In convex quadrilateral $A B C D$,
\[
\begin{array}{l}
\angle A B D=\angle C B D=45^{\circ}, \\
\angle A D B=30^{\circ}, \angle C D B=15^{\circ} .
\end{array}
\]

Find the ratio of the sides of quadrilateral $A B C D$.
Connect $A C$, take the incenter $I$ of $\triangle A B C$, connect $A I 、 C I$, and draw $A E \perp D C$ at point $E$.
From $\angle A B D=\angle C B D$ we know the incenter $I$ lies on $B D$.
According to the properties of the incenter of a triangle, we have
$$
\begin{array}{l}
\angle A I C=90^{\circ}+\frac{1}{2} \angle A B C \\
=90^{\circ}+\frac{1}{2} \times 90^{\circ}=135^{\circ} .
\end{array}
$$

And $\angle A D C$
$$
\begin{array}{l}
=\angle A D B+\angle C D B \\
=45^{\circ},
\end{array}
$$

Thus, $\angle A I C+\angle A D C$
$$
=180^{\circ} \text {. }
$$

Therefore, $A 、 I 、 C 、 D$ are concyclic. At this point,
$$
\begin{array}{l}
\angle I A C=\angle I D C \\
=\angle C D B=15^{\circ}, \\
\angle B A C=2 \angle I A C=30^{\circ} .
\end{array}
$$

In the right triangle $\triangle A B C$, let $B C=1$, it is easy to know
$$
\begin{array}{l}
A C=2, A B=\sqrt{A C^{2}-B C^{2}}=\sqrt{3} . \\
\text { Also, } \angle D C A=\angle D I A=\angle I A B+\angle A B I=60^{\circ},
\end{array}
$$

Thus, in the right triangle $\triangle A C E$, we have
$$
C E=1, A E=\sqrt{3} \text {. }
$$

Since $\triangle A E D$ is an isosceles right triangle, we have
$$
\begin{array}{l}
D E=A E=\sqrt{3}, \\
A D=\sqrt{2} D E=\sqrt{2} \times \sqrt{3}=\sqrt{6}, \\
C D=C E+D E=1+\sqrt{3} .
\end{array}
$$

Therefore, $A B: B C: C D: D A$
$$
=\sqrt{3}: 1:(1+\sqrt{3}): \sqrt{6} \text {. }
$$