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?