Parallel lines

$\angle BCD=45+90=135^\circ$

Let $AC=CD=1$ . Then by pythagorean theorem , $AD=BD=\sqrt{2}$ .

Apply cosine law on $\triangle BCD$ .

$(\sqrt{2})^2=(BC)^2+1^2-2(BC)(1)(\cos~135)$ $\implies$ $(BC)^2+\sqrt{2}BC-1=0$

By the quadratic formula , $BC=\dfrac{-\sqrt{2}+\sqrt{6}}{2}$

Apply again cosine law on $\triangle BCD$ .

$\left(\dfrac{-2\sqrt{2}+\sqrt{6}}{2}\right)^2=1^2+(\sqrt{2})^2-2(1)(\sqrt{2})(\cos~x)$

After calculation, we get

$\color{#D61F06}\large \boxed{x=15^\circ}$

We drop perpendiculars from $C$ and $D$ to $AB$ . The $ECDF$ is a rectangle, so $CE=DF$ . Since $\angle CAB=45°, DF=CE=AE=EB$ . From that $DB=AB=2*EB$ and $DF=\dfrac{1}{2}DB$ . Since $\angle DFE=90°, \angle DBF=30°$ . Therefore $\angle CBD=\angle CBE-\angle DBF=45°-30°=\boxed{15°}$