Giải thích các bước giải:

a.Xét $\Delta AMB,\Delta AMC$ có:

Chung $AM$

$\widehat{MAB}=\widehat{MAC}$

$AB=AC$

$\to \Delta AMB=\Delta AMC(c.g.c)$

$\to \widehat{AMB}=\widehat{AMC}$

Do $\widehat{AMB}+\widehat{AMC}=180^o$

$\to \widehat{AMB}=\widehat{AMC}=90^o$

$\to AM\perp BC$

b.Xét $\Delta ABM,\Delta CMN$ có:

$MA=MN$

$\widehat{AMB}=\widehat{CMN}$

$MB=MC$

$\to \Delta AMB=\Delta NMC(c.g.c)$

c.Từ b $\to \widehat{MAB}=\widehat{MNC}$

$\to AB//NC$

d.Xét $\Delta AMC,\Delta NMC$ có:

Chung $MC$

$\widehat{AMC}=\widehat{NMC}$

$MA=MN$

$\to \Delta AMC=\Delta NCM(c.g.c)$

e.Từ d $\to \widehat{ACM}=\widehat{MCN}$

$\to CB$ là phân giác $\widehat{ACN}$