1. Draw $K M$ until it intersects with $C B$ at point $N$. Let $A M=2 y, M D=3 y, A K=x$, $K B=3 x$. Then $B L=4 y, L C=y$. Triangles $A K M$ and $N K B$ are similar, hence $N B=6 y$, since $A K: K B=1: 3$. Therefore, $N L=6 y+4 y=10 y$. Triangles $N Q L$ and $A Q M$ are similar, so $A Q: Q L=A M: N L=2 y: 10 y=1: 5$. At the same time, $N Q: Q M=5: 1$ and $N K: K M=3: 1$. Let $N K=3 t, K M=t, N Q=5 p, Q M=p$. From this,

$$
4 t=6 p \Rightarrow t=\frac{3}{2} p
$$

Thus, $K Q=N Q-N K=5 p-3 t=5 p-\frac{9}{2} p=\frac{1}{2} p, Q M=p$. Therefore, $K Q: Q M=1: 2$.
Answer: 1. 1:5, 1:2. 2. 3:13, 1:3. 3. 2:7, 1:8. 4. 3:10, $1: 12$

## Grading Criteria:

All solved correctly - 20 points.

Additional construction made, similarity of triangles proven - from 5 to 10 points. In all other cases - 0 points.

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