# Show that two segments are the same measure.

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?

## 1 Answer I. The proof by congruent triangles also works if $$G$$ is within but $$F$$ is outside triangle $$ABC$$.

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.

• In the first case, how did you calculate that angle EBF is equal to angle FCD – 1qwertyyyy Dec 17 '20 at 7:32
• By your method: $\angle EBF+30^o+(30^o-\angle FCD)=180^o-120^o=60^o$. So $\angle EBF+60^o-\angle FCD=60^o$, i.e. $\angle EBF-\angle FCD=0$, making $\angle EBF=\angle FCD$. – Edward Porcella Dec 17 '20 at 17:44