smc

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. $\text{(A):}$ Set point $C$ anywhere on the perimeter of the semicircle except on $AB$. By triangle inequality, $AC+BC>AB$, so $\text{(A)}$ is wrong. $\text{(B):}$ Set point $C$ on the perimeter of the semicircle infinitesimally close to $AB$, and so $AC+BC$ almost equals $AB$, therefore $\text{(B)}$ is wrong. $\text{(C):}$ Because we proved that $AC+BC$ can be very close to $AB$ in case $\text{(B)}$, it follows that $\text{(C)}$ is wrong. $\text{(E):}$ Because we proved that $AC+BC$ can be very close to $AB$ in case $\text{(B)}$, it follows that $\text{(E)}$ is wrong. Therefore, the only possible case is $\boxed{\le AB\sqrt{2}}$