3.2. In the circle $\omega$, an isosceles triangle $ABC$ with base $AC$ is inscribed. It turns out that the radius of the inscribed circle of $ABC$ is equal to the radius of the circle that touches the smaller arc $BC$ of the circle $\omega$ and the side $BC$ at its midpoint (see Fig. 2). Find the ratio of the sides of triangle $ABC$.
3.2. Answer: $A B: B C: A C=2: 2: 1$.

Let $I$ be the incenter of triangle $A B C, P$ the point of tangency of this circle with side $B C, M$ the midpoint of $B C, Q$ the midpoint of arc $B C$ of the circumcircle of triangle $A B C$ (see Fig. 4). Since $\angle I C P=1 / 2 \angle A C B=$ $\angle Q C M$ and $\angle I P C=90^{\circ}=\angle Q M C$, then $\frac{P C}{M C}=\frac{I P}{Q M}=1 / 2$. Also $\frac{A C}{P C}=2$ and $\frac{B C}{M C}=2$, therefore, $\frac{A B}{A C}=2$.