1. As shown in Figure $3, A D$ is the angle bisector of $\triangle A B C$,
$$
\begin{array}{l}
\angle C=2 \angle B, A B=a^{2}- \\
4 b+4, A C=8 c-27- \\
2 b^{2}, C D=9+4 a-c^{2}
\end{array}
$$

Then $B C=$ . $\qquad$
II. $1 \cdot \frac{7}{3}$. 
Extend $AC$ to point $E$ such that $CE = CD$, and connect $DE$. Then we have $\angle E = \angle CDE = \frac{1}{2} \angle ACD = \angle B$.
Since $AD$ is the angle bisector of $\angle BAC$, then
$$
\angle BAD = \angle EAD, \frac{AB}{BD} = \frac{AC}{CD}.
$$

Therefore, $\triangle ABD \cong \triangle AED$.
Hence, $AB = AE = AC + CE = AC + CD$.
Given $AB = a^2 - 4b + 4$, $AC = 8c - 27 - 2b^2$,
$$
CD = 9 + 4a - c^2,
$$

we have $a^2 - 4b + 4 = (8c - 27 - 2b^2) + (9 + 4a - c^2)$,
which simplifies to $(a-2)^2 + 2(b-1)^2 + (c-4)^2 = 0$.
Thus, $a = 2$, $b = 1$, $c = 4$.
Therefore, $AB = 4$, $AC = 3$, $CD = 1$.
It is easy to see that $BD = \frac{4}{3}$, so
$$
BC = BD + CD = \frac{4}{3} + 1 = \frac{7}{3}.
$$