$1 \cdot 18$ If $\quad \frac{a+b}{b+c}=\frac{c+d}{d+a}$, then
(A) $a=c$.
(B) $a+b+c+d=0$.
(C) Either $a=c$, or $a+b+c+d=0$, or both.
(D) If $a=c$, then $a+b+c+d \neq 0$.
(E) $a(b+c+d)=c(a+b+d)$.
(15th American High School Mathematics Examination, 1964)
[Solution] From $\frac{a+b}{b+c}=\frac{c+d}{d+a}$, we have $\frac{a+b}{c+d}=\frac{b+c}{d+a}$, and
$$
\frac{a+b+c+d}{c+d}=\frac{b+c+d+a}{a+d} .
$$

If $a+b+c+d \neq 0$, then $a=c$;
If $a+b+c+d=0$, then the relationship between $a$ and $c$ is undefined.
Therefore, the answer is $(C)$.