jmc

Since $BC=8$ and $M$ is the midpoint of $BC$, $BM=CM=4$. But $AM=4$, so $M$ is the circumcenter of triangle $ABC$. Furthermore, $BC$ is a diameter of the circle, so $\angle BAC=90^{\circ }$.



Then by Pythagoras on right triangle $ABC$, $AC=\sqrt{B{C}^2-A{B}^2}=\sqrt{8^2-5^2}=\sqrt{64-25}=\boxed{\sqrt{39}}$.