3.23. In a right triangle $ABC$, the acute angle $A$ is equal to $\alpha$ radians. An arc of a circle centered at the vertex of the right angle $C$ touches the hypotenuse at point $D$ and intersects the legs $AC$ and $BC$ at points $E$ and $F$, respectively. Find the ratio of the areas of the curvilinear triangles $ADE$ and $BDF$.
3.23. By the condition, $D$ is the point of tangency of the arc $FE$ of the circle with center $C$ inscribed in the right triangle $ABC$ (Fig. 3.25). This means that $CD \perp AB$ and $CD=R$, where $R$ is the radius of the circle. We need to find

$$
\frac{S_{A D E}}{S_{B D F}}=\frac{S_{\triangle D A C}-S_{\text {cext. EDC }}}{S_{\triangle D B C}-S_{\text {cext. FDC }}}
$$

Since $\angle B A C=\alpha$ (by the condition), then

$\angle A C D=\frac{\pi}{2}-\alpha, \angle B C D=\alpha$, from which $A D=C D \operatorname{ctg} \alpha=R \operatorname{ctg} \alpha, B D=$ $=C D \operatorname{tg} \alpha=R \operatorname{tg} \alpha$. Then $S_{\triangle D A C}=\frac{1}{2} R^{2} \operatorname{ctg} \alpha, S_{\triangle D B C}=\frac{1}{2} R^{2} \operatorname{tg} \alpha$. Further, $S_{\text {cext.EDC }}=\frac{1}{2} R^{2}\left(\frac{\pi}{2}-\alpha\right), S_{\text {cext } \cdot F D C}=\frac{1}{2} R^{2} \alpha$. Therefore,

$$
S_{A D E}=\frac{1}{2} R^{2}\left(\operatorname{ctg} \alpha-\frac{\pi}{2}+\alpha\right) S_{B D F}=\frac{1}{2} R^{2}(\operatorname{tg} \alpha-\alpha)
$$

and finally we get $\frac{S_{A D E}}{S_{B D F}}=\frac{\operatorname{ctg} \alpha-\frac{\pi}{2}+\alpha}{\operatorname{tg} \alpha-\alpha}$.