41. Given a rectangular trapezoid. It is known that a certain line, parallel to the bases, divides it into two trapezoids, each of which can have a circle inscribed in it. Determine the bases of the original trapezoid if its lateral sides are equal to $c$ and $d(d>c)$.
41. Extend the non-parallel sides of the trapezoid until they intersect, we will get three similar triangles, and the similarity coefficient between the middle and the larger triangle and between the smaller and the middle one is the same. Let this coefficient be $\lambda$, the larger base - $x$, the radius of the larger circle - $R$. Then the segments parallel to $x$ will be $\lambda x$ and $\lambda^{2} x$, the larger lateral side of the lower trapezoid $2 R \frac{d}{c}$, the second radius $\lambda R$. Therefore, $R+\lambda R=\frac{c}{2}$. By the property of a circumscribed quadrilateral, $x+\lambda x=2 R+2 R \frac{d}{c}$. Finally, dropping a perpendicular from the end of the smaller base of the entire trapezoid to the larger base, we get a right triangle with legs $c, x-\lambda^{2} x$ and hypotenuse $d$. Thus, we have the system

$$
\left\{\begin{aligned}
x(1+\lambda) & =2 R \frac{c+d}{c} \\
x\left(1-\lambda^{2}\right) & =\sqrt{d^{2}-c^{2}} \\
R(1+\lambda) & =\frac{c}{2}
\end{aligned}\right.
$$

from which $\lambda=\frac{d-\sqrt{\overline{d^{2}-c^{2}}}}{c}$.

Answer: the bases are $\frac{d-\sqrt{\overline{d^{2}-c^{2}}}}{2}$ and $\frac{d+\sqrt{\overline{d^{2}-c^{2}}}}{2}$.