7. In a cyclic quadrilateral $ABCD$, there is a point $X$ on side $AB$ such that diagonal $BD$ bisects $CX$ and $AC$ bisects $DX$. Find the minimum value of $\frac{AB}{CD}$.

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7. As shown in Figure 1, let the midpoints of $C X$ and $D X$ be $M$ and $N$ respectively,
$$
\begin{array}{l}
\angle B A C=\angle B D C=\alpha, \angle C A D=\angle C B D=\beta . \\
\quad \text { By } S_{\triangle A X N}=S_{\triangle A D N} \\
\Rightarrow A X \cdot A N \sin \alpha=A D \cdot A N \sin \beta \\
\Rightarrow \frac{A D}{A X}=\frac{\sin \alpha}{\sin \beta}=\frac{B C}{C D} \\
\Rightarrow \frac{A X}{C D}=\frac{A D}{B C} .
\end{array}
$$

Similarly, $\frac{B X}{C D}=\frac{B C}{A D}$.
By the AM-GM inequality,
$$
\frac{A B}{C D}=\frac{A X}{C D}+\frac{B X}{C D}=\frac{A D}{B C}+\frac{B C}{A D} \geqslant 2 \text {. }
$$

When $\frac{A B}{C D}=2, A D=B C$, quadrilateral $A B C D$ is an isosceles trapezoid, satisfying $A B=2 C D$, and $X$ is the midpoint of $A B$.