If $\triangle ABC$ is inscribed in a semicircle whose diameter is $AB$, then $AC+BC$ must be 
$\textbf{(A)}\ \text{equal to }AB\qquad\textbf{(B)}\ \text{equal to }AB\sqrt{2}\qquad\textbf{(C)}\ \geq AB\sqrt{2}\qquad\textbf{(D)}\ \leq AB\sqrt{2}$ $\textbf{(E)}\ AB^{2}$
Because $AB$ is the diameter of the semi-circle, it follows that $\angle C = 90$. Now we can try to eliminate all the solutions except for one by giving counterexamples.
$\textbf{(A):}$ 
Set point $C$ anywhere on the perimeter of the semicircle except on $AB$. By triangle inequality, $AC+BC>AB$, so $\textbf{(A)}$ is wrong.
$\textbf{(B):}$
Set point $C$ on the perimeter of the semicircle infinitesimally close to $AB$, and so $AC+BC$ almost equals $AB$, therefore $\textbf{(B)}$ is wrong.
$\textbf{(C):}$
Because we proved that $AC+BC$ can be very close to $AB$ in case $\textbf{(B)}$, it follows that $\textbf{(C)}$ is wrong.
$\textbf{(E):}$
Because we proved that $AC+BC$ can be very close to $AB$ in case $\textbf{(B)}$, it follows that $\textbf{(E)}$ is wrong.
Therefore, the only possible case is $\boxed{\textbf{(D)}\ \leq AB\sqrt{2}}$