a)
Trong \(\Delta ABC\) cân tại A \(\Rightarrow\left\{{}\begin{matrix}\widehat{ABC}=\widehat{ACB}\\AB=AC\end{matrix}\right.\)
Vì AB = AC mà AE = AG nên BE = CG
Xét \(\Delta BEC\) và \(\Delta CGB\)
có : \(BC\) là cạnh chung
\(\widehat{ABC}=\widehat{ACB}\left(cmt\right)\)
\(BE=CG\left(cmt\right)\)
\(\Rightarrow\Delta BEC=\Delta CGB\left(c-g-c\right)\)
\(\Rightarrow CE=BG\)( hai cạnh tương ứng )
b)
Vì \(\Delta BEC=\Delta CGB\) nên \(\widehat{C_2}=\widehat{B_2}\)
Mà \(\widehat{C_1}+\widehat{C_2}=\widehat{ACB}\)
Và \(\widehat{B_1}+\widehat{B_2}=\widehat{ABC}\)
Nên \(\widehat{C_1}+\widehat{C_2}=\widehat{B_1}+\widehat{B_2}\) \(\Rightarrow\widehat{C_1}=\widehat{B_1}\)
Vì \(\widehat{C_2}=\widehat{B_2}\)(cmt)
Ta lại có
\(\widehat{H_1}=\widehat{C_1}+\widehat{G_1}=\widehat{B_1}+\widehat{E_1}\)(T/c góc ngoài)
\(\Rightarrow\widehat{G_1}=\widehat{E_1}\) vì \(\widehat{C_1}=\widehat{B_1}\) (cmt)
Xét \(\Delta BHE\) và \(\Delta CHG\)
có \(\widehat{B_1}=\widehat{C_1}\left(cmt\right)\)
\(BE=CG\left(cmt\right)\)
\(\widehat{E_1}=\widehat{G_1}\left(cmt\right)\)
\(\Rightarrow\Delta BHE=\Delta CHG\left(g-c-g\right)\)