Let the sides of the equilateral triangle ABC be extended. If AD=BE=CF prove that D,E and F are the vertices of an equilateral triangle.

Question

Let the sides of the equilateral triangle $\triangle ABC$ be extended as shown in the image. If $AD=BE=CF$. Prove that $D,E$ and $F$ are the vertices of an equilateral triangle.

Answer 1

We can start with constructing segments $DF$, $FE$, and $ED$. Thus, $\triangle {DAF}$, $\triangle {FCE}$, and $\triangle {DBE}$ are formed. We can use SAS to show that these three triangles are all congruent to each other. More specifically $DA$, $FC$, and $EB$ are congruent, $AF$, $CE$, and $DB$ are congruent, and $\angle {DAF}$, $\angle {FCE}$, $\angle {DBE}$ are congruent. Having proved the congruency of these three triangles, we can now see that $FD$=$FE$=$DE$. These three segments also form the lengths of $\triangle {DFE}$, which is equilateral from the equality stated before. Since this triangle is equilateral and has vertices $D$, $F$, and $E$, we have finished our proof.