Problem 5. In $\triangle A B C$, a point $E$ is marked on side $B C$ such that $\overline{A E}=\overline{B E}$. Calculate the perimeter of $\triangle A E C$, if $\overline{A C}=5 \mathrm{~cm}$ and $\overline{B C}=9 \mathrm{~cm}$.
Solution. Let $\overline{E C}=x$. Since $\overline{B C}=9 \mathrm{~cm}$, we get

$$
\overline{B E}=\overline{B C}-\overline{E C}=9-x
$$

However, $\overline{A E}=\overline{B E}$, so $\overline{A E}=9-x$. The perimeter of $\triangle A E C$ will be

$$
\begin{aligned}
L & =\overline{A E}+\overline{E C}+\overline{A C} \\
& =9-x+x+5=14 \mathrm{~cm}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_1b044dba0a21c223c4c1g-3.jpg?height=351&width=671&top_left_y=1317&top_left_x=799)

## VI Section