2. In quadrilateral $A B C D$, $\angle A: \angle B: \angle C$ : $\angle D=9: 12: 10: 17, A B=B C, C D=2$. Then the length of $B C$ is $\qquad$.
$2.1+\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2}$.
From the given information, we have
$$
\begin{array}{l}
\angle A=\frac{9}{9+12+10+17} \times 360^{\circ}=67.5^{\circ}, \\
\angle B=\frac{12}{9+12+10+17} \times 360^{\circ}=90^{\circ}, \\
\angle C=\frac{10}{9+12+10+17} \times 360^{\circ}=75^{\circ} .
\end{array}
$$

As shown in Figure 8, connect $A C$.
Since $A B=B C$, we have
$$
\begin{array}{l}
\angle B A C=\angle B C A \\
=45^{\circ} .
\end{array}
$$

Therefore, $\angle D A C$
$$
\begin{array}{l}
=\angle B A D-\angle B A C \\
=22.5^{\circ} \text {, } \\
\angle A C D=\angle B C D-\angle B C A=30^{\circ} .
\end{array}
$$

Draw $D E \perp A C$ at $E$. Then
$$
\begin{array}{l}
D E=\frac{1}{2} C D=1, \\
C E=\sqrt{C D^{2}-D E^{2}}=\sqrt{3} .
\end{array}
$$

Take a point $F$ on $A E$ such that $E F=D E=1$.
Thus, $\angle D F E=\angle F D E=45^{\circ}$,
$$
D F=\sqrt{E F^{2}+D E^{2}}=\sqrt{2} \text {. }
$$

Then $\angle A D F=\angle D F E-\angle D A C$
$$
=22.5^{\circ}=\angle D A F \text {. }
$$

Therefore, $A F=D F=\sqrt{2}$.
Thus, $A C=A F+E F+C E=\sqrt{2}+1+\sqrt{3}$.
From $\sin \angle B A C=\frac{B C}{A C}$, we get
$$
B C=(\sqrt{2}+1+\sqrt{3}) \sin 45^{\circ}=1+\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2} .
$$