$17 \cdot 124$ On line segment $AD$ there are two points $B, C$, and the lengths of line segments $AB, AC, AD$ are $a, b, c (a<b<c)$, respectively. If line segments $AB, CD$ are rotated around $B, C$ respectively, so that point $A$ coincides with point $D$ at point $E$, forming $\triangle BCE$, then which of the following three inequalities
( I ) $a<\frac{c}{2}$;
( II ) $b<a+\frac{c}{2}$;
( III ) $b<\frac{c}{2}$,

can be true?
(A) Only (I) is true.
(B) Only (II) is true.
(C) Only (I) and (II) are true.
(D) Only (II) and (III) are true.
(China Junior High School Mathematics Correspondence Competition, 1987)
[Solution] As shown in the figure, by the triangle inequality theorem, which states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side, we have:
$$
\begin{array}{l}
\left\{\begin{array}{l}
a+(b-a)>c-b, \\
(b-a)+(c-b)>a, \\
a+(c-b)>b-a .
\end{array}\right. \\
\left\{\begin{array}{l}
2 b>c, \\
2 a<c, \\
2 b<2 a+c .
\end{array}\right.
\end{array}
$$

Therefore, only (I) and (II) are correct.
Hence, the answer is (C).