a, Xét \(\Delta ADM;\Delta CBM\) có :
\(\left\{{}\begin{matrix}MD=MB\\\widehat{M1}=\widehat{M2}\\AM=CM\end{matrix}\right.\)
\(\Leftrightarrow\Delta ADM=\Delta CBM\left(c-g-c\right)\)
\(\Leftrightarrow AD=BC\left(1\right)\)
Xét \(\Delta AEN;\Delta BCN\) có :
\(\left\{{}\begin{matrix}AN=BM\\\widehat{N1}=\widehat{N2}\\EN=CN\end{matrix}\right.\)
\(\Leftrightarrow\Delta AEN=\Delta BCN\left(c-g-c\right)\)
\(\Leftrightarrow AE=BC\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow AD=AE\left(đpcm\right)\)
b, \(\Delta ADM=\Delta BCM\left(cmt\right)\)
\(\Leftrightarrow\widehat{D}=\widehat{BCM}\)
mà đây là 2 góc so le trong
\(\Leftrightarrow AD\) // \(BC\left(đpcm\right)\) \(\left(3\right)\)
c, Ta có :
\(\Delta AEN=\Delta BCN\left(cmt\right)\)
\(\Leftrightarrow\widehat{AEN}=\widehat{BCN}\)
mà đây là 2 góc so le trong
\(\Leftrightarrow AE\) // \(BC\left(đpcm\right)\left(4\right)\)
c, Từ \(\left(3\right)+\left(4\right)\Leftrightarrow A;D;E\) thẳng hàng \(\left(đpcm\right)\)
