smc

In triangle $ABC$, $AL$ bisects angle $A$, and $CM$ bisects angle $C$. Points $L$ and $M$ are on $BC$ and $AB$, respectively. The sides of $△ABC$ are $a$, $b$, and $c$. Then $\frac{AM}{MB}=k\frac{CL}{LB}$ where $k$ is: