Find DC/DA

Let $AB=1$ . By sine rule , $\dfrac {DA}{AB} = \dfrac {\sin \angle DBA}{\sin \angle BDA} \implies DA = \dfrac {\sin 30^\circ}{\sin 135^\circ} = \dfrac 1{\sqrt 2}$ . Also $\dfrac {AC}{AB} = \dfrac {\sin \angle CBA}{\sin \angle BCA} \implies AC = \dfrac {\sin 105^\circ}{\sin 45^\circ} = \sqrt 2 \sin 75^\circ$ . By cosine rule .

$\begin{aligned} DC^2 & = DA^2 + AC^2 - 2DA \cdot AC \cos \angle DAC \\ \left(\frac {DC}{DA}\right)^2 & = 1 + 4\sin^2 75^\circ - 4 \sin 75^\circ \cos 15^\circ & \small \blue{\text{Note that }\sin (90^\circ - \theta) = \cos \theta} \\ & = 1 + 4\cos^2 15^\circ - 4\cos^2 15^\circ \\ \implies \frac {DC}{DA} & = \boxed 1 \end{aligned}$

Let $\angle {ACD}=α, |\overline {CD}|=x, |\overline {AD}|=y, |\overline {BD}|=z$ . Then $\dfrac{x}{\sin 15\degree}=\dfrac{y}{\sin α}, \dfrac{y}{\sin 30\degree}=\dfrac{z}{\sin 15\degree},\dfrac{z}{\sin (45\degree-α)}=\dfrac{x}{\sin 75\degree}$ . So, y= $2z\cos 15\degree, z=x\sin α, x=\dfrac{y\sin 15\degree}{\sin α}\implies \tan α=2-\sqrt 3\implies α=15\degree\implies |\overline {CD}|=|\overline {AD}|$ . Hence the required ratio is $\boxed 1$ .