SOLUTION. First, we will show that the center $D$ of the incircle of triangle $ABC$ lies on the given circle $k$.

From the symmetry of the tangents $AB, AC$ of circle $k$ with respect to the axis $OA$, it follows that the chord $BC$ of the circle $k$ makes triangle $ABC$ isosceles, and the center $D$ of its incircle lies on the line connecting the main vertex $A$ and the midpoint $E$ of the base $BC$, with $DE$ being its radius (Fig. 3). Since the center $D$ also lies on the angle bisector of $\angle ABC$, the marked angles $\angle DBA$ and $\angle DBE$ are equal. The marked angles $\angle DAB$ and $\angle EBO$ are also equal, as both complement the same angle $\angle AOB$ to $90^\circ$. For the exterior angle $\angle ODB$ of triangle $DAB$, we have

$$
|\angle ODB| = |\angle DBA| + |\angle DAB| = |\angle DBE| + |\angle EBO| = |\angle DBO|,
$$

so triangle $OBD$ is isosceles with base $BD$, and thus $|OD| = |OB| = r$. Therefore, point $D$ indeed lies on circle $k$, and $|OE| = |OD| - |DE| = r - \varrho$.

![](https://cdn.mathpix.com/cropped/2024_04_17_c3ca9e219c63c0b48733g-6.jpg?height=591&width=920&top_left_y=704&top_left_x=568)

Fig. 3

Now, from the similarity of triangles $BOE$ and $AOB$ guaranteed by the theorem $u$, we get

$$
\frac{r-\varrho}{r} = \frac{|OE|}{|OB|} = \frac{|OB|}{|OA|} = \frac{r}{d} \quad \text{or} \quad \frac{r^2}{d} = r - \varrho,
$$

thus

$$
\varrho = r - \frac{r^2}{d} = \frac{rd - r^2}{d} = \frac{r(d - r)}{d}.
$$
5. A circle $k(O, r)$ and a point $A$ are given, where $|AO|=d>r$. Tangents from point $A$ touch the circle $k$ at points $B, C$. A circle is inscribed in triangle $ABC$. Express its radius @ in terms of the given lengths $d$ and $r$.

