$32 \cdot 68$ Points $A, B, C, D$ are on a line in the given order and do not coincide. The lengths of segments $A B, A C$, and $A D$ are $x, y$, and $z$ respectively. If segments $A B, C D$ can rotate around points $B, C$ respectively until points $A$ and $D$ coincide, forming a triangle with a definite area, then among the following three inequalities:
I. $x<\frac{z}{2} ; \quad$ II. $y<x+\frac{z}{2}$;
III. $y<\frac{\tilde{z}}{2}$, which must be satisfied are
(A) I only.
(B) II only.
(C) I and II only.
(D) II and III only.
(E) I, II, and III.
(30th American High School Mathematics Examination, 1979)
[Solution] Since one side of a triangle is less than the sum of the other two sides, we have
$$
\begin{array}{ll}
x < y + z, \\
y < x + z, \\
z < x + y, \\
x < \frac{z}{2}.
\end{array}
$$

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