10.205. A circle with a radius of 5 cm is inscribed in some angle. The length of the chord connecting the points of tangency is 8 cm. Two tangents, parallel to the chord, are drawn to the circle. Find the sides of the resulting trapezoid.
Solution.

The circle with center $O$ (Fig. 10.15) touches the sides of angle $E$ at points $K$ and $L, K L=8 \text{ cm}, O K=5 \text{ cm}, A B\|C D\| K L$. Draw the diameter of the circle $M N \| K L, P$ is the intersection point of $E O$ and $K L, \angle K P O=90^{\circ}, \triangle K P O \sim$ $\sim \triangle O K M-$ by two angles ( $\left.\angle P K O=\angle K O M, \angle K P O=\angle O K M=90^{\circ}\right)$. Then $\frac{K P}{K O}=\frac{K O}{M O}, M O=\frac{K O^{2}}{K P}=\frac{25}{4}$ cm. $M N=2 M O=\frac{25}{2}$ cm. $M N-$ is the midline of trapezoid $A B C D$ and, since a circle is inscribed in this trapezoid, then $B C=A D=M N=\frac{25}{2}$ cm. $B B_{1}$ is the height of trapezoid $A B C D$, $B B_{1}=2 O K=10 \text{ cm}$. From $\triangle B B_{1} C\left(\angle B B_{1} C=90^{\circ}\right): B_{1} C=\sqrt{B C^{2}-B B_{1}^{2}}=7.5 \text{ cm}$. $C D-A B=2 B_{1} C=15$ cm, $C D+A B=2 B C=25$ cm. Then $C D=20$ cm, $A B=5$ cm.

Answer: 20 cm; 5 cm; 12.5 cm; 12.5 cm.