Problem 5. On the segment $A B$, points $C, D, E, F$ are arranged in the given order, such that $\overline{C D}=\overline{D E}=\overline{E F}$. The distance between the midpoints of segments $C D$ and $E F$ is $12 \mathrm{~cm}$, and the distance between the midpoints of segments $A C$ and $F B$ is $23 \mathrm{~cm}$. Find the length of segment $A B$.
Solution. Let $G, H, K, L$ be the midpoints of segments $A C$,

![](https://cdn.mathpix.com/cropped/2024_06_05_d4fd26fa5c16caf575adg-3.jpg?height=79&width=577&top_left_y=1133&top_left_x=882)
$C D, E F, F B$, respectively (see diagram).

Since $\overline{C D}=\overline{E F}$, it follows that $\overline{C H}=\overline{E K}$ and thus $\overline{H D}+\overline{E K}=\overline{C D}=\overline{D E}$. Then we have $12=\overline{H D}+\overline{D E}+\overline{E K}=2 \overline{D E}$, so $\overline{D E}=6 \mathrm{~cm}$. From this, we get $\overline{C F}=3 \overline{D E}=18 \mathrm{~cm}$. On the other hand,

$$
\overline{G L}=\overline{G C}+\overline{C F}+\overline{F L}, \text { and } \overline{G C} \equiv \overline{A G}=\frac{\overline{A C}}{2} \text { and } \overline{F L}=\overline{L B}=\frac{\overline{F B}}{2} \text {, }
$$

so we have $23=\frac{\overline{A C}}{2}+18+\frac{\overline{F B}}{2}$. From the last equality, we get $\frac{\overline{A C}}{2}+\frac{\overline{F B}}{2}=5 \mathrm{~cm}$. According to this,

$$
\overline{A B}=\overline{A G}+\overline{G L}+\overline{L B}=\frac{\overline{A C}}{2}+\frac{\overline{F B}}{2}+\overline{G L}=5+23=28 \mathrm{~cm}
$$

## VI section