The pre-rmo question

$\angle {ADB}=45\degree,\angle {ACD}=30\degree\implies \angle {DAC}=15\degree$

Let $\angle {BAD}=α$

Then

$\dfrac {|\overline {AD}|}{|\overline {BD}|}=\dfrac {\sin (45\degree+α)}{\sin α}$

$\dfrac {|\overline {AD}|}{|\overline {DC}|}=\dfrac {\sin 30\degree}{\sin 15\degree}$

Solving these two we get

$(\sqrt 3-1)\cos α=(3-\sqrt 3)\sin α$

$\implies \tan α=\dfrac {1}{\sqrt 3}\implies α=\boxed {30\degree}$ .