Giải thích các bước giải:

a.Ta có: $\hat C=180^o-\hat A-\hat B=40^o$

$\to \hat C<\hat B<\hat A$

$\to AB<AC<BC$

b.Xét $\Delta BAD,\Delta BMD$ có:

Chung $BD$

$\widehat{ABD}=\widehat{MBD}$

$BA=BM$

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

c.Từ câu a $\to \widehat{BAD}=\widehat{BMD}, DA=DM$

$\to \widehat{DAH}=180^o-\widehat{BAD}=180^o-\widehat{BMD}=\widehat{DMC}$

Xét $\Delta DAH,\Delta DMC$ có:

$\widehat{DAH}=\widehat{DMC}$

$DA=DM$

$\widehat{ADH}=\widehat{MDC}$

$\to \Delta DAH=\Delta DMC(g.c.g)$

$\to DH=DC$

$\to \Delta DHC$ cân tại $D$

d.Ta có: $BD$ là phân giác $\hat B\to \widehat{DBA}=\widehat{DBM}=\dfrac12\hat B=30^o$

$\to \widehat{BDM}=\widehat{BDA}=180^o-\hat A-\widehat{ABD}=70^o$

$\to \widehat{ADM}=\widehat{ADB}+\widehat{BDM}=140^o$

$\to \widehat{HDC}=140^o$

$\to \widehat{DHC}=\widehat{DCH}=90^o-\dfrac12\widehat{HDC}=20^o$

Ta có:

$\widehat{MDC}=180^o-\widehat{HDC}=40^o$

$\to \widehat{DMC}=180^o-\widehat{MDC}-\widehat{MCD}=180^o-40^o-40^o=100^o>\widehat{MCD}$

$\to DC>DM$

Mà $DM=DA$

$\to AD<DC$