10.415 A line parallel to the bases of a given rectangular trapezoid divides it into two trapezoids, into each of which a circle can be inscribed. Find the bases of the original trapezoid, if its lateral sides are equal to $c$ and $d$, where $c<d$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-251.jpg?height=337&width=276&top_left_y=220&top_left_x=187)

Let in the given rectangular trapezoid $ABCD$, $c=AB$, $d=CD$, $MN \parallel BC$, and in $MBCN$ and $AMND$ inscribed circles with radii $r_1$ and $r_2$ can be inscribed. Extend the lateral sides of the trapezoid to intersect at point $P$. Let $BC=x$, $MN=y$, $AD=z$. $\triangle MPN \sim \triangle APD$.

Then $\frac{y}{z}=\frac{r_1}{r_2}=\frac{PM}{PA}$ (in similar triangles, the radii of the inscribed circles are proportional to the corresponding sides).

But $BM=2r_1$, $AM=2r_2 \Rightarrow \frac{PM}{PA}=\frac{BM}{AM} \Rightarrow \frac{y}{z}=\frac{PM}{PA}=\frac{PM-BM}{PA-AM}$ (property of equal ratios), i.e., $\frac{y}{z}=\frac{PB}{PM}$. But $\frac{PB}{PM}=\frac{x}{y}$, $\triangle BPC \sim \triangle MPN$.

In the end, $\frac{y}{z}=\frac{x}{y} \Rightarrow y^2=xz$. By theorem 2, $BM+CN=x+y$, $AM+DN=y+z \Rightarrow$ $\Rightarrow c+d=x+2y+z=x+2\sqrt{xz}+z=(\sqrt{x}+\sqrt{z})^2$, $\sqrt{x}+\sqrt{z}=\sqrt{c+d}$. If $CK$ is the height of the trapezoid, then $z-x=KD=\sqrt{d^2-c^2} \Rightarrow \sqrt{z}-\sqrt{x}=\frac{\sqrt{d^2-c^2}}{\sqrt{c+d}}=\sqrt{d-c} \Rightarrow$

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

Answer: $\frac{(\sqrt{d+c} \pm \sqrt{d-c})^2}{4}$.