65th Czech and Slovak Mathematical Olympiad

Since $ABC$ is an external angle of the isosceles triangle $XKB$ with the apex $B$ (see the picture), a line $KX$ is parallel to a bisectrix of the angle $ABC$.



Fig. 4

The ratio $LB:LK=2:3$ yields, that the bisectrix of $ABC$ meets the centroid of the triangle $KLM$. If we denote $L{L}_1$ its median and $L_2$ its intersection with the bisectrix of $ABC$ then we obtain $$\frac{L{L}_2}{L{L}_1}=\frac{LB}{LK}=\frac{2}{3}.$$ from similarity of the triangles $△LB{L}_2\sim △LK{L}_1$ (by A-A). So the point $L_2$ divides the median $L{L}_1$ in the same ratio as the centroid and therefore it is the centroid of the triangle $KLM$.

It follows from the symmetry of the problem, that bisectrix of the angle $BCA$ meets the centroid of the triangle $KLM$. And the fact, that intersection of the bisectrices is the incentre, proves the claim of the problem.