3. In triangle $A B C$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn, $L$ is the intersection point of segments $B_{1} C_{1}$ and $A A_{1}$, $K$ is the intersection point of segments $B_{1} A_{1}$ and $C C_{1}$. Find the ratio $L M: M K$, if $M$ is the intersection point of the angle bisector $B B_{1}$ with segment $L K$, and $A B: B C: A C=2: 3: 4$. (16 points)
# Solution.

$A B: B C: A C=2: 3: 4$,

$a=B C=3 x, b=A C=4 x$,

$$
c=A B=2 x
$$

Draw the lines

$L L_{1}\left\|B B_{1}, K K_{1}\right\| B B_{1}$,

$L_{1} \in C_{1} B, K_{1} \in A_{1} B$.

$\angle C_{1} L_{1} L=\angle A_{1} K_{1} K=\angle B / 2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_98d1c66abeb3b26d6d6dg-02.jpg?height=586&width=1243&top_left_y=498&top_left_x=772)

By Thales' theorem and the properties of angle bisectors, we have the ratios: $\frac{C_{1} L_{1}}{L_{1} B}=\frac{C_{1} L}{L B_{1}}=\frac{A C_{1}}{A B_{1}}, \frac{A_{1} K_{1}}{K_{1} B}=\frac{A_{1} K}{K B_{1}}=\frac{C A_{1}}{C B_{1}}$, or $\frac{C_{1} L_{1}}{L_{1} B}=$ $\frac{A C_{1}}{A B_{1}}, \frac{K_{1} B}{A_{1} K_{1}}=\frac{C B_{1}}{C A_{1}}$

$$
\begin{aligned}
& \frac{C_{1} L_{1}}{L_{1} B} \cdot \frac{K_{1} B}{A_{1} K_{1}}=\frac{A C_{1}}{A B_{1}} \cdot \frac{C B_{1}}{C A_{1}}, \frac{K_{1} B}{L_{1} B}=\frac{A C_{1}}{A B_{1}} \cdot \frac{C B_{1}}{C A_{1}} \cdot \frac{A_{1} K_{1}}{C_{1} L_{1}}=\frac{A C_{1}}{B C_{1}} \cdot \frac{C B_{1}}{A B_{1}} \cdot \frac{B A_{1}}{C A_{1}} \cdot \frac{B C_{1}}{B A_{1}} \cdot \frac{A_{1} K_{1}}{C_{1} L_{1}}=\frac{b}{a} \cdot \frac{a}{c} \cdot \frac{c}{b} \cdot \frac{B C_{1}}{B A_{1}} \cdot \frac{A_{1} K_{1}}{C_{1} L_{1}}, \\
& \frac{K_{1} B}{L_{1} B}=\frac{A_{1} K_{1}}{B A_{1}} \cdot \frac{B C_{1}}{C_{1} L_{1}}=\frac{K K_{1}}{B B_{1}} \frac{B B_{1}}{L L_{1}},\left(\Delta A_{1} K K_{1} \sim \Delta A_{1} B_{1} B, \Delta C_{1} L L_{1} \sim \Delta C_{1} B_{1} B\right), \quad \frac{K_{1} B}{L_{1} B}=\frac{K K_{1}}{L L_{1}}
\end{aligned}
$$

Since $\angle B L_{1} L=\angle B K_{1} K, \frac{L L_{1}}{L_{1} B}=\frac{K K_{1}}{K_{1} B}$, the triangles $B L_{1} L$ and $B K_{1} K$ are similar,

$\angle L B M=\angle L_{1} L B=\angle K_{1} K B=\angle K B M$. Therefore, $B M-$ is the angle bisector of triangle $L B K$, $L M: M K=B L: B K$.

By Menelaus' theorem for triangle $A B O$, we have

$$
\begin{aligned}
& \frac{A L}{L O} \cdot \frac{O B_{1}}{B B_{1}} \cdot \frac{B C_{1}}{A C_{1}}=1, \frac{A L}{L O}=\frac{b}{a} \cdot \frac{B O+O B_{1}}{O B_{1}}=\frac{b}{a} \cdot\left(\frac{a+c}{b}+1\right)=\frac{a+c+b}{a} \\
& \Delta A L L_{1} \sim \Delta A O B, \quad \frac{A L}{A O}=\frac{L L_{1}}{B O}, \frac{A L}{A L+L O}=\frac{L L_{1}}{B O}, \quad L L_{1}=\frac{a+b+c}{2 a+b+c} \cdot B O
\end{aligned}
$$

Similarly, $\frac{C K}{K O}=\frac{a+c+b}{c}, K K_{1}=\frac{a+b+c}{a+b+2 c} \cdot B O$.

$$
\frac{L M}{M K}=\frac{L B}{K B}=\frac{L L_{1}}{K K_{1}}=\frac{a+b+2 c}{2 a+b+c}=\frac{3 x+4 x+4 x}{6 x+4 x+2 x}=\frac{11}{12} .
$$

Answer: $\frac{11}{12}$.