In $\triangle A B C$, $A B=37, A C=58$. With $A$ as the center and $A B$ as the radius, an arc is drawn intersecting $B C$ at point $D$, and $D$ is between $B$ and $C$. If the lengths of $B D$ and $D C$ are both integers, find the length of $B C$.
Solve as shown in the figure, first prove
$$
\begin{array}{l}
B C \cdot C D \\
=A C^{2}-A B^{2} .
\end{array}
$$

Draw $A H \perp B C$ at $H$, then
$$
\begin{array}{l}
A C^{2}=A H^{2}+C H^{2}, \\
A B^{2}=A H^{2}+B H^{2},
\end{array}
$$

we have
$$
\begin{array}{l}
A C^{2}-A B^{2}=C H^{2}-B H^{2} \\
\quad=(C H+B H)(C H-B H)=B C \cdot C D .
\end{array}
$$

Also, $\because A B=37, A C=58$.
$$
\therefore B C \cdot C D=58^{2}-37^{2}=3 \times 5 \times 7 \times 19 \text {. }
$$

Since $A C - A B < B C, D$ is not between $B, C$, so it should be discarded, hence we should take $B C=57$, at this time $C D$ $=35, B D=24$.