Giải thích các bước giải:
a.Xét $\Delta BEA, \Delta BEM$
$\begin{cases}chung \quad BE\\\widehat{EBA}=\widehat{EBM}\text{(BE là phân giác $\widehat{ABC}$)}\\ BA=BM\end{cases}$
$\rightarrow \Delta BEA=\Delta BEM(c.g.c)$
$\rightarrow đcpm$
b.Ta có $\Delta BEA=\Delta BEM$
$\rightarrow \widehat{BAE}=\widehat{BME}$
$\rightarrow \widehat{BME}=90^o$
$\rightarrow BM\perp ME$
$\rightarrow EM\perp BC$
c.Ta có:
$\widehat{ABC}+\widehat{BCA}=90^o$
$\widehat{MEC}+\widehat{BCA}=90^o$
$\rightarrow \widehat{ABC}+\widehat{BCA}=\widehat{MEC}+\widehat{BCA}$
$\rightarrow \widehat{ABC}=\widehat{MEC}$