Hạ EG vuông góc DC (G thuộc DC)
Ta có:
\(\widehat{FEG}+\widehat{AEM}=90^0\\ \widehat{MAE}+\widehat{AEM}=90^0\)
\(\rightarrow\widehat{FEG}=\widehat{MAE}\)
Xét \(\Delta ABM\) và \(\Delta EGF\) có:
\(EG=AB\left(=BC\right)\)
\(\widehat{FEG}=\widehat{MAB}\)
\(\rightarrow\Delta ABM=\Delta EGF\\ \rightarrow AM=EF\)