a) Xét \(\Delta AEF:AE=AF\left(gt\right).\)
\(\Rightarrow\Delta AEF\) cân tại A.
\(\Rightarrow\widehat{AEF}=\widehat{AFE}=\dfrac{180^o-\widehat{A}}{2}.\left(1\right)\)
Xét \(\Delta ABC\) cân tại A:
\(\widehat{B}=\widehat{C}=\dfrac{180^o-\widehat{A}}{2}.\left(2\right)\)
Từ (1); (2) \(\Rightarrow\widehat{AEF}=\widehat{AFE}=\widehat{B}=\widehat{C}=\dfrac{180^o-\widehat{A}}{2}.\)
Ta có: \(\widehat{AEF}=\widehat{B}\left(cmt\right).\)
Mà 2 góc này ở vị trí đồng vị.
\(\Rightarrow EF//BC\left(dhnb\right).\)
Xét tứ giác BEFC:
\(EF//BC\left(cmt\right).\)
\(\Rightarrow\) Tứ giác BEFC là hình thang (dhnb).
Mà \(\widehat{B}=\widehat{C}\left(cmt\right).\)
\(\Rightarrow\) Tứ giác BEFC là hình thang cân (dhnb).
b) Ta có:
\(\widehat{AEF}=\widehat{AFE}=\widehat{B}=\widehat{C}=\dfrac{180^o-\widehat{A}}{2}\left(cmt\right).\)
Mà \(\widehat{A}=40^o\left(gt\right).\)
\(\Rightarrow\widehat{AEF}=\widehat{AFE}=\widehat{B}=\widehat{C}=\dfrac{180^o-40^o}{2}=70^o.\)
Mà: \(\left\{{}\begin{matrix}\widehat{AEF}+\widehat{BEF}=180^o.\\\widehat{AFE}+\widehat{CFE}=180^o.\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{BEF}=110^o.\\\widehat{CFE}=110^o.\end{matrix}\right.\)