Two circles touch in $M$, and lie inside a rectangle $ABCD$. One of them touches the sides $AB$ and $AD$, and the other one touches $AD,BC,CD$. The radius of the second circle is four times that of the first circle. Find the ratio in which the common tangent of the circles in $M$ divides $AB$ and $CD$.
1. **Define the radii and positions of the circles:**
   Let the radius of the first circle be \( r \). Since the second circle's radius is four times that of the first circle, its radius is \( 4r \).

2. **Position the circles within the rectangle:**
   - The first circle touches sides \( AB \) and \( AD \). Therefore, its center is at \( (r, r) \).
   - The second circle touches sides \( AD \), \( BC \), and \( CD \). Therefore, its center is at \( (r, h - 4r) \), where \( h \) is the height of the rectangle.

3. **Determine the height of the rectangle:**
   Since the second circle touches \( AD \) and \( CD \), the distance from its center to \( CD \) is \( 4r \). Therefore, the height of the rectangle \( h \) is:
   \[
   h = r + 4r = 5r
   \]

4. **Find the common tangent at point \( M \):**
   The common tangent at \( M \) is horizontal because both circles are tangent to the vertical sides \( AD \) and \( BC \). The tangent line \( t \) touches the first circle at \( (r, r) \) and the second circle at \( (r, h - 4r) = (r, r) \).

5. **Determine the points of intersection of the tangent line with \( AB \) and \( CD \):**
   - Let \( t \cap AB = H \). Since the first circle touches \( AB \) at \( (r, r) \), the tangent line \( t \) is at height \( r \) from \( AB \). Therefore, \( H \) is the midpoint of \( AB \), and \( AH = HB \).
   - Let \( t \cap CD = I \). Since the second circle touches \( CD \) at \( (r, h - 4r) = (r, r) \), the tangent line \( t \) is at height \( r \) from \( CD \). Therefore, \( I \) is the midpoint of \( CD \), and \( ID = DC \).

6. **Determine the ratio in which the common tangent divides \( AB \) and \( CD \):**
   - Since \( H \) is the midpoint of \( AB \), the ratio \( \frac{AH}{HB} = 1 \).
   - Since \( I \) is the midpoint of \( CD \), the ratio \( \frac{ID}{DC} = 1 \).

Therefore, the common tangent divides both \( AB \) and \( CD \) in the ratio \( 1:1 \).

The final answer is \( \boxed{ 1:1 } \).