Problem 19. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$. In what ratio does the point $E$, which lies on the plane passing through vertex $A$ and the centers $K$ and $H$ of the faces $A_{1} B_{1} C_{1} D_{1}$ and $B_{1} C_{1} C B$, divide the edge $B_{1} C_{1}$?
Solution. We represent the lower base of the given cube as square $A B C D$, and the line $A K$ as point $A=K$ (Fig. 10). On this diagram, the plane $A K H$ is depicted as the line $A H$. The lines $B_{1} C_{1}$ and $H A$ intersect at point $E$. For triangle $B_{1} C A$, the segments $A H$ and $B_{1} C_{1}$ are medians, so $B_{1} E: E C_{1}=2: 1$.