3. Given a parallelogram $D A B C$. On the side $D C$ is a point $L$ such that 3. $\overline{D L}=\overline{D C}$, and on the side $D A$ is a point $K$ such that $4 \cdot \overline{D K}=\overline{D A}$. Through the points $L$ and $K$ is drawn a line $p$ which intersects the diagonal $D B$ at point $M$. What fraction of the diagonal $D B$ is the segment $D M$?
Solution. We draw the lines $A A^{\prime}$ and $C C^{\prime}$ parallel to the line $p$, with the points $A^{\prime}$ and $C^{\prime}$ lying on the diagonal $D B$. Then the triangles $D L M$ and $D C C^{\prime}$ are similar, and also, the triangles $D L M$ and $D A A^{\prime}$ are similar. From the similarity of

![](https://cdn.mathpix.com/cropped/2024_06_05_a994af73711f6cb21cd1g-2.jpg?height=286&width=662&top_left_y=787&top_left_x=839)
these triangles, we get

$$
3 \cdot \overline{D M}=\overline{D C^{\prime}} \text { and } 4 \cdot \overline{D M}=\overline{D A^{\prime}}
$$

The triangles $D C C^{\prime}$ and $B A A^{\prime}$ are congruent, as they have equal sides $\overline{D C}=\overline{B A}$ and the corresponding angles at these sides are equal (the lines $C C^{\prime}$ and $A A^{\prime}$ are parallel). Therefore, $\overline{D C^{\prime}}=\overline{A^{\prime} B}$.

From the above, we have

$$
\overline{D B}=\overline{D A^{\prime}}+\overline{A^{\prime} B}=\overline{D A^{\prime}}+\overline{D C^{\prime}}=4 \cdot \overline{D M}+3 \cdot \overline{D M}=7 \cdot \overline{D M}
$$

Thus, the segment $D M$ is one-seventh of the diagonal $D B$.