Giải thích các bước giải:

a.Xét $\Delta AMB,\Delta AMD$ có:

Chung $AM$

$\widehat{MA}=\widehat{MAD}$

$AB=AD$

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

$\to MB=DM$

b.Từ a $\to \widehat{ADM}=\widehat{ABM}$

$\to \widehat{ADK}=\widehat{ABC}$

Xét $\Delta ADK,\Delta ABC$ có:

Chung $\hat A$

$AD=AB$

$\widehat{ADK}=\widehat{ABC}$

$\to \Delta ADK=\Delta ABC(g.c.g)$

c.Từ b $\to AK=AC$

$\to \Delta AKC$ cân tại $A$

d.Ta có:

$\widehat{MDC}=180^o-\widehat{ADM}=180^o-\widehat{ABM}=\widehat{MBK}$

Xét $\Delta MDC,\Delta MBK$ có:

$\widehat{MDC}=\widehat{MBK}$

$MD=MB$

$\widehat{DMC}=\widehat{BMK}$

$\to \Delta MDC=\Delta MBK(g.c.g)$

$\to MC=MK$