Giải thích các bước giải:

a.Xét $\Delta ABE,\Delta ACD$ có:

$AE=AD$

Chung $\hat A$

$AB=AC$

$\to \Delta ABE=\Delta ACD(c.g.c)$

$\to CD=BE,\widehat{ABE}=\widehat{ACD},\widehat{AEB}=\widehat{ADC}$

Xét $\Delta MDB,\Delta CME$ có:

$\widehat{MDB}=180^o-\widehat{ADC}=180^o-\widehat{AEB}=\widehat{MEC}$

$BD=AB-AD=AC-AE=CE$

$\widehat{MBD}=\widehat{ABE}=\widehat{ACD}=\widehat{MCE}$

$\to \Delta BMD=\Delta CME(g.c.g)$

c.Từ a $\to MD=ME$

Xét $\Delta AMD,\Delta AME$ có:
Chung $AM$

$AD=AE$

$MD=ME$

$\to \Delta AMD=\Delta AME(c.c.c)$

$\to \widehat{MAD}=\widehat{MAE}$

$\to AM$ là phân giác $\hat A$