4. As shown in Figure 3, $A B$ and $C D$ are two chords of $\odot O$, $\angle A O B$ and $\angle O C D$ are supplementary, $\angle O A B$ and $\angle C O D$ are equal. Then the value of $\frac{A B}{C D}$ is $\qquad$
4. $\frac{3+\sqrt{5}}{2}$.

It is easy to get $\angle A O B=108^{\circ}, \angle O C D=72^{\circ}$, $\angle A=\angle B=\angle C O D=36^{\circ}$. As shown in Figure 9, take $B M=O B$. Then $\triangle B O M \cong \triangle O C D$. Thus, $C D=O M, A M=O M$. Also, $\triangle M A O \sim \triangle O A B$, so $O A^{2}=A B \cdot A M$, which means
$$
\begin{array}{l}
B M^{2}=A M \cdot A B . \\
\text { Therefore, } \frac{A M}{B M}=\frac{\sqrt{5}-1}{2}, \\
\frac{A M}{A M+B M}=\frac{\sqrt{5}-1}{\sqrt{5}+1}=\frac{3-\sqrt{5}}{2} . \\
\text { Then } \frac{A B}{C D}=\frac{2}{3-\sqrt{5}}=\frac{3+\sqrt{5}}{2} .
\end{array}
$$