Problem 2. In an isosceles triangle $ABC (\overline{AB} = \overline{AC})$, the leg $AC$ is extended through vertex $A$ to point $D$, such that the perimeter of triangle $BAD$ is $16 \text{~cm}$. Calculate the base $BC$ of triangle $ABC$, if the perimeter of triangle $BCD$ is $29 \text{~cm}$.
Solution. We have

$$
\begin{aligned}
& L_{\triangle B A D}=\overline{B A}+\overline{A D}+\overline{D B}=16 \text{~cm} \\
& L_{\triangle B C D}=\overline{B C}+\overline{C A}+\overline{A D}+\overline{D A}=29 \text{~cm} \\
& \overline{C A}=\overline{B A} \\
& \overline{B C}=L_{\triangle B C D}-L_{\triangle B A D}=29-16=13 \text{~cm}
\end{aligned}
$$

Thus, $\overline{B C}=13 \text{~cm}$.

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