Answer: B

Small Hint:

Let $\angle DAB=\alpha$, so $\angle CAB=3\alpha$ and $\cos2\alpha=\frac{AC}{AD}$

Big Hint:

Set $AC=1$ and express $CD$ and $CB$ using $\tan2\alpha$ and $\tan3\alpha$

Solution:

Let $\angle DAB=\alpha.$ Since $\frac{AC}{AD}=\cos2\alpha=\frac{2}{3},$ the identity for $\cos2\alpha$ gives $\tan\alpha=\frac{1}{\sqrt5}.$ Taking $AC=1,$ $$\begin{aligned} CD&=\tan2\alpha=\frac{\sqrt5}{2},\\ CB&=\tan3\alpha=\frac{7\sqrt5}{5}. \end{aligned}$$ Thus $BD=CB-CD=\frac{9\sqrt5}{10},$ and $\frac{CD}{BD}=\frac{5}{9}.$ Therefore $m+n=14,$ so B is correct.