Vì \(\Delta ABC\) đều
\(\Rightarrow\left\{{}\begin{matrix}AB=BC=AC\\\widehat{BAC}=\widehat{ABC}=\widehat{ACB}\left(t/c\right)\Leftrightarrow\widehat{DAF}=\widehat{DBE}=\widehat{FCE}\end{matrix}\right.\)
Vì \(\left\{{}\begin{matrix}AB=BC=AC\left(cmt\right)\\AD=BE=CF\left(gt\right)\end{matrix}\right.\)
\(\Rightarrow AB-AD=BC-BE=AC-CF\)
\(\Rightarrow BD=CE=AF\)
Xét \(\Delta ADF\&\Delta CFE\) có:
\(AD=CF\left(gt\right)\)
\(\widehat{DAF}=\widehat{FCE}\left(cmt\right)\)
\(AF=CE\left(cmt\right)\)
Nên \(\Delta ADF=\Delta CFE\left(c.g.c\right)\)
\(\Rightarrow DF=FE\) (2 cạnh tương ứng) (1)
Xét \(\Delta CFE\&\Delta BED\) có:
\(CF=BE\left(gt\right)\)
\(\widehat{FCE}=\widehat{DBE}\left(cmt\right)\)
\(BD=CE\left(cmt\right)\)
Nên \(\Delta CFE=\Delta BED\left(c.g.c\right)\)
\(\Rightarrow FE=ED\) (2 cạnh tương ứng) (2)
Từ (1) và (2) \(\Rightarrow DF=FE=ED\)
\(\Rightarrow\Delta DEF\) đều.