17. (10 points) In $\triangle A B C$, let the sides opposite to $\angle A$, $\angle B$, and $\angle C$ be $a$, $b$, and $c$, respectively, satisfying
$a \cos C = (2 b - c) \cos A$.
(1) Find the size of $\angle A$;
(2) If $a = \sqrt{3}$, and $D$ is the midpoint of side $B C$, find the range of $A D$.
17. (1) By the sine rule, we have
$$
\begin{array}{l}
\sin A \cdot \cos C=2 \sin B \cdot \cos A-\sin C \cdot \cos A \\
\Rightarrow \sin (A+C)=2 \sin B \cdot \cos A \\
\Rightarrow \sin B=2 \sin B \cdot \cos A .
\end{array}
$$

Since $\sin B \neq 0$, then $\cos A=\frac{1}{2}$.
Given $0<\angle A<\pi$, thus, $\angle A=\frac{\pi}{3}$.
(2) By the sine rule, we have
$$
\frac{b}{\sin B}=\frac{a}{\sin A}=2 \Rightarrow b=2 \sin B \text {. }
$$

By the cosine rule, we have
$$
\begin{array}{l}
A D^{2}=b^{2}+\left(\frac{a}{2}\right)^{2}-2 \times \frac{a}{2} b \cos C \\
=4 \sin ^{2} B+\frac{3}{4}-2 \sqrt{3} \sin B \cdot \cos C \\
=4 \sin ^{2} B+\frac{3}{4}-2 \sqrt{3} \sin B \cdot \cos \left(\frac{2 \pi}{3}-B\right) \\
=\sin ^{2} B+\sqrt{3} \sin B \cdot \cos B+\frac{3}{4} \\
=\frac{\sqrt{3}}{2} \sin 2 B-\frac{1}{2} \cos 2 B+\frac{5}{4} \\
=\sin \left(2 B-\frac{\pi}{6}\right)+\frac{5}{4} .
\end{array}
$$

Given $\angle B \in\left(0, \frac{2 \pi}{3}\right)$, we know
$$
\begin{array}{l}
2 \angle B-\frac{\pi}{6} \in\left(-\frac{\pi}{6}, \frac{7 \pi}{6}\right) \\
\Rightarrow \sin \left(2 B-\frac{\pi}{6}\right) \in\left(-\frac{1}{2}, 1\right] \\
\Rightarrow A D \in\left(\frac{\sqrt{3}}{2}, \frac{3}{2}\right] .
\end{array}
$$