Points $A , B, C$, and $D$ are distinct and lie, in the given order, on a straight line. 
Line segments $AB, AC$, and $AD$ have lengths $x, y$, and $z$, respectively. 
If line segments $AB$ and $CD$ may be rotated about points $B$ and $C$, respectively, 
so that points $A$ and $D$ coincide, to form a triangle with positive area, 
then which of the following three inequalities must be satisfied?
$\textbf{I. }x<\frac{z}{2}\qquad\\ \textbf{II. }y<x+\frac{z}{2}\qquad\\ \textbf{III. }y<\frac{z}{2}\qquad$

$\textbf{(A) }\textbf{I. }\text{only}\qquad \textbf{(B) }\textbf{II. }\text{only}\qquad \textbf{(C) }\textbf{I. }\text{and }\textbf{II. }\text{only}\qquad \textbf{(D) }\textbf{II. }\text{and }\textbf{III. }\text{only}\qquad \textbf{(E) }\textbf{I. },\textbf{II. },\text{and }\textbf{III. }$
Solution by e_power_pi_times_i
We know that this triangle has lengths of $x$, $y-x$, and $z-y$. Using the Triangle Inequality, we get $3$ inequalities: $2y>z, z>2x, 2x+z>0$. Therefore, we know that $\textbf{I}$ is true and $\textbf{III}$ is false. In $\textbf{II}$, we have to prove $2yz$, so we have to prove $z<2y<2x+z$. $2x<z$, so we have to prove that $z<2z$, which is true for all positive $z$. Therefore the answer is $\boxed{\textbf{(C) } \textbf{I. }\text{and }\textbf{II. }\text{only}}$.