a. Xét \(\Delta ABMand\Delta ACM\)
\(\left\{{}\begin{matrix}B=C\left(\Delta ABCcantaiA\right)\\BM=CM\left(gt\right)\\AB=AC\left(\Delta ABCcantaiA\right)\end{matrix}\right.\)
\(\Rightarrow\Delta ABM=\Delta ACM\left(c-g-c\right)\)
b. Xét \(\Delta AEMand\Delta AFM\)
\(\left\{{}\begin{matrix}E=F=90^o\left(gt\right)\\A_1=A_2\left(\Delta ABM=\Delta ACM\right)\\AMchung\end{matrix}\right.\)
\(\Rightarrow\Delta AEM=\Delta AFM\left(ch-gn\right)\)
\(\Rightarrow AE=AF\left(2canhtuongung\right)\)
c. Ta có: \(AE=AF\left(cmt\right)\)
\(\Rightarrow\Delta AEFcantaiA\)
\(\Rightarrow E=\dfrac{180^o-A}{2}\left(1\right)\)
Ta có: \(\Delta ABCcantaiA\left(gt\right)\)
\(\Rightarrow B=\dfrac{180^o-A}{2}\left(2\right)\)
Từ \(\left(1\right)and\left(2\right)\)
\(\Rightarrow E=B\)
Mà 2 góc này ở vị trí đồng vị
\(\Rightarrow EF//BC\)