Giải thích các bước giải:
a.Xét $\Delta ADE,\Delta ABC$ có:
$AD=AB$
$\widehat{EAD}=\widehat{BAC}$
$AE=AC$
$\to \Delta ADE=\Delta ABC(c.g.c)$
$\to DE=BC$
Mà $M,N$ là trung điểm $BC, DE$
$\to 2DN=2BM$
$\to DN=BM$
b.Từ câu a $\to\widehat{EDA}=\widehat{ABM}\to \widehat{NDA}=\widehat{ABM}$
Xét $\Delta ADN,\Delta ABM$ có:
$DN=BM$
$\widehat{NDA}=\widehat{ABM}$
$AD=AB$
$\to\Delta ADN=\Delta ABM(c.g.c)$
$\to \widehat{BAM}=\widehat{DAN}$
c.Ta có:
$ \widehat{BAM}=\widehat{DAN}$
$\to \widehat{MAD}+ \widehat{BAM}=\widehat{MAD}+\widehat{DAN}$
$\to\widehat{BAD}=\widehat{MAN}$
$\to \widehat{MAN}=180^o\to M,A,N$ thẳng hàng
