4. On the side $AC$ of triangle $ABC$, a point $K$ is taken which divides $AC$ in the ratio $1:3$, and on the side $BC$ a point $L$ is taken which divides $BC$ in the ratio $1:4$. Let $M$ be the intersection point of segments $AL$ and $BK$. Determine the ratio of segments $KM$ and $MB$.
Solution. We extend the segment $C A$ through vertex $A$. Let $N$ be the intersection point of the line $A C$ and the line through $B$ parallel to the line $A L$. Then, from Thales' theorem, it follows that $\frac{A N}{A C}=\frac{B L}{C L}=\frac{1}{4}$. Therefore, $A N=\frac{1}{4} A C$. Since $A K=\frac{1}{4} A C$ as well, we get $A N=A K$, i.e.

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point $A$ is the midpoint of segment $K N$ (diagram on the right). In triangle $B N K$, the line $A M$ is parallel to side $B N$, and since this line bisects side $K N$, it must also bisect side $B K$. Therefore, $K M: M B=1: 1$.