Example 44. In triangle $A B C$, the ratio $|B C|:|A C|=3$ and $\widehat{A C B}=\alpha$. Points $D$ and $K$ are taken on side $A B$ such that $\widehat{A C D}=\widehat{D C K}=\widehat{K C B}$. Find: $|C D|:|C K|$.
S o l u t i o n (Fig. 48). The problem under consideration is of type $M$. We will solve it by introducing an auxiliary parameter. Let, for example, $|A C|=b$, then $|B C|=3 b$.

Using the area $S_{\triangle A B C}$ as a reference element (see example 9, p. 22), we will express it in two ways. On the one hand,

$$
\begin{aligned}
S_{\triangle A B C}=S_{\triangle A C D} & +S_{\triangle D C B}=\frac{1}{2} b|C D| \sin \frac{\alpha}{3}+\frac{1}{2} 3 b|C D| \sin \frac{2 \alpha}{3}= \\
& =\frac{1}{2} b|C D|\left(\sin \frac{\alpha}{3}+3 \sin \frac{2 \alpha}{3}\right)
\end{aligned}
$$

On the other hand,

$$
\begin{aligned}
S_{\triangle A B C}=S_{\triangle A C K} & +S_{\triangle K C B}=\frac{1}{2} b|C K| \sin \frac{2 \alpha}{3}+\frac{1}{2} 3 b|C K| \sin \frac{\alpha}{3}= \\
& =\frac{1}{2} b|C K|\left(\sin \frac{2 \alpha}{3}+3 \sin \frac{\alpha}{3}\right)
\end{aligned}
$$

Thus,

$$
|C D| \cdot \frac{b}{2}\left(\sin \frac{\alpha}{3}+3 \sin \frac{2 \alpha}{3}\right)=|C K| \cdot \frac{b}{2}\left(\sin \frac{2 \alpha}{3}+3 \sin \frac{\alpha}{3}\right)
$$

from which

$$
\frac{|C D|}{|C K|}=\frac{\sin \frac{2 \alpha}{3}+3 \sin \frac{\alpha}{3}}{\sin \frac{\alpha}{3}+3 \sin \frac{2 \alpha}{3}}=\frac{2 \sin \frac{\alpha}{3} \cos \frac{\alpha}{3}+3 \sin \frac{\alpha}{2}}{\sin \frac{\alpha}{3}+6 \sin \frac{\alpha}{3} \cos \frac{\alpha}{3}}=\frac{2 \cos \frac{\alpha}{3}+3}{1+6 \cos \frac{\alpha}{3}}
$$

Investigation. By the problem's meaning, $b=|A C|>0$
$$

The last inequality is valid because, by the problem's meaning, $0^{\circ}<\alpha<180^{\circ}$, hence $\cos \frac{\alpha}{3}>-\frac{1}{2}$, and $\cos \frac{\alpha}{3}+\frac{1}{6}>0$
$$

Thus,

$$
|C D|:|C K|=\frac{2 \cos \frac{\alpha}{3}+3}{1+6 \cos \frac{\alpha}{3}}
$$