What a ratio!

This solution calculates the ratio for only one possible configuration of points $A, B, C,$ and $D$ . In this configuration, shown on the left, $\angle ACB=30^\circ$ making $\angle ADB=120^\circ$ . Triangles $ACB$ and $ADB$ are chosen to be isosceles with $AC=AB$ and $AD=DB$ , so that $AC\times BD=AD\times BC$ automatically. The distance $EB$ is set to be $1$ .

The fraction to be calculated is $\dfrac{AB \times CD}{ AC \times BD}$

$AB=2$

$CD=CE-DE=\frac{1}{\tan15^\circ}-\tan30^\circ=\sqrt{3}+2-\frac{1}{\sqrt{3}}=\frac{2(\sqrt{3}+1)}{\sqrt{3}}$

$AC=BC=\frac{1}{\sin15^\circ}=(\sqrt{3}+1)\sqrt2$

$BD=\frac{1}{\cos30^\circ}=\frac{2}{\sqrt{3}}$

So the ratio

$\frac{AB\times CD}{AC\times BD}=\frac{2(\sqrt{3}+1)}{(\sqrt{3}+1)\sqrt{2}}=\sqrt{2}$

Construct a line $\perp BD$ at $D.$
Introduce a point $E$ on the line $\perp BD$ such that $BD = BE.$

Now, $AC \times BD = AD \times BC => \dfrac{AC}{BC} = \dfrac{AD}{BD} =\dfrac{AD}{BE}$ .

Also, $\angle CAD + \angle CBD = \angle CBD + \angle CBE = 90^{\circ} => \angle CAD = \angle CBE$

$=> \Delta ADC \sim \Delta BEC => \angle ACD = \angle BCE => \angle ACB = \angle DCE ; \dfrac{AC}{BC} = \dfrac{DC}{EC}$

$=> \Delta CAB\sim \Delta CDE => AB \times CD = AC \times DE --(I)$

But, the construction of $E$ gives a right isosceles $\Delta DBE => DE = \sqrt2 \times BD.$

Substituting $DE = \sqrt2 \times BD.$ in $[I]$ we get: $AB \times CD = AC \times \sqrt2 \times BD => \dfrac{AB\times CD}{AC\times BD}= \sqrt2$