a/ Xét \(\Delta EBD;\Delta EIB\) có :
\(\left\{{}\begin{matrix}\widehat{EDF}=\widehat{BIE}=90^0\\\widehat{DEF}=\widehat{BEI}\\EBchung\end{matrix}\right.\)
\(\Leftrightarrow\Delta EDB=\Delta EIB\left(ch-gn\right)\)
b/ \(\Delta EDB=\Delta EIB\left(cmt\right)\)
\(\Leftrightarrow DB=BI\)
Xét \(\Delta DBH;\Delta IBF\) có :
\(\left\{{}\begin{matrix}\widehat{BDH}=\widehat{BIF}=90^0\\DB=BI\\\widehat{DBH}=\widehat{IBF}\end{matrix}\right.\)
\(\)\(\Leftrightarrow\Delta DBH=\Delta IBF\left(g-c-g\right)\)
\(\Leftrightarrow BH=BF\)
c/ \(\Delta EDB=\Delta EIB\left(cmt\right)\)
\(\Leftrightarrow ED=EI\left(1\right)\)
\(\Delta DBH=\Delta IBF\left(cmt\right)\)
\(\Leftrightarrow DH=IF\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow ED+EH=IE+IF\)
\(\Leftrightarrow EH=EF\)
Xét \(\Delta EHK;\Delta EFK\) có :
\(\left\{{}\begin{matrix}DH=DF\\EKchung\\HK=HF\end{matrix}\right.\)
\(\Leftrightarrow\Delta EHK=\Delta EFK\left(c-c-c\right)\)
\(\Leftrightarrow\widehat{HEK}=\widehat{FEK}\)
Mà EK nằm giữa EH; EF
\(\Leftrightarrow EK\) là tia phân giác của \(\widehat{HEF}\left(3\right)\)
\(\Delta EBD=\Delta EBI\left(cmt\right)\)
\(\Leftrightarrow\widehat{BED}=\widehat{BEI}\)
Mà EB nằm giữa ED; EI
\(\Leftrightarrow EB\) là tia phân giác của \(\widehat{DEI}\left(4\right)\)
Từ \(\left(3\right)+\left(4\right)\Leftrightarrow E;B;K\) thằng hàng
d/ \(ED=IE\left(cmt\right)\)
\(\Leftrightarrow\Delta EID\) cân tại E
\(\Leftrightarrow\widehat{DEI}=180^0-2.\widehat{EDI}\left(5\right)\)
\(EH=EF\)
\(\Leftrightarrow\Delta EHF\) cân tại E
\(\Leftrightarrow\widehat{HEF}=180^0-2.\widehat{EHF}\left(6\right)\)
Từ \(\left(5\right)+\left(6\right)\Leftrightarrow\widehat{EDI}=\widehat{EHF}\)
Mà đây là 2 góc so le trong
\(\Leftrightarrow DI\backslash\backslash HF\left(đpcm\right)\)
