In triangle $ABC$ on side $AC$ are points $K$, $L$ and $M$ such that $BK$ is an angle bisector of $\angle ABL$, $BL$ is an angle bisector of $\angle KBM$ and $BM$ is an angle bisector of $\angle LBC$, respectively. Prove that $4 \cdot LM <AC$ and $3\cdot \angle BAC - \angle ACB < 180^{\circ}$