10.033. 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.
## Solution.

Let's find the length of the side $B C$ (Fig. 10.33): $B C=\sqrt{B M^{2}+M C^{2}}=$ $=\sqrt{6^{2}+8^{2}}=10$ (cm). Considering that $A O$ is the bisector of $\triangle A B M$, we have $M O / O B=A M / A B$ or $r /(8-r)=6 / 10$, from which $r=3$ (cm). Since $D E \| A C$, then $\triangle D B E \sim \triangle A B C$, i.e., $D E / A C=B N / B M$ or $D E / 12=(8-2 r) / 8$, from which $D E=3$ cm.

Answer: 3 cm.