62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour

Mark on the extension of median $BM$ by point $M$ such point $D$, that $BM=MD$ (fig. 11). Then triangles $ABM$, $CDM$ are equal by two sides and the angle between them, and also $BD=2BM=BC$. Then triangle $BDC$ is isosceles, and $$\begin{array}{rl}\angle ABM& =\angle CDM=\angle BCD=\angle BCA+\angle DCA\\ & =\angle BCA+\angle BAC,\end{array}$$ as desired.