A Slim Right Triangle

Let $\angle CAD = \theta$ , then $\angle BAD = 2\theta$ and $\angle BAC = 3 \theta$ .

We note that:

$\begin{aligned} \frac{BC}{AC} & = \tan \angle BAC \\ & = \tan (3 \theta) \\ & = \frac{3 \tan \theta - \tan ^3 \theta }{1-3 \tan^2 \theta} \quad \quad \small \color{#3D99F6}{\text{We also note that } \tan \angle CAD = \tan \theta = \frac{CD}{AC} = \frac{1}{2} } \\ & = \frac{\frac{3}{2} - \frac{1}{8}}{1- \frac{3}{4}} = \frac{11}{2} \\ \Rightarrow BC & = AC \times \tan \theta = 2 \times \frac{11}{2} = 11 \\ \Rightarrow BD & = BC - CD = 11 - 1 = \boxed{10} \end{aligned}$

Given: AC=2, BC=1

In ∆ACD tanx= $\dfrac{CD}{AC}=\dfrac{1}{2}$ ........(1.) $\small{\color{goldenrod}{\text{Using sine rule and equating AD in triangles ACD and ADB:}}}$ $\dfrac{CD }{\sin x}=\cos 3x\times \dfrac{BD}{\sin 2x}$ $\small{\color{goldenrod}{\text{Using sin2x=2(sinx)(cosx) and cos3x=}4\cos^3x-3 \cos x}}$ BD= $\dfrac{2}{4\cos^2 x-3}=\dfrac{2}{\dfrac{4}{1+\tan^2x}-3}$ $\Large {\color{forestgreen}{=10}}~~~\small{(Using~(1.))}$

Let's chuck this triangle on to the Argand plane, with $A$ being the origin. Thus, $C=2$ and $D=2+i$ . We also have that $\angle CAD$ is $\frac {1}{3} \angle CAB$ . Thus,

$3 \arg{D} = \arg{B}$

$\arg{D^3} = \arg{B}$

$D^3 = B$

$(2+i)^3 = B$

$B = 2+11i$

From this, we get $BC=11$ , which implies $BD=10$ .

Construct $D'$ on $\overline CB$ such that $CD'=CD=1$ and $D'\ne D$ . Since $AC\perp DD'$ , $AC$ perpendicularly bisects $DD'$ and thus $\triangle D'AD$ is isosceles, which implies $\angle DAD'=2\angle CAD=\angle DAB$ . By Pythagorean Theorem $AD'=\sqrt {5}$ and by angle bisector theorem $\frac {DB}{AB}=\frac {DD'}{AD'}=\frac {2}{\sqrt {5}}$

Applying Pythagorean Theorem on $ABC$ : $4+(DB+1)^2=AB^2$ .

By substitution: $BD^2-8BD-20=(BD-10)(BD+2)=0\implies BD=\boxed {10}$ .