Giải thích các bước giải:

a.Xét $\Delta AHN,\Delta AHC$ có:
Chung $\hat A$

$\widehat{ANH}=\widehat{AHC}(=90^o)$

$\to \Delta AHN\sim\Delta ACH(g.g)$

$\to \dfrac{AH}{AC}=\dfrac{AN}{AH}$

$\to AH^2=AN.AC$

b.Tương tự câu a chứng minh được $AH^2=MA.AB$

$\to AM.AB=AN.AC$

$\to\dfrac{AM}{AC}=\dfrac{AN}{AB}$

Mà $\widehat{MAN}=\widehat{BAC}$

$\to \Delta ANM\sim\Delta ABC(c.g.c)$ 

c.Xét $\Delta AMC, \Delta ANB$ có:

Chung $\hat A$

$\dfrac{AM}{AN}=\dfrac{AC}{AB}$ vì $AM.AB=AN.AC$

$\to \Delta AMC\sim\Delta ANB(c.g.c)$

$\to \dfrac{MC}{NB}=\dfrac{AC}{AB}$

$\to AB.CM=AC.BN$

d.Từ câu c $\to \widehat{ACM}=\widehat{ABN}$

$\to \widehat{NCK}=\widehat{MBK}$

Mà $\widehat{NKC}=\widehat{MKB}$

$\to \Delta KBM\sim\Delta KCN(g.g)$

$\to \dfrac{KB}{KC}=\dfrac{KM}{KN}$

$\to \dfrac{KB}{KM}=\dfrac{KC}{KN}$

Mà $\widehat{MKN}=\widehat{BKC}$

$\to \Delta MKN\sim\Delta BKC(c.g.c)$

e.Trên $MC$ lấy $D$ sao cho $\widehat{MNB}i=\widehat{CND}$

Mà $\widehat{MBN}=\widehat{MBK}=\widehat{KCN}=\widehat{NCD}$

$\to \Delta NMB\sim\Delta NDC(g.g)$

$\to\dfrac{MB}{DC}=\dfrac{NB}{NC}$

$\to MB.NC=DC.NB$

Xét $\Delta NMD,\Delta NBC$ có:

$\widehat{NMD}=\widehat{NMK}=\widehat{NBC}$

$\widehat{MND}=\widehat{MNB}+\widehat{BND}=\widehat{CND}+\widehat{BND}=\widehat{BNC}$

$\to \Delta NMD\sim\Delta NBC(g.g)$
$\to \dfrac{NM}{NB}=\dfrac{MD}{BC}$

$\to MN.BC=NB.MD$

$\to MN.BC+MB.NC=NC.NB+NB.MD=MC.NB$