Be $\triangle CAB$ right in $A$ such that $AB=a$ and $\angle CBA = \alpha$.
Extend $BC$ to $D$ such that $\angle CAD=2\alpha$ and $\angle ADC=x$.
If $M$ is a point $\in BC$ such that $BM=MC$ and $MD=a$, then compute $\angle ADC=x$
I tried drawing the altitude $h$ from $A$ to $BC$ such that $AH=h$, and then i tried some relations between similar triangles, but found nothing.
Any hints?
PS: If anyone has a solution that involves trigonometry, i would like to see it, but i highly prefer a solution that doesn't involve trigonometry.
Naturally, the first step is to extend the diagram. We have $AB=MD$; let's take advantage of that by constructing $E$ below $BD$ so that triangles $MED$ and $ACB$ are congruent.
From the right angle at $A$, the midpoint of $BC$ is the circumcenter of $ABC$, and $AM=BM=CM$. Similarly, if we mark the midpoint $F$ of $DE$, $DF=EF=MF$. Since $BC=DE$, all six of these lengths are equal; call that length $b$.
Now, it's time to use the angle marked in red. We also have $\angle AMC=2\alpha$ from the isosceles triangle $BAM$, and that makes triangles $MAD$ and $ACD$ similar. In particular, $\angle MAD=\angle ACD=90^\circ+\alpha$. Combine that with $\angle MED = 90^\circ-\alpha$, and $EDAM$ is a cyclic quadrilateral. From our earlier note that $DF=EF=MF$, then, $F$ is the circumcenter of $EDAM$.
So now, we've marked all of the segments of length $y$ in purple. Of particular note are the newly drawn segments $AM$, $FM$, and $AF$. That's an equilateral triangle $FAM$. On top of that, we know angles $AMD$ and $DMF$, so we can deduce that $60^\circ=\angle AMF=3\alpha$.
We were asked to find $\angle CDA$. From the angles of $2\alpha$ at $A$ and $90^\circ+\alpha$ at $C$, this third angle of the triangle is $90^\circ-3\alpha=90^\circ-60^\circ=30^\circ$. Done.