a) Xét \(\Delta AMB\) và \(\Delta AMC\) có:
\(AB=AC\left(gt\right)\)
\(MB=MC\) ( M là trung điểm BC )
\(AM:\) cạnh chung
\(\Rightarrow\Delta AMB=\Delta AMC\left(c.c.c\right)\)
\(\Rightarrow\widehat{AMB}=\widehat{AMC}\) ( 2 góc tương ứng )
b) Ta có: \(\widehat{AMB}+\widehat{AMC}=180^o\) ( Tổng 3 góc tam giác )
Mà \(\widehat{AMB}=\widehat{AMC}\left(cmt\right)\)
\(\Rightarrow\widehat{AMB}=\widehat{AMC}=\dfrac{180^o}{2}=90^o\)
Hay \(AM\perp BC\)
c) Ta có: \(\Delta AMB=\Delta AMC\left(cmt\right)\)
\(\Rightarrow\widehat{ABM}=\widehat{ACM}\) ( 2 góc tương ứng )
Hay \(\widehat{EBC}=\widehat{FCB}\)
Xét \(\Delta EBC\) và \(\Delta FCB\) có:
\(EB=FC\left(gt\right)\)
\(BC:\) cạnh chung
\(\widehat{EBC}=\widehat{FCB}\) ( vừa cm )
\(\Rightarrow\Delta EBC=\Delta FCB\left(c.g.c\right)\)