I want to determine the lenght of $x$ by making equations for each triangle ACD and ADB.
Let's recall $\angle DAB = \alpha $, $\angle DBA = \beta $ in triangle ADB
$$\alpha + \beta + 90 = 180$$
$$\alpha + \beta =180 \tag{1}$$
In isosceles triangle ACD, $\angle CAD = \angle CDA = \alpha$ and let $\angle ACD = \theta $
$$2\alpha + \theta = 180 \tag{2}$$
From two equations, we can conclude that $\alpha = 2\beta$. However, this does not actually help me at all. What am I missing?
Regards
@Enzo Call $t=AC=CD=DB$. From Pythagorean Theorem you have $$ 10^2=t^2+x^2 $$ From the law of cosines in triangle $ACD$ you get $$ 2t^2-2t^2\cos\theta=x^2 $$ But $\cos\theta=\cos(180-2\alpha)=-\cos(2\alpha)=1-2\cos^2\alpha=1-2(t/10)^2$. If you plug this value in the second displayed equation above you get a system for $x$ and $t$ that reduces to a quadratic equation for either $t^2$ or $x^2$
You can conclude by observing that $\angle ADB$ is inscribed in a semicircle and is therefore right. This means that $\alpha + \beta = 90$, so $\alpha = 30$ and $\beta = 60$. It follows that $x = \frac{\sqrt{3}}{2}(10) = 5 \sqrt{3}$.
$$\alpha + \beta + 90 = 180, \, \beta= 2 \alpha,\,3 \alpha= 90, \, \alpha=30,\, \beta=60$$
Use Rule of Sines on right triangle $ \Delta ADB$ inside semi-circle $ACDB$
$$ \dfrac{x}{\sin 60}= \dfrac{10}{\sin 90} \rightarrow x= 10 \sqrt{3}/2 $$
