1.20. A tangent, parallel to the base, is drawn to the circle inscribed in an isosceles triangle with a base of 12 cm and a height of 8 cm. Find the length of the segment of this tangent, enclosed between the sides of the triangle.
1.20. Let's find the length of the side $B C$ (Fig. 1.18); $B C=\sqrt{B M^{2}+M C^{2}}=\sqrt{6^{2}+8^{2}}=10$ (cm). Considering that $A O$ is the bisector of $\triangle A B M$, we have $\frac{M O}{O B}=\frac{A M}{A B}$, or $\frac{r}{8-r}=\frac{6}{10}$, from which $r=3$ (cm). Since $D E \| A C$, then $\triangle D B E \sim \triangle A B C$. That is, $\frac{D E}{A C}=\frac{B N}{B M}$, or $\frac{D E}{12}=\frac{8-2 r}{8}$, from which $D E=3 \text{ cm}$.

Answer: 3 cm.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-423.jpg?height=525&width=687&top_left_y=107&top_left_x=272)

Fig. 1.18

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-423.jpg?height=546&width=481&top_left_y=1238&top_left_x=44)