4. The segments $\overline{A C}$ and $\overline{B D}$ intersect at point $O$. The perimeter of triangle $A B C$ is equal to the perimeter of triangle $A B D$, and the perimeter of triangle $A C D$ is equal to the perimeter of triangle $B C D$. Find the length of segment $\overline{A O}$, if $\overline{B O}=10$ cm.
Solution. Since $L_{\triangle A B C}=L_{\triangle A B D}$, we have

$$
\overline{A C}+\overline{B C}=\overline{A D}+\overline{B D}
$$

and since $L_{\triangle A C D}=L_{\triangle B C D}$, we have

$$
\overline{A C}+\overline{A D}=\overline{B C}+\overline{B D}
$$

We rewrite the second equation as

$$
\overline{A C}-\overline{B C}=\overline{B D}-\overline{A D}
$$

and add it to the first equation, then we get

![](https://cdn.mathpix.com/cropped/2024_06_05_5d3747084f372969d7bdg-2.jpg?height=366&width=466&top_left_y=920&top_left_x=995)
$2 \overline{A C}=2 \overline{B D}$, which means $\overline{A C}=\overline{B D}$. Therefore, $\overline{A D}=\overline{B C}$, so $\triangle A B D \cong \triangle A B C$, from which $\measuredangle A B D=\measuredangle C A B$. Hence, $\triangle A O B$ is isosceles, so $\overline{A O}=\overline{B O}=10$ cm.