2. In $\triangle A B C$, it is known that the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively, and the altitude from $A$ to side $B C$ is $A D=\frac{1}{2} a$. Then the range of $\frac{(b+c)^{2}}{2 b c}$ is . $\qquad$
2. $[2,1+\sqrt{2}]$.

Notice,
$$
\begin{array}{l}
S_{\triangle A B C}=\frac{1}{2} b c \sin A=\frac{1}{2} a A D=\frac{1}{4} a^{2} \\
\Rightarrow 2 b c \sin A=a^{2}=b^{2}+c^{2}-2 b c \cos A \\
\Rightarrow(b+c)^{2}=2 b c(1+\sin A+\cos A) \\
\Rightarrow \frac{(b+c)^{2}}{2 b c}=1+\sqrt{2} \sin \left(A+\frac{\pi}{4}\right) \leqslant 1+\sqrt{2} .
\end{array}
$$

When $\angle A=\frac{\pi}{4}$, the equality holds.
Also, $\frac{(b+c)^{2}}{2 b c} \geqslant 2$, the equality holds when $b=c, \angle A=\frac{\pi}{2}$.
Therefore, the range of $\frac{(b+c)^{2}}{2 b c}$ is $[2,1+\sqrt{2}]$.