7. Given that the midpoint of line segment $A B$ is $C$, a circle is drawn with point $A$ as the center and the length of $A B$ as the radius. On the extension of line segment $A B$, take point $D$ such that $B D = A C$; then, with point $D$ as the center and the length of $D A$ as the radius, draw another circle, which intersects $\odot A$ at points $F$ and $G$. Connect $F G$ to intersect $A B$ at point $H$. Then the value of $\frac{A H}{A B}$ is $\qquad$
7. $\frac{1}{3}$.

As shown in Figure 7, extend $A D$ to intersect $\odot D$ at point $E$, and connect $A F, E F$.
From the given conditions, we have
$$
\begin{array}{l}
A C=\frac{1}{3} A D, \\
A B=\frac{1}{3} A E .
\end{array}
$$

In $\triangle F H A$ and $\triangle E F A$,
$$
\begin{array}{l}
\angle E F A=\angle F H A=90^{\circ}, \\
\angle F A H=\angle E A F,
\end{array}
$$

thus, Rt $\triangle F H A \backsim \mathrm{Rt} \triangle E F A \Rightarrow \frac{A H}{A F}=\frac{A F}{A E}$. Since $A F=A B$, we have $\frac{A H}{A B}=\frac{1}{3}$.