12.178. In a right triangle $ABC$, the bisector $AD$ of the acute angle $A$, equal to $\alpha$, is drawn. Find the ratio of the radii of the circles inscribed in triangles $ABD$ and $ADC$.
Solution.

Let $O$ be the center of the circle inscribed in $\triangle A B D$, $F$ be the point of tangency of this circle with $A D$, $O_{1}$ be the center of the circle inscribed in $\triangle A D C$, and $E$ be the point of tangency of this circle with $A D$ (Fig. 12.48).

\[
\begin{aligned}
& \angle B A D=\frac{\alpha}{2}, \angle A D B=90^{\circ}-\frac{\alpha}{2}, \angle O_{1} A E=\angle F A O=\frac{1}{2} \angle B A D=\frac{\alpha}{4}, \\
& \angle F D O=45^{\circ}-\frac{\alpha}{4} .
\end{aligned}
\]

$\angle O_{1} D O=90^{\circ}$, since it is formed by the angle bisectors of adjacent angles $A D C$ and $A D B$. Then $\angle E O_{1} D=\angle F D O$, as angles with mutually perpendicular sides.

If $O F=r, O_{l} E=R$, then from $\triangle A F O\left(\angle A F O=90^{\circ}\right): A F=r \operatorname{ctg} \frac{\alpha}{4} ;$

from $\triangle O F D\left(\angle O F D=90^{\circ}\right): D F=r \operatorname{ctg}\left(45^{\circ}-\frac{\alpha}{4}\right)$;

from $\triangle O_{1} E D\left(\angle O, E D=90^{\circ}\right): D E=R \operatorname{tg}\left(45^{\circ}-\frac{\alpha}{4}\right)=R \operatorname{ctg}\left(45^{\circ}+\frac{\alpha}{4}\right)$;

from $\triangle O, E A\left(\angle O, E A=90^{\circ}\right): A E=R \operatorname{ctg} \frac{\alpha}{4}$.

\[
\begin{aligned}
& A F+F D=A E+E D \Rightarrow r \operatorname{ctg} \frac{\alpha}{4}+r \operatorname{ctg}\left(45^{\circ}-\frac{\alpha}{4}\right)=R \operatorname{ctg} \frac{\alpha}{4}+ \\
& +R \operatorname{ctg}\left(45^{\circ}+\frac{\alpha}{4}\right) \Leftrightarrow r \cdot \frac{\sin 45^{\circ}}{\sin \frac{\alpha}{4} \sin \left(45^{\circ}-\frac{\alpha}{4}\right)}=R \cdot \frac{\sin \left(45^{\circ}+\frac{\alpha}{2}\right)}{\sin \frac{\alpha}{4} \sin \left(45^{\circ}+\frac{\alpha}{4}\right)} \Leftrightarrow \\
& \Leftrightarrow \frac{R}{r}=\frac{\sin \left(45^{\circ}+\frac{\alpha}{4}\right) \sin 45^{\circ}}{\sin \left(45^{\circ}-\frac{\alpha}{4}\right) \sin \left(45^{\circ}+\frac{\alpha}{2}\right)}=\frac{\sqrt{2} \sin \left(45^{\circ}+\frac{\alpha}{4}\right)}{2 \cos \left(45^{\circ}+\frac{\alpha}{4}\right) \sin \left(45^{\circ}+\frac{\alpha}{2}\right)}= \\
& =\frac{\sqrt{2} \operatorname{tg}\left(45^{\circ}+\frac{\alpha}{4}\right)}{2 \sin \left(45^{\circ}+\frac{\alpha}{2}\right)} \\
& \text { Answer: } \frac{\sqrt{2} \operatorname{tg}\left(45^{\circ}+\frac{\alpha}{4}\right)}{2 \sin \left(45^{\circ}+\frac{\alpha}{2}\right)}
\end{aligned}
\]

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0931.jpg?height=503&width=509&top_left_y=1093&top_left_x=125)

Fig. 12.49