Theo đề ra ta có :AB = BC = AC ; AD = BE = CF
\(\Rightarrow AB-AD=BC-BE=AC-CF\\ \Leftrightarrow BD=EC=AF\)
\(Xét\Delta ADFvà\Delta BEDcó:\\ \left\{{}\begin{matrix}AD=BE\left(gt\right)\\\widehat{EBD}=\widehat{DAF}\left(gt\right)\\AF=BD\left(cmt\right)\end{matrix}\right.\\ \Rightarrow\Delta ADF=\Delta BED\left(c-g-c\right)\)
\(\Rightarrow DE=DF\) ( hai cạnh tương ứng ) [1]
\(Xét\Delta ADFvà\Delta CFEcó:\\ \left\{{}\begin{matrix}AD=CF\left(gt\right)\\\widehat{ECF}=\widehat{DAF}\left(gt\right)\\AF=AF\left(cmt\right)\end{matrix}\right.\\ \Rightarrow\Delta ADF=\Delta CFE\left(c-g-c\right)\)
\(\Rightarrow FE=DF\) ( hai cạnh tương ứng ) [2]
Từ [1] và [2] \(\Rightarrow\) Đpcm
