10.246. The center of the circle inscribed in a right trapezoid is at a distance of 8 and $4 \mathrm{cm}$ from the ends of the lateral side. Find the midline of the trapezoid.
## Solution.

In trapezoid $ABCD$ (Fig. 10.53) $BC \| AD, AB \perp AD, O$ is the center of the inscribed circle, $OC=4 \text{ cm}, OD=8 \text{ cm}$. Since by the condition the circle with center $O$ touches the sides $\angle BCD$ and $\angle ADC$, then $CO$ and $OD$ are the bisectors of these angles. But $\angle BCD + \angle ADC = 180^\circ$. Then

$$
\begin{aligned}
& \angle OCD + \angle ODC = \frac{1}{2} \angle BCD + \frac{1}{2} \angle ADC = 90^\circ \\
& \text{and } \angle COD = 180^\circ - (\angle OCD + \angle ODC) = 90^\circ.
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0745.jpg?height=405&width=507&top_left_y=76&top_left_x=449)

Fig. 10.53

From $\triangle COD (\angle COD = 90^\circ): CD = \sqrt{OC^2 + OD^2} = 4\sqrt{5} \text{ cm}$. Let $M, N$, $K$ be the points of tangency of the inscribed circle with the sides $BC, AD, CD$ of the trapezoid. Then $MN$ is the diameter of the circle, $OK \perp CD, OK = r$, where $r$ is the radius of the inscribed circle. From $\triangle COD (\angle COD = 90^\circ)$:

$$
OK = \frac{OC \cdot OD}{CD} = \frac{8\sqrt{5}}{5} \text{ cm}; AB = MN = 2r = 2OK = \frac{16\sqrt{5}}{5} \text{ cm}
$$

By the property of a trapezoid in which a circle can be inscribed,

$$
BC + AD = AB + CD = \frac{16\sqrt{5}}{5} + 4\sqrt{5} = \frac{36\sqrt{5}}{5} \text{ cm. Then the length of the midline of the trapezoid is } \frac{18\sqrt{5}}{5} \text{ cm.}
$$

Answer: $\frac{18\sqrt{5}}{5}$ cm.