$17 \cdot 122$ As shown in the figure, in isosceles $\triangle ABC$, $CD$ is the altitude on the base $AB$, $E$ is the midpoint of the leg $BC$, and $AE$ intersects $CD$ at $F$. Now, three routes are given:
(a) $A \rightarrow F \rightarrow C \rightarrow E \rightarrow B \rightarrow D \rightarrow A$;
(b) $A \rightarrow C \rightarrow E \rightarrow B \rightarrow D \rightarrow F \rightarrow A$;
(c) $A \rightarrow D \rightarrow B \rightarrow E \rightarrow F \rightarrow C \rightarrow A$;
Let their lengths be $L(a), L(b), L(c)$, respectively. Then, among the following three relations:
$$
L(a)<L(b), \quad L(a)<L(c), \quad L(b)<L(c)
$$

the number of those that must hold is
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(China Junior High School Mathematics League, 1985)
[Solution] From the given conditions, we know that $F$ is the centroid of $\triangle ABC$, thus $CF=2DF$, $AF=2EF$.
Since
$$
\begin{array}{l}
L(c)=AB+BE+EF+FC+CA, \\
L(a)=AF+FC+CB+BA,
\end{array}
$$

Therefore, $L(c)-L(a)=BE+EF-AF=BE+EF-BF>0$,
which means
$$
L(c)>L(a) \text {. }
$$

Also, $L(b)=AC+CE+EB+BD+DF+FA$,
then
$$
\begin{aligned}
& L(c)-L(b) \\
= & (AB-BD)+(EF-FA)+(FC-DF)-CE \\
= & AD+DF-CE-EF .
\end{aligned}
$$

When $\triangle ABC$ is an equilateral triangle, $AD=CE, DF=EF$, then $L(c)=$ $L(b)$, which indicates that $L(b)0$.
This means $L(a)>L(b)$. Therefore, the inequality $L(a)<L(b)$ does not always hold.
In summary, the correct choice is $(B)$.