Giải thích các bước giải:

a.Xét $\Delta KAC,\Delta DCK$ có:

Chung $KC$

$\widehat{AKC}=\widehat{DKC}(=90^o)$

$KA=DK$

$\to \Delta AKC=\Delta DKC(c.g.c)$

b.Xét $\Delta KAC,\Delta DKM$ có:

$\widehat{KAC}=\widehat{KDM}$ vì $AC//DM$

$KA=DK$

$\widehat{AKC}=\widehat{DKM}$

$\to \Delta KAC=\Delta KDM(g.c.g)$

$\to DM=AC$

Vì $AB<AC, AK\perp BC\to AK<AB$

$\to DK<AB$

$\to DK<AB<AC$

$\to DK<AB<DM$

c.Ta có: $AC//DM, AB\perp AC\to AB\perp DM$

                $AK\perp BC\to MB\perp AD$

$\to B$ là trực tâm $\Delta AMD$

$\to DB\perp AM$

Mà $BN\perp AM$

$\to D, B, N$ thẳng hàng

Xét $\Delta AKB,\Delta DKB$ có:

Chung $KB$

$\widehat{AKB}=\widehat{DKB}(=90^o)$

$KA=KD$

$\to \Delta AKB=\Delta DKB(c.g.c)$

$\to \widehat{ABK}=\widehat{KBD}$

$\to \widehat{ABC}=\widehat{DBC}=\widehat{NBM}$