15) As shown in the figure, the center of the hyperbola is at the origin $O$, and the foci are on the $x$-axis. The two asymptotes are $l_{1}$ and $l_{2}$. A line passing through the right focus $F$ and perpendicular to $l_{1}$ intersects $l_{1}$ and $l_{2}$ at points $A$ and $B$, respectively. It is also known that the eccentricity of the hyperbola $e=\frac{\sqrt{5}}{2}$.
(1) Prove that $|\overrightarrow{O A}|$, $|\overrightarrow{A B}|$, and $|\overrightarrow{O B}|$ form an arithmetic sequence;
(2) If $F(\sqrt{5}, 0)$, find the length of the chord $C D$ intercepted by the line $A B$ on the hyperbola.
(15) (1) Given $e^{2}=\frac{5}{4}$, i.e., $\frac{c^{2}}{a^{2}}=\frac{5}{4}$, hence
$$
a^{2}=\frac{4}{5} c^{2},
$$

thus
$$
b^{2}=c^{2}-a^{2}=\frac{1}{5} c^{2},
$$

so $\frac{b}{a}=\frac{\frac{c}{\sqrt{5}}}{\frac{2 c}{\sqrt{5}}}=\frac{1}{2}$.
Let $\angle A O F=\angle B O F=\theta$, then $\tan \theta=\frac{1}{2}$,
$$
\tan \angle A O B=\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^{2} \theta}=\frac{4}{3},
$$

i.e., $\frac{|\overrightarrow{A B}|}{|\overrightarrow{O A}|}=\frac{4}{3}$.
Let $|\overrightarrow{O A}|=3 m(m>0)$, then $|\overrightarrow{A B}|=4 m,|\overrightarrow{O B}|=5 m$, satisfying $|\overrightarrow{O A}|+|\overrightarrow{O B}|=2|\overrightarrow{A B}|$. Therefore, $|\overrightarrow{O A}|, |\overrightarrow{A B}|, |\overrightarrow{O B}|$ form an arithmetic sequence.
(2) Given $c^{2}=5$, substituting into (1) and (2) gives $a^{2}=4, b^{2}=1$, thus the equation of the hyperbola is $\frac{x^{2}}{4}-y^{2}=1$. Let the slope of line $A B$ be $k$, then
$$
k=\tan \angle A F O=\cot \theta=2 .
$$

Thus the equation of line $A B$ is: $y=2(x-\sqrt{5})$, solving the system $\left\{\begin{array}{l}y=2(x-\sqrt{5}), \\ \frac{x^{2}}{4}-y^{2}=1\end{array}\right.$, eliminating $y$ gives $15 x^{2}-32 \sqrt{5} x+84=0$, hence the length of chord $C D$
$$
\begin{aligned}
|C D| & =\sqrt{1+k^{2}} \cdot \frac{\sqrt{\Delta}}{15} \\
& =\sqrt{5} \times \frac{\sqrt{(-32 \sqrt{5})^{2}-4 \times 15 \times 84}}{15}=\frac{4}{3} .
\end{aligned}
$$