Find $x$

$\angle BAC=\angle BCA =45^\circ$ . Let $AB=BC=AD=1$ , then $AC=\sqrt{2}$

$\angle ADC=180-x-(45-x)=135^\circ$

Extend $CD$ to $E$ such that $ED \perp AE$ . Since $\angle ADC$ and $\angle ADE$ are supplementary angles, $\angle ADE=180-135=45^\circ$ . It follows that $\triangle AED$ is an isosceles triangle. By ratio and proportion, we have

$\dfrac{AE}{1}=\dfrac{1}{\sqrt{2}}$ $\color{#D61F06}\large \implies$ $AE=\dfrac{\sqrt{2}}{2}$ , since $AC$ is twice $AE$ , $\triangle AEC$ is a $30-60-90$ special right triangle. It means that $\angle ACD=30^\circ$ .

Finally, $45-x=30$ or $\color{#3D99F6}\boxed{\large x=15^\circ}$

Let $AB=BC=AD=1$ . Then $AC = \sqrt{AB^2+BC^2} = \sqrt 2$ . We note that $\angle ADC = 180^\circ - x - (45^\circ - x) = 135^\circ$ . Using sine rule :

$\begin{aligned} \frac {\sin \angle ACD}{AD} & = \frac {\sin \angle ADC}{AC} \\ \frac {\sin (45^\circ -x)}1 & = \frac {\sin 135^\circ}{\sqrt 2} \\ \sin (45^\circ -x) & = \frac 12 \\ 45^\circ -x & = 30^\circ \\ \implies x & = \boxed{15}^\circ \end{aligned}$

Join B to D. From Quad ABCD, we see that <ADC=135°. Larger <ABC = 270°. Moreover, because Tr. ABC is isosceles with BA=BC, B lies on the perpendicular bisector of AC as also Larger <ABC = 270°=2*(<ADC). Then, B is the circumcentre of Tr. ADC and BD=BA=BC=AD. In other words, Tr. BAD is equilateral and hence 45°+x=60° or x=15°