\(a,\widehat{AMB}=90^0\) (góc nt chắn nửa đg tròn)
\(\Rightarrow\widehat{FCB}+\widehat{FMB}=90^0+90^0=180^0\)
Do đó: BCFM nội tiếp
\(b,\left\{{}\begin{matrix}\widehat{BAM}\text{ chung}\\\widehat{FCB}=\widehat{AMB}=90^0\end{matrix}\right.\Rightarrow\Delta ACF\sim\Delta AMB\left(g.g\right)\\ \Rightarrow AC.AB=AF.AM\\ c,\widehat{EMF}+\widehat{FMO}=\widehat{EMO}=90^0\)
Mà \(OA=OM=R\Rightarrow\widehat{FMO}=\widehat{MAO}\)
\(\Rightarrow\widehat{EMF}+\widehat{MAO}=90^0\left(1\right)\\ \widehat{EFM}=\widehat{AFC}\left(đđ\right);\widehat{AFC}+\widehat{MAO}=90^0\left(\Delta ACF\bot C\right)\\ \Rightarrow\widehat{EFM}+\widehat{MAO}=90^0\left(2\right)\\ \left(1\right)\left(2\right)\Rightarrow\widehat{EFM}=\widehat{EMF}\\ \Rightarrow\Delta EMF\text{ cân tại }E\\ \Rightarrow EM=EF\)
