55rd Ukrainian National Mathematical Olympiad - Third Round (Second Tour)

On the segment $BC$ put a point such as $BD=BL$ (fig. 16). Then $\Delta ABL=\Delta BLD$, mark $\angle LAB=\angle BLA=\angle BLD=\angle BDL$. On the segment $BD$ choose a point $K$, such that $\angle ALM=\angle DLK$. Then $\Delta ALM=\Delta DLK$ because of $LD=AL$, but then $\angle AML=\angle LKD=\angle BCA$, that's why $KL=AM=LC$, and it is exactly what we have to prove.