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:
$\widehat{NMD}=\widehat{NMK}=\widehat{NBC}$
$\widehat{MND}=\widehat{MNB}+\widehat{BND}=\widehat{CND}+\widehat{BND}=\widehat{BNC}$
$\to \Delta NMD\sim\Delta NBC(g.g)$
$\to \dfrac{NM}{NB}=\dfrac{MD}{BC}$
$\to MN.BC=NB.MD$
$\to MN.BC+MB.NC=NC.NB+NB.MD=MC.NB$
