Example 39. An isosceles triangle is inscribed in a circle, with the base equal to $a$, and the angle at the base equal to $\beta$. We will find the radius of the circle that is tangent to the lateral sides of the triangle and the first circle.

Given (Fig. 43): $\omega_{1}\left(O_{1},\left|O_{1} A\right|\right), \triangle A B C,\{A, B, C\} \subset \omega_{1}$, $|A B|=|A C|,|B C|=a, \widehat{A B C}=\beta, \omega_{2}$ - a circle tangent to $[A B],[A C]$ and $\omega_{1}$.

Find: the radius of the circle $\omega_{2}$.
Solution. Construct point $D$ - the midpoint of the base $BC$ of triangle $ABC$. Since $|AB|=|AC|$, $(AD)$ is the perpendicular bisector of segment $BC$. Since $\omega_{1}$ is circumscribed around $\triangle ABC$, its center - point $O_{1} \in (AD)$.

It is clear that the bisector of angle $BAC$ lies on line $AD$. The center of circle $\omega_{2}$ - point $O_{2}$ - belongs to the bisector of angle $BAC$, i.e., $O_{2} \in (AD)$. Thus, points $O_{1}$ and $O_{2}$ belong to line $AD$.

Let circle $\omega_{2}$ touch circle $\omega_{1}$ at point $E$. Then points $O_{1}, O_{2}$, and $E$ belong to the same line. Therefore, $E \in (AD)$ and $\left|O_{2}E\right|$ is the length of the radius of circle $\omega_{2}$.

Since by the Law of Sines $\frac{|BC|}{\sin B \widehat{AC}}=2\left|O_{1}A\right|$, we have

$$
\left|O_{1}A\right|=\frac{a}{2 \sin \widehat{AC}}
$$

Since $\widehat{BAC}=180^{\circ}-2\beta$, we have

$$
\left|O_{1}A\right|=\frac{a}{2 \sin (180^{\circ}-2\beta)} \text{, or } \left|O_{1}A\right|=\frac{a}{2 \sin 2\beta} \text{. }
$$

Let $\omega_{2}$ touch $[AB]$ at point $K$, then $\left[O_{2}K\right] \perp [AB]$ and $\left|O_{2}K\right|$ is the length of the radius of circle $\omega_{2}$. Let $\left|O_{2}E\right|=\left|O_{2}K\right|=x$.

Since $\widehat{BAD}=90^{\circ}-\beta$, from the right triangle $\triangle AO_{2}K$ we get: $x=\left|AO_{2}\right| \sin (90^{\circ}-\beta)$. Considering that

$$
\left|AO_{2}\right|=|AE|-\left|O_{2}E\right|=2\left|O_{1}A\right|-\left|O_{2}E\right|=\frac{a}{\sin 2\beta}-x
$$

we form the equation

$$
x=\left(\frac{a}{\sin 2\beta}-x\right) \cos \beta
$$

solving which we find:

$$
x=\frac{a}{4 \cos^2 \frac{\beta}{2} \sin \beta}
$$

Investigation. By the meaning of the problem $00$ and $0^{\circ}<\beta<90^{\circ}$, is equivalent to the system

$$
0<\frac{1}{\cos^2 \frac{\beta}{2}}<\frac{1}{\cos \beta}
$$

which in turn is equivalent to the system $0<\cos \beta<\cos^2 \frac{\beta}{2}$, or $0<\cos \beta<\frac{1+\cos \beta}{2}$,

i.e.,

$$
0<2 \cos \beta<\cos^2 \frac{\beta}{2}, \text{ or } 0<\cos \beta<1,
$$

from which $0^{\circ}<\beta<90^{\circ}$.

Thus,

$$
\left|O_{2}K\right|=\frac{a}{4 \cos^2 \frac{\beta}{2} \sin \beta}, \text{ where } 0^{\circ}<\beta<90^{\circ}
$$