On side $AB$ and $AC$ on the outside of any triangle $ABC$ there are built equilateral triangles $ABD$ and $ACE$. Let $F$ be point such that $\sphericalangle CBF=\sphericalangle FCB=30° $. Prove that $|DF|=|FE|$.
--- Notice that triangles $DAC$ and $BAE$ are congruent, so $|DC|=|BE|$ and $\sphericalangle AEB=\sphericalangle ACD$. From this we have: $$\sphericalangle BGC=180°-\sphericalangle EGC=180°-(180°-(60°-\sphericalangle AEB)-(60°+\sphericalangle ACD))=120°$$ So, $$\sphericalangle EBF + 30°+(30°-\sphericalangle FCD)=180°-120°=60° \Rightarrow \sphericalangle EBF=\sphericalangle FCD$$ Since $|FB|=|FC|$, triangles $DFC$ and $BEF$ are congruent so $|DF|=|EF|$.
This proof seems fine, but it only works when point $F$ is inside triangle $ABC$. Are there any other more universal ways to prove this statement using congruency of triangles?
We still have $\angle BGC=120^o$. And since, as OP calculates$$\angle EBF+30^o+(30^o-\angle FCD)=\angle EGC=180^o-120^o=60^o$$then$$\angle EBF+60^o-\angle FCD=60^o$$making$$\angle EBF-\angle FCD=0$$and therefore$$\angle EBF=\angle FCD$$and$$DF=FE$$
II. Further, if $F$ is within triangle $ABC$ but $G$ is outside, as in the next figure
we have$$\angle EBC=\angle EGC+\angle BCG$$i.e.$$\angle EBF+30^o=60^o+\angle FCD-30^o$$or$$\angle EBF+60^o=60^o+\angle FCD$$so that again$$\angle EBF=\angle FCD$$and$$DF=FE$$
III. Lastly, even if $F$ and $G$ are both outside triangle $ABC$
again we get$$\angle EBC=\angle EGC+\angle BCG$$i.e.$$\angle EBF+30^o=60^o+\angle FCD-30^o$$or$$\angle EBF+60^o=60^o+\angle FCD$$so that again$$\angle EBF=\angle FCD$$and$$DF=FE$$
Hence it seems the proof by congruent triangles that $DF=FE$ applies generally.