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 $LL_1$ its median and $L_2$ its intersection with the bisectrix of $ABC$ then we obtain
$$\frac{LL_2}{LL_1} = \frac{LB}{LK} = \frac{2}{3}.$$
from similarity of the triangles $\triangle LBL_2 \sim \triangle LKL_1$ (by A-A). So the point $L_2$ divides the median $LL_1$ 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.