Belarusian Mathematical Olympiad

We mark the point $L$ on the extension of $BA$ over $A$ so that $AL = AB$. Let $M$ be the point of intersection of the medians of $LBC$. Since $AK : KC = 1 : 2$, point $K$ of the lines $BK$ and $CL$. Then $CA$ is the point of intersection of the medians of $LBC$. Therefore, $BM$ is also the median of $LBC$. Since $BK = KC$, the medians $BM$ and $CA$ are equal. Hence the triangle $LBC$ is isosceles and $LB = LC$. Since $BK$ is the bisector of the triangle of $ABC$, we have $AB : BC = AK : KC = 1 : 2$, so $BC = 2AB = BL$. Therefore, the triangle $LBC$ is equilateral and $\angle ABC = 60^\circ$. Since $\angle ABC = 60^\circ$ and $BC = 2AB$, we see that $ABC$ is a right-angled triangle, hence $\angle CAB = 90^\circ$, $\angle BCA = 30^\circ$.