a) Xét \(\Delta BEA\)và \(\Delta BEM\)có:
\(BA=BM\left(gt\right)\)
\(\widehat{ABE}=\widehat{MBE}\)( do BE là tia phân giác \(\widehat{ABC}\))
BE là cạnh chung
\(\Rightarrow\Delta BEA=\Delta BEM\left(c.g.c\right)\)
b) Vì \(\Delta BEA=\Delta BEM\left(cmt\right)\)
\(\Rightarrow\widehat{BAE}=\widehat{BME}\left(=90^0\right)\)
\(\Rightarrow EM\perp BC\)
c) Theo định lý tổng 3 góc trong 1 tam giác ta có:
\(\hept{\begin{cases}\widehat{MEC}+\widehat{ECM}+\widehat{EMC}=180^0\\\widehat{BAC}+\widehat{ABC}+\widehat{BCA}=180^0\end{cases}}\)
Mà \(\widehat{BAC}=\widehat{EMC}\left(=90^0\right)\)
\(\Rightarrow\widehat{ABC}=\widehat{MEC}\)