Cứu em bài 6 với em cảm ơn nha

Giải thích các bước giải:

a.Xét $\Delta AMB,\Delta AMD$ có:

Chung $AM$

$\widehat{MAB}=\widehat{MAD}$

$AB=AD$

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

$\to BM=MD$

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:

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

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

$\to \widehat{MDC}=180^o-\widehat{ABC}$

$\to \widehat{MDC}=\widehat{BAC}+\widehat{ACB}$

$\to \widehat{MDC}>\widehat{ACB}$

$\to \widehat{MDC}>\widehat{DCM}$

$\to MC>MD$

$\to MC>MB$ vì $BM=DM$