2026/09/13

Points $D,E,F$ are on side $AB, BC, CA$ of $\triangle{ABC}$ respectively, and $AD=BE=CF$, $\triangle{DEF}$ is equilateral. Show that $\triangle{ABC}$ is also equilateral.

Prove:

\(\begin{multline} \shoveleft \text{Make }\triangle{GHI} \text{ such that points }A,B,C \text{ are on side }HI, IG, GH \text{ respectively}\\ \shoveleft \text{and }DE \parallel IG, EF \parallel GH, FD \parallel HI \implies \triangle{GHI} \text{ is also equilateral}\\ \shoveleft \text{Extend }DE \text{ and meet }HI, GH \text{ at }N,K\\ \shoveleft \text{Extend }EF \text{ and meet }IG, HI \text{ at }J,M\\ \shoveleft \text{Extend }FD \text{ and meet }IG, GH \text{ at }O,L\\ \shoveleft \implies \triangle{DON}, \triangle{ION}, \triangle{FML},\triangle{HML},\triangle{GKJ},\triangle{EKJ} \text{ are equilateral triangles}\\ \shoveleft \text{and their are congruent to each other, let their side length equls to }a\\ \shoveleft \implies LK=JO=MN=ME, \text{let its length be }t \implies DE=EF=FD=t-a \ne 0\\ \shoveleft \angle{AMF}=\angle{FLC}=60^\circ, \angle{MFA}=\angle{LCF}\implies \triangle{MFA}\sim\triangle{LCF}\\ \shoveleft \text{Similarly, }\triangle{CEK \sim \triangle{EBJ}, \triangle{BDO}\sim\triangle{DAN}}\\ \shoveleft \implies MA \cdot CL = a^2 = CK \cdot BJ = BO \cdot AN\\ \shoveleft \text{Let }MA=b \implies CL=\dfrac{a^2}{b} \implies CK=t-\dfrac{a^2}{b}\implies BJ=\dfrac{a^2}{CK}=\dfrac{a^2b}{bt-a^2}\\ \shoveleft \implies BO=t-BJ=\dfrac{bt^2-a^2t-a^2b}{bt-a^2}\implies AN\cdot BO=(t-b)\dfrac{bt^2-a^2t-a^2b}{bt-a^2}=a^2\\ \shoveleft \implies (a^2-t^2)(b^2-bt+a^2)=0 \implies b^2-bt+a^2=0\implies CK=t-\dfrac{a^2}{b}=b=MA\\ \shoveleft \implies \triangle{CEK}\cong\triangle{AFM}\implies CE=AF, \text{Similarly, }BD=CE\\ \shoveleft \implies AF=CE=BD \implies AB=BC=CA \implies \triangle{ABC} \text{ is equilateral}\blacksquare \end{multline}\)