jmc

We add the edges of the pyramid to our diagram below.



Taking $ABC$ to be the base of pyramid $ABCH$, the height is $HD$. Since $ABC$ is half unit square $ABCD$, the area of $ABC$ is $\frac{1}{2}$. Therefore, the volume of pyramid $ABCH$ is $$\frac{[ABC]\cdot HD}{3}=\frac{(1/2)(1)}{3}=\boxed{\frac{1}{6}}.$$