In $\triangle A B C$; $A B=A C, C D$ is the altitude, $D M \perp B C, D N \perp A C$, $M, N$ are the feet of the perpendiculars. When $C D=D M+D N$, find $\frac{A B}{B C}$ (do not use trigonometric functions).
Solution: It is obvious that $\angle A \neq 90^{\circ}$. We discuss in two cases:
(1) As shown in Figure 1, when $\angle A$ is an acute angle, let $BC=a, AB=$
$AC=b$. We have
$$
\begin{array}{l}
b \cdot CD=2 S_{\triangle ABC} \\
=2 S_{\triangle DBC}+2 S_{\triangle ACC} \\
=a \cdot DM+b \cdot DN \\
\Rightarrow b(CD-DN) \\
=a \cdot DM . \\
\because CD=DM+DN, \\
\therefore CD-DN=DM .
\end{array}
$$

Thus $b=a$, i.e., $\frac{AB}{BC}=\frac{b}{a}=1$.
At this point, $\triangle ABC$ is an equilateral triangle.
(2) As shown in Figure 2, when
$\angle A$ is an obtuse angle, extend $BA$ to $F$, such that $AF=$ $AB$, it is easy to know that $\triangle BCF$ is a right triangle. Since $CD \perp$ $BF$, we have $\angle BCD=\angle F$.

Take $CE=$ $b$ on $BC$, and connect $DE$.
$$
\begin{array}{l}
\because CD=DM+DN, \\
\therefore \frac{1}{2} b \cdot CD=\frac{1}{2} b \cdot DM+\frac{1}{2} b \cdot DN,
\end{array}
$$

i.e., $S_{\triangle ABC}=S_{\triangle DFC}+S_{\triangle DAC}$.
But $S_{\triangle ABC}=S_{\triangle ACF}=S_{\triangle FDC}+S_{\triangle DAC}$,
thus $S_{\triangle DEC}=S_{\triangle ADC}$.
From the proven $\angle ECD=\angle BCD=\angle F$, we know
$$
EC \cdot CD=CF \cdot DF \text {. }
$$

In the right $\triangle BCF$,
$$
\begin{array}{l}
CF=\sqrt{4b^2-a^2}, \\
DF=\frac{CF^2}{BF}=\frac{4b^2-a^2}{2b} \\
CD=\frac{BC \cdot CF}{BF}=\frac{a \sqrt{4b^2-a^2}}{2b},
\end{array}
$$

Substituting into (1) and simplifying, we get $4b^2-ab-a^2=0$,
i.e., $4\left(\frac{b}{a}\right)^2-\frac{b}{a}-1=0$.
Solving, we get $\frac{b}{a}=\frac{1+\sqrt{17}}{8}$, i.e., $\frac{AB}{BC}=\frac{1+\sqrt{17}}{8}$.
(Huang Quanfu, Huaining Jiangzhen Middle School, Anhui, 246142)