12.167. The radius of the sector's arc is $R$, the central angle $A O B$ is $\alpha$. Through the midpoint $C$ of the radius $O A$, a line is drawn parallel to the radius $O B$ and intersects the arc $A B$ at point $D$. Find the area of triangle $O C D$.
Solution.

Let $\angle B O D=\beta$ (Fig. 12.38). Then $\angle C D O=\beta, \angle D O C=\alpha-\beta$, $\angle D C O=180^{\circ}-\alpha$.

In $\triangle C O D: \frac{D O}{\sin \angle D C O}=\frac{C O}{\sin \angle C D O} \Leftrightarrow \frac{R}{\sin \left(180^{\circ}-\alpha\right)}=\frac{R}{2 \sin \beta} ; \sin \beta=\frac{\sin \alpha}{2}$.

Since $O C=\frac{1}{2} O D$, then $\angle O D C=\beta$ is not the largest angle of triangle $C O D$ and $\cos \beta>0$,

$$
\begin{aligned}
& \quad \cos \beta=\sqrt{1-\sin ^{2} \beta}=\sqrt{1-\frac{\sin ^{2} \alpha}{4}}=\frac{\sqrt{4-\sin ^{2} \alpha}}{2} \\
& S_{\triangle O C D}=\frac{1}{2} C O \cdot D O \sin \angle C O D=\frac{1}{2} \cdot \frac{R}{2} \cdot R \sin (\alpha-\beta)= \\
& =\frac{R^{2}}{4}(\sin \alpha \cos \beta-\cos \alpha \sin \beta)=\frac{R^{2}}{4}\left(\sin \alpha \cdot \frac{\sqrt{4-\sin ^{2} \alpha}}{2}-\cos \alpha \cdot \frac{\sin \alpha}{2}\right)= \\
& =\frac{R^{2} \sin \alpha}{8}\left(\sqrt{4-\sin ^{2} \alpha}-\cos \alpha\right) \\
& \\
& \text { OmBem: } \frac{R^{2} \sin \alpha}{8}\left(\sqrt{4-\sin ^{2} \alpha}-\cos \alpha\right)
\end{aligned}
$$