3. The bisectors of the external angles of triangle $ABC$, $AA_{1}$, $BB_{1}$, and $CC_{1}$, intersect the extensions of the opposite sides of the triangle at points $A_{1}$, $B_{1}$, and $C_{1}$, respectively. Find the angle $A_{1} C_{1} B_{1}$ and the length of the segment $A_{1} B_{1}$, if $AC=5$, $BC=2$, and the angle $ACB$ is $\arccos \frac{13}{20}$. (16 points)
Solution.

$b=AC=5, a=BC=2, c=AB$,

$$
\gamma=\angle ACB=\arccos \frac{13}{20}
$$

1) We will prove that $\frac{BA_{1}}{CA_{1}}=\frac{c}{b}$. For the areas of triangles $AA_{1}B$ and $AA_{1}C$, we have the following relationships:

$$
\begin{gathered}
\frac{S_{AA_{1}B}}{S_{AA_{1}C}}=\frac{BA_{1}}{CA_{1}}, \\
S_{AA_{1}B}=c \cdot A_{1}E, S_{AA_{1}C}=b \cdot A_{1}D,
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_bfa14092a69dadb13b59g-02.jpg?height=694&width=1077&top_left_y=418&top_left_x=889)

$A_{1}E=A_{1}D$. Segments $A_{1}E$ and $A_{1}D$ are the heights of triangles $AA_{1}B$ and $AA_{1}C$. Since $AA_{1}$ is the bisector of angle $DAE$, the right triangles $AA_{1}D$ and $AA_{1}E$ are equal, and $A_{1}E=A_{1}D$. Therefore, $\frac{BA_{1}}{CA_{1}}=\frac{c}{b}$. Similarly, we get $\frac{BC_{1}}{AC_{1}}=\frac{a}{b}, \frac{AB_{1}}{CB_{1}}=\frac{c}{a}$. Then $\frac{BA_{1}}{CA_{1}} \cdot \frac{CB_{1}}{AB_{1}} \cdot \frac{AC_{1}}{BC_{1}}=\frac{c}{b} \cdot \frac{a}{c} \cdot \frac{b}{a}=1$. By Menelaus' theorem, points $A_{1}, B_{1}, C_{1}$ lie on the same line. $\angle A_{1}C_{1}B_{1}=180^{\circ}, A_{1}B_{1}=A_{1}C_{1}+C_{1}B_{1}$.

2) Using the cosine theorem, we find $c=AB$.

$$
c^{2}=a^{2}+b^{2}-2ab \cos \gamma=4+25-13=16, c=4
$$

3) $\frac{BA_{1}}{CA_{1}}=\frac{4}{5}, \frac{BA_{1}}{2+BA_{1}}=\frac{4}{5}, BA_{1}=8 ; \frac{BC_{1}}{AC_{1}}=\frac{2}{5}, \frac{BC_{1}}{4+BC_{1}}=\frac{2}{5}, BC_{1}=\frac{8}{3} ; \quad \frac{AB_{1}}{CB_{1}}=\frac{4}{2}, \frac{5+CB_{1}}{CB_{1}}=\frac{4}{2}, CB_{1}=5$. $\alpha=\angle BAC, \cos \alpha=\frac{16+25-4}{40}=\frac{37}{40}, \beta=\angle ABC, \cos \beta=\frac{16+4-25}{16}=-\frac{5}{16}$.
4) $C_{1}B_{1}^{2}=AC_{1}^{2}+AB_{1}^{2}-2AC_{1} \cdot AB_{1} \cos \alpha=\frac{400}{9}+100-2 \cdot \frac{200}{3} \cdot \frac{37}{40}=\frac{190}{9}, C_{1}B_{1}=\frac{\sqrt{190}}{3}$.

$$
A_{1}C_{1}^{2}=BA_{1}^{2}+BC_{1}^{2}-2BA_{1} \cdot BC_{1} \cos \beta=64+\frac{64}{9}+2 \cdot \frac{64}{3} \cdot \frac{5}{16}=\frac{4 \cdot 190}{9}, A_{1}C_{1}=\frac{2 \sqrt{190}}{3}
$$

$$
A_{1}B_{1}=A_{1}C_{1}+C_{1}B_{1}=\sqrt{190}
$$

Answer: $180^{\circ}, \sqrt{190}$.