jmc

Let $O$ be the intersection of the diagonals of $ABEF$. Since the octagon is regular, $△AOB$ has area $1/8$. Since $O$ is the midpoint of $\overline{AE}$, $△OAB$ and $△BOE$ have the same area. Thus $△ABE$ has area $1/4$, so $ABEF$ has area $\boxed{\frac{1}{2}}$.