Tri-Angle

Let $AC=2$ , $AD=3$ , $CD=a$ , $BD=b$ , $\angle BAD = \theta$ and $\angle CAD = 2 \theta$ .

By Pythagorean theorem: $CD = \sqrt{AD^2-AC^2} \implies a = \sqrt{3^2-2^2} = \sqrt 5$

From $\cos 2 \theta = \dfrac {AC}{AD} = \dfrac 23 \implies \tan 2 \theta = \dfrac {\sqrt 5}2$

Using the identity $\cos 2 \theta = \dfrac {1-\tan^2 \theta}{1+\tan^2 \theta}$ :

$\begin{aligned} \cos 2 \theta & = \frac 23 \\ \frac {1-\tan^2 \theta}{1+\tan^2 \theta} & = \frac 23 \\ 3 - 3 \tan^2 \theta & = 2 + 2 \tan^2 \theta \\ 5 \tan^2 \theta & = 1 \\ \implies \tan \theta & = \frac 1{\sqrt 5} \end{aligned}$

We note that:

$\begin{aligned} BC & = AC \tan 3 \theta \\ a + b & = 2 \tan 3 \theta = \frac {2 (\tan \theta + \tan 2 \theta)}{1 - \tan \theta \tan 2 \theta} = \frac {2\left(\frac 1{\sqrt 5} + \frac {\sqrt 5}2\right)}{1 - \frac 1{\sqrt 5} \cdot \frac {\sqrt 5}2} = \frac {14}{\sqrt 5} \\ \implies b & = \frac {14}{\sqrt 5} - a = \frac {14}{\sqrt 5} - \sqrt 5 = \frac {9\sqrt 5}5 \end{aligned}$

$\implies \dfrac {CD}{BD} = \dfrac ab = \dfrac {\sqrt 5}{\frac {9\sqrt 5}5} = \dfrac 59$

$\implies m + n = 5+9 = \boxed{14}$