1. In a convex quadrilateral $ABCD$, $\angle B=\angle D=90^{\circ}, AB=4, AD=5, \angle A=60^{\circ}$, $\frac{BC}{CD}=$ $\qquad$ .
1. 2 .

Connect $A C$, let $A C=R$, then $\sin \angle D A C=\frac{D C}{A C}=\frac{\sqrt{R^{2}-25}}{R}, \cos \angle D A C=\frac{5}{R}, \sin \angle B A C=$
$$
\begin{array}{l}
\frac{B C}{A C}=\frac{\sqrt{R^{2}-16}}{R}, \cos \angle B A C=\frac{A B}{A C}=\frac{4}{R}, \frac{1}{2}=\cos 60^{\circ}=\cos (\angle D A C+\angle B A C)=\frac{20}{R^{2}}- \\
\frac{\sqrt{R^{2}-25} \cdot \sqrt{R^{2}-16}}{R^{2}}, \text { i.e., } \sqrt{\left(R^{2}-25\right)\left(R^{2}-16\right)}=20-\frac{1}{2} R^{2}, \frac{3}{4}\left(R^{2}\right)^{2}-21 R^{2}=0, \frac{3}{4} R^{2}- \\
21=0 \Rightarrow \frac{3}{4} R^{2}=21 \Rightarrow R^{2}=28, D C=\sqrt{R^{2}-25}=\sqrt{3}, B C=\sqrt{R^{2}-16}=2 \sqrt{3} . \text { Therefore, } \frac{B C}{C D} \\
=2 .
\end{array}
$$