Angle Measurement 

What is the value of $x$ ?
I know that it will be less than 20.
 A: 
Let $l$ be the line bisecting $\angle ACD$, and reflect the figure across this line. Notice that the reflection preserves $C$ and switches $A$ and $D$. Let $B'$ be the image of $B$ under this reflection.
Since $\triangle ABC$ is isoceles, with $\angle ACB = \angle ABC$ and $\angle BAC = 20^{\circ}$, it follows that $\angle ACB = 80^{\circ}$, and hence $\angle DCB' = 80^{\circ}$. Thus,
\begin{align} \angle ACD + \angle ACB + \angle BCB' + \angle DCB' &= 360^{\circ} \\\implies 140^{\circ} + 80^{\circ} + \angle BCB' + \angle 80^{\circ} &= 360^{\circ}\\\implies \angle BCB' &= 60^{\circ}.\end{align}
Now, $BC = B'C$ since $B'$ and $C$ are the images of $B$ and $C$ under the reflection. Hence, $\triangle BCB'$ is a triangle with $BC = B'C$ and $\angle BCB' = 60^{\circ}$, so it must be equilateral (by SAS congruency). Thus, $BB' = BC$. Since $AB = DB'$ due to reflection and $AB = CD$ by hypothesis, it follows that $CD = DB'$. Thus, by SSS congruence, we have $\triangle BCD\cong \triangle BB'D$, so $\angle CDB = \angle B'DB$. In particular, $\angle CDB$ bisects $\angle CDB'$. By reflection, $\angle CDB' = \angle CAB = 20^{\circ}$, so $\angle CDB = \boxed{10^{\circ}}$.
(Thanks to @zar for the picture!)
A: Lets first work out $|AD|$ using the isosceles triangle $ACD$:
\begin{align}
|AD|&=2z\sin(70°),\\
\end{align}
where I define $z\equiv |AB|=|AC|=|CD|$.
Form the isosceles triangle  $ABC$ we can work out  $|BC|:$
\begin{align}
|BC|&=2z\sin(10°).\\
\end{align}
For the triangle $ABD$ one now knows the angle at $A$ which is $20°+20°=40°$ and the two sides $|AB|=z$ and $|AD|$ from the equation above. One can calculate the missing side with the cosine law/rule:
\begin{align}
|BD|^2&=z^2 +(2z\sin(70°))^2-2z(2z\sin(70°))\sin(40°).\\
\end{align}
At last we can use the sine law/rule in the triangle $BCD$ to get $x$ using that the missing angle at $B$ is $360°-140°-80°=140°$:
\begin{align}
\frac{|BD|}{\sin(140°)}&=\frac{|BC|}{\sin(x)},\\
\sin(x)&=\frac{|BC|\sin(140°)}{|BD|},\\
\sin(x)&=\frac{\sin(10°)\sin(140°)}{\sqrt{1/4+\sin(70°)^2-\sin(70°)\sin(40°)}}.
\end{align}
This results in an angle $x\sim8.8°$, which looks plausible to me. I might have screwed up a factor two or a half at some point/angle but I am sure that this way works. There might be another way to get to this or to simplify the last equation but I stopped there.
