4. $\triangle A B C$ is inscribed in $\odot O, F$ is a point on arc $\overparen{B C}$, $O G \perp B F$ at point $G$, and $O G=\frac{1}{2} A C$. Then the positional relationship between $A F$ and $B C$ is $\qquad$ .
4. $A F \perp B C$.

As shown in Figure 3, draw the diameter $F M$, and connect $B M, A M$. Then $\angle M A F=90^{\circ}$.
By $O G \perp B F$
$$
\Rightarrow B G=G F \text {. }
$$

In $\triangle F B M$, by $O F=O M$, we get
$$
O G=\frac{1}{2} B M .
$$

Also, $O G=\frac{1}{2} A C$, then
$$
\begin{array}{l}
B M=A C \Rightarrow \overparen{B M}=\overparen{A C} \\
\Rightarrow M A / / B C \Rightarrow A F \perp B C
\end{array}
$$