1998 AHSME Problem 28

Let $\angle DAB = \theta$ . then $\angle CAD = 2\theta$ , Let $AC = 2$ and $AD=3$ . By Pythagorean theorem

$\begin{aligned} CD^2 & = AD^2 - AC^2 = 3^2 - 2^2 = 5 \\ \implies CD & = \sqrt 5 \end{aligned}$

Then we have:

$\begin{aligned} \tan 2\theta & = \frac {CD}{AC} = \frac {\sqrt 5}2 \\ \frac {2\tan \theta}{1-\tan^2 \theta} & = \frac {\sqrt 5}2 \\ \sqrt 5 \tan^2 \theta + 4\tan \theta - \sqrt 5 & = 0 \\ (\sqrt 5 \tan \theta - 1)(\tan \theta + \sqrt 5) & = 0 \\ \implies \tan \theta & = \frac 1{\sqrt 5} & \small \color{#3D99F6} \text{Since }\theta < \frac \pi 2 \end{aligned}$

And that $\tan 3 \theta = \dfrac {\tan \theta + \tan 2\theta}{1-\tan \theta \tan 2\theta} = \dfrac {\frac 1{\sqrt 5}+\frac{\sqrt 5}2}{1-\frac 1{\sqrt 5}\times \frac {\sqrt 5}2} = \dfrac 7{\sqrt 5}$ .

We note that $BC = AC \tan 3 \theta = 2 \tan 3 \theta = \dfrac {14}{\sqrt 5}$ . Then $BD = BC - CD = \dfrac {14}{\sqrt 5} - \sqrt 5 = \dfrac 9{\sqrt 5}$ and $\dfrac {CD}{BD} = \frac {\sqrt 5}{\frac 9{\sqrt 5}} = \dfrac 59$ .

Therefore, $m+n = 5+9 = \boxed{14}$ .