jmc

Since $\angle C=45^{\circ }$, triangle $ACD$ is a $45^{\circ }$-$45^{\circ }$-$90^{\circ }$ triangle, which means $AD=CD=AC/\sqrt{2}=3$. Then $BD=BC-CD=4-3=1$.



Also, $\angle EBC=90^{\circ }-\angle BCE=45^{\circ }$, so triangle $BHD$ is a $45^{\circ }$-$45^{\circ }$-$90^{\circ }$ triangle. Hence, $HD=BD=1$. Then $AH=AD-HD=3-1=2$, so $AH:HD=\boxed{2}$.