Let $ABC$ be a triangle. The inside bisector of the angle $\angle BAC$ cuts $[BC]$ in $L$ and the circle $(C)$ circumsbribed to the triangle $ABC$ in $D$. The perpendicular to $(AC)$ going through $D$ cuts $[AC]$ in $M$ and the circle $(C)$ in $K$. Find the value of $\frac{AM}{MC}$ knowing that $\frac{BL}{LC}=\frac{1}{2}$.