The right triangle $ABC$ with a right angle at $C$. From all the rectangles $CA_1MB_1$, where $A_1\in BC, M\in AB$ and $B_1\in AC$ which one has the biggest area?
1. **Define the problem and variables:**
   - We are given a right triangle \(ABC\) with a right angle at \(C\).
   - We need to find the rectangle \(CA_1MB_1\) with the largest area, where \(A_1 \in BC\), \(M \in AB\), and \(B_1 \in AC\).

2. **Set up the variables:**
   - Let \(B_1M = a\), \(A_1M = b\), \(AM = x\), \(BM = y\), and \(\angle B_1AM = \alpha\).

3. **Express the area of the rectangle:**
   - The area of the rectangle \(CA_1MB_1\) is given by \(S_{CA_1MB_1} = ab\).
   - Using trigonometric identities, we can express this area as:
     \[
     S_{CA_1MB_1} = ab = \frac{1}{2}xy \sin 2\alpha
     \]
   - Note that \(\frac{1}{2} \sin 2\alpha\) is a constant.

4. **Maximize the product \(xy\):**
   - We need to maximize \(xy\) under the constraint \(x + y = \text{constant} = c\).
   - Express \(y\) in terms of \(x\):
     \[
     y = c - x
     \]
   - Substitute \(y\) into the product \(xy\):
     \[
     xy = x(c - x) = xc - x^2
     \]
   - This is a quadratic function in terms of \(x\):
     \[
     xy = -x^2 + cx
     \]

5. **Find the maximum value of the quadratic function:**
   - The quadratic function \(f(x) = -x^2 + cx\) reaches its maximum at the vertex.
   - The vertex of a quadratic function \(ax^2 + bx + c\) is given by \(x = -\frac{b}{2a}\).
   - Here, \(a = -1\) and \(b = c\), so:
     \[
     x = \frac{c}{2}
     \]
   - Substitute \(x = \frac{c}{2}\) back into the expression for \(y\):
     \[
     y = c - \frac{c}{2} = \frac{c}{2}
     \]

6. **Conclusion:**
   - The maximum area is achieved when \(x = \frac{c}{2}\) and \(y = \frac{c}{2}\).
   - This means that \(M\) is the midpoint of \(AB\).

The final answer is \( \boxed{ \text{When} \, M \, \text{is a midpoint of} \, AB } \)