Dễ thôi
ta có\(\Delta HBE\infty\Delta ABF\)(\(\widehat{BHE}=\widehat{BAF}=90^0\);\(\widehat{EBH}=\widehat{ABF}\))
\(\Rightarrow\widehat{BEH}=\widehat{AFB}\)
Lại có:\(\widehat{BEH}=\widehat{AEF}\)
\(\Rightarrow\widehat{AFE}=\widehat{AEF}\)
Vậy tam giác AEF cân tại A