3. A line parallel to the bases of a given rectangular trapezoid divides it into two trapezoids, each of which can have a circle inscribed in it. Find the bases of the original trapezoid if its lateral sides are equal to \( c \) and \( d \), where \( c < d \).
Solution. Consider the given rectangular trapezoid $ABCD$, where $c=AB, d=CD, MN \| BC$, and in $MBC L$ and $AMLD$ inscribed circles with radii $r_{1}$ and $r_{2}$. Extend the lateral sides of the trapezoid to intersect at point $O$.

Let $BC=x, MN=y, AD=z$.

Notice that $\triangle MOL \sim \triangle AOD$.

Then $\frac{y}{z}=\frac{r_{1}}{r_{2}}=\frac{OM}{OA}$ (in similar triangles, the radii of the inscribed circles are proportional to the corresponding sides). Further,

$$
\begin{array}{r}
BM=2r_{1}, AM=2r_{2} \Rightarrow \frac{OM}{OA}=\frac{MB}{AM} \Rightarrow \\
\Rightarrow \frac{y}{z}=\frac{OM}{OA}=\frac{OM-BM}{OA-AM}
\end{array}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_9b22d083e3fa0031316bg-07.jpg?height=499&width=346&top_left_y=390&top_left_x=752)

i.e., $\frac{y}{z}=\frac{OB}{OM}$. Since $\frac{OB}{OM}=\frac{x}{y} (\triangle BOC \sim \triangle MOL)$, then $\frac{y}{z}=\frac{x}{y} \Rightarrow y^{2}=xz$.

We have $BM+CL=x+y, AM+DL=y+z$. Therefore,

$$
c+d=x+2y+z=x+2\sqrt{xz}+z=(\sqrt{x}+\sqrt{z})^{2}, \sqrt{x}+\sqrt{z}=\sqrt{c+d}
$$

Let $CK-$ be the height of the trapezoid. Then $z-x=KD=\sqrt{d^{2}-c^{2}}$, and thus,

$$
\sqrt{z}-\sqrt{x}=\frac{\sqrt{d^{2}-x^{2}}}{\sqrt{c+d}}=\sqrt{d-c} \Rightarrow \sqrt{z}=\frac{\sqrt{d+c}+\sqrt{d-c}}{2}
$$

and

$$
\sqrt{x}=\frac{\sqrt{d+c}-\sqrt{d-c}}{2}
$$

Thus, the bases of the trapezoid are

$$
\frac{(\sqrt{d+c} \pm \sqrt{d-c})^{2}}{2}
$$

Answer. $\frac{(\sqrt{d+c} \pm \sqrt{d-c})^{2}}{2}$