jmc

Since $AC=4$ and $H$ divides $\overline{AC}$ into two pieces for which $AH=3(HC)$ we deduce that $AH=3$ and $HC=1$. We can now employ the Pythagorean Theorem in triangle $ABH$ to compute $$BH=\sqrt{(AB{)}^2-(AH{)}^2}=\sqrt{4^2-3^2}=\sqrt{7}.$$ Finally, we use the Pythagorean Theorem in triangle $BHC$ to find that $$BC=\sqrt{(BH{)}^2+(HC{)}^2}=\sqrt{(\sqrt{7}{)}^2+1^2}=\sqrt{8}=\boxed{2\sqrt{2}}.$$