Example 3. Given: $\frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{b}}{\mathrm{c}}=\frac{\mathrm{c}}{\mathrm{d}}=\frac{\mathrm{d}}{\mathrm{a}}$, find the value of $\frac{a+b+c+d}{b+a+c-d}$.
Solution 1: By the ratio theorem, we have:
$$
\begin{array}{c}
\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a} \\
=\frac{a+b+c+d}{a+b+c+d}=1, \text { so } a=b=c=d, \\
\therefore \frac{a+b+c+d}{a+b+c-d}=\frac{4 d}{2 d}=2 .
\end{array}
$$

Solution 2: Let $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=k$, then
$a=bk, b=ck, c=dk, d=ak$,
$$
\therefore a=ak^{4}, k^{4}=1, k= \pm 1 \text{. }
$$

When $k=1$, we have $a=b=c=d$,
so $\frac{a+b+c+d}{a+b+c-d}=2$,
When $k=-1$, we have $a=-b=c=-d$,
$$
\therefore \frac{a+b+c+d}{a+b+c-d}=0 \text{. }
$$

Thus, we see that Solution 1 is not comprehensive, as it only considers the case $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=1$.