\(Xét\Delta BDEvà\Delta BAEcó:\\ \left\{{}\begin{matrix}BD=BA\left(gt\right)\\\widehat{DBE}=\widehat{ABE}\left(gt\right)\\EBlàcạnhchung\end{matrix}\right.\\ \Rightarrow\Delta BDE=\Delta BAE\left(c-g-c\right)\\ \Rightarrow\widehat{DEB}=\widehat{AEB}\left(haigóctươngứng\right)\)
Ta có : BD=BA; CD=FA
\(\Rightarrow CD+BD=FA+AB\left(vìD\in BC;F\in AB\right)\\ \Leftrightarrow BC=BF\)
\(Xét\Delta BECvà\Delta BEFcó:\\ \left\{{}\begin{matrix}BC=BF\left(gt\right)\\\widehat{CBE}=\widehat{FBE}\left(gt\right)\\EBlàcạnhchung\end{matrix}\right.\\ \Rightarrow\Delta BEC=\Delta BEF\left(c-g-c\right)\\ \Rightarrow\widehat{CEB}=\widehat{FEB}\left(haigóctươngứng\right)\)
Ta lại có :
\(\widehat{CEB}+\widehat{BEA}=180^0\)
Mà : \(\widehat{CEB}=\widehat{FEB}\) ; \(\widehat{DEB}=\widehat{AEB}\)
\(\Rightarrow\widehat{FEB}+\widehat{BED}=180^0\\ \Rightarrow F;E;Dthẳnghàng\)
d/ Chứng minh: BE FC
