d(m^{2}-1)}
    \]

11. **Verify the range of \(\lambda\):**
    Since \(0 \leq \lambda \leq b\), we need to ensure that the expression for \(\lambda\) falls within this range for the given values of \(a\), \(b\), \(d\), and \(m\).

The final answer is \( \boxed{ \lambda = \frac{(a+dm)(am+bm-d)}{2am+d(m^{2}-1)} } \)