Nối E với D.\(EK\cap BC=\left\{M\right\}\).
Xét tam giác DME và tam giác DMK ta có:
\(\left\{{}\begin{matrix}EM=KM\left(gt\right)\\\widehat{DME}=\widehat{DMK}\left(=90^o\right)\\DM:chung\end{matrix}\right.\)
Do đó \(\Delta DME=\Delta DMK\left(c.g.c\right)\)
\(\Rightarrow\widehat{MDE}=\widehat{MDK}\left(cgtu\right)\)(1)
mà \(\widehat{MDK}=\widehat{BDF}\left(d.d\right)\)
\(\Rightarrow\widehat{MDE}=\widehat{BDF}\)
Ta có:
\(\widehat{HDM}=\widehat{HDB}\left(=90^o\right)\)
\(\Rightarrow\widehat{EDM}+\widehat{EDH}=\widehat{FDB}+\widehat{FDH}\)
mà \(\widehat{EDM}=\widehat{FDB}\left(cmt\right)\)
Do đó \(\widehat{EDH}=\widehat{FDH}\)(2)
Từ (1) và (2) suy ra:
\(\widehat{EDH}+\widehat{EDM}=\widehat{FDH}+\widehat{KDM}\)
\(\Rightarrow\widehat{FDH}+\widehat{KDM}=90^o\)
Do đó: \(\widehat{EDH}+\widehat{EDM}+\widehat{FDH}+\widehat{KDM}=90^o+90^o\)
\(\Rightarrow\widehat{FDK}=180^o\)
Vậy ba điểm F;D;K thẳng hàng
Chúc bạn học tốt!!!
