6. Given a triangle with side lengths of 7 cm, 12 cm, and 9 cm respectively. Explain how to construct a segment connecting a vertex and the opposite side with a length of 9 cm so that the perimeters of the two resulting triangles are equal.
6. First method of solution. Since in the obtained triangles one side will be common ( $A D$ ), in order for the perimeters to be equal, it is necessary that the side of 9 cm be divided into parts, the difference in lengths of which would equal the difference of the other two sides ( $12-7=5 \text{ cm}$ ). Based on this, the number 9 should be represented as the sum of two addends, one of which is 5 more than the other. The numbers that satisfy this condition are 2 and 7. Thus, we get the following solution:

![](https://cdn.mathpix.com/cropped/2024_05_21_e53d4ff4f2278a1c0d12g-067.jpg?height=414&width=814&top_left_y=1843&top_left_x=587)

Second method of solution. The solution to this problem will be easily found if the perimeter is represented as the sum of segments:

$$
\begin{aligned}
& P_{\triangle A D C}=A C+C D+D A \\
& P_{\triangle A B D}=A B+B D+D A
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_e53d4ff4f2278a1c0d12g-068.jpg?height=304&width=1010&top_left_y=474&top_left_x=521)

1) $12-7=5$ (cm) - by this much $B D$ should be longer than $D C$;
2) $9-5=4$ (cm) - two segments $C D$;
3) $4: 2=2$ (cm) - the length of segment $C D$;
4) $2+5=7$ (cm) - the length of segment $B D$.