Ta có: \(\widehat{ABD}=\widehat{AEH}+\widehat{BHE}\)(Theo tính chất góc ngoài)
Lại có: BE=BH
=> tam giác BHE cân tại B
=> \(\widehat{BHE}=\widehat{BEH}\)
\(\Rightarrow\widehat{ABD}=2\widehat{AEH}\)
\(\Rightarrow\widehat{AEH}=\widehat{ADB}\)
Ta có: \(\widehat{EHB}=\widehat{FHD}\left(đ^2\right)\)
\(\Rightarrow\widehat{FHD}=\widehat{FDH}\)
=> tam giác FDH cân tại F
=> FH=FD=\(\widehat{HAF}+\widehat{ADH}=90^o\)
\(\Rightarrow\widehat{HAF}=90^0-\widehat{ADH}\)
\(\widehat{AHF}+\widehat{FHD}=90^o\)
\(\Rightarrow\widehat{AHF}=90^o-\widehat{FHD}\)
\(\Rightarrow\widehat{HAF}=\widehat{AHF}\)
=> tam giác AFH cân tại F
=> FA=FH