Question 28
In triangle $ABC,$ $\overline{AL}$ bisects angle $A$ and $\overline{CM}$ bisects angle $C.$ Points $L$ and $M$ are on $\overline{BC}$ and $\overline{AB},$ respectively. The sides of triangle $ABC$ are $a,$ $b,$ and $c.$ Then $\dfrac{AM}{MB}=k\dfrac{CL}{LB},$ where $k$ is: