2. In $\triangle A B C$, the angle bisector of $\angle B$ intersects with the external angle bisector of $\angle C$ at point $D$. If $\angle A=27^{\circ}$, then, $\angle B D C=$ $\qquad$
2.13.5 ${ }^{\circ}$.

As shown in Figure 4, since $\angle A=27^{\circ}$, $\angle B C E=\frac{\angle A+\angle A B C}{2}$, then
$$
\begin{array}{l}
\angle B D C=\angle B C E-\angle C B D \\
=\frac{\angle A+\angle A B C}{2}-\frac{\angle A B C}{2} \\
=\frac{\angle A}{2} .
\end{array}
$$