10.368 Calculate the length of the angle bisector of angle $A$ in triangle $A B C$ with side lengths $a=18$ cm, $b=15$ cm, $c=12$ cm.
Solution. Let $A K$ be the bisector of $\angle A$ of the given $\triangle A B C$. By Theorem 4, $\frac{C K}{B K}=\frac{15}{12} \Rightarrow \frac{18-B K}{B K}=\frac{15}{12} \Rightarrow B K=8, C K=10$.

Applying the Law of Cosines to $\triangle A B K$ and $\triangle A C K$, we have:

$\left\{\begin{array}{l}8^{2}=12^{2}+A K^{2}-24 \cdot A K \cdot \cos \angle K A B, \\ 10^{2}=15^{2}+A K^{2}-30 \cdot A K \cdot \cos \angle K A B\end{array} \Leftrightarrow\right.$

$\Leftrightarrow\left\{\begin{array}{l}A K^{2}-24 \cdot A K \cdot \cos \angle K A B+80=0, \\ A K^{2}-30 \cdot A K \cdot \cos \angle K A B+125=0\end{array} \Rightarrow A K=10\right.$.

Answer: 10 cm