60th 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 $\angle CAB=90^{\circ }$, $\angle BCA=30^{\circ }$.