In right triangle $ABC$ with the right angle at $B$ and the $30$ degree angle at $A$, bisect the $60$ degree angle at $C$, and let the point on segment $AB$ where this bisector crosses be $D$. Now drop a perpendicular from $D$ to meet segment $AC$ at $E$.
We now have three smaller $30-60-90$ triangles which are all congruent to each other. Giving each with the ordering "vertex angle 30, vertex angle 90, vertex angle 60", these are
$$\Delta AED, \ \Delta CED, \ \Delta CBD.$$
The angles match so they are similar, while the first two share side $ED$ and the second two share side $CD$, so they are in fact congruent.
Now noting that $AC=AE+EC,$ we have
using from corresponding parts of congruent triangles that $AE=CB=BC$ and $EC=CE=CB=BC.$
I think this argument uses only congruent triangles, and the (fairly simple) ideas of angle bisectors and perpendiculars.