Giải thích các bước giải:

1.Xét $\Delta AHB,\Delta AHC$ có:

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

$\widehat{BAH}=90^o-\widehat{CAH}=\widehat{ACH}$

$\to \Delta AHB\sim\Delta CHA(g.g)$

2.Từ câu 1 $\to \dfrac{AH}{CH}=\dfrac{BH}{AH}\to AH^2=HB\cdot HC$

Vì $HC=BC-BH=9\to AH^2=36\to AH=6$

3. Gọi $AC\cap HM=D$

Xét $\Delta DHN,\Delta DAM$ có:

Chung $\hat D$

$\widehat{DHN}=\widehat{DAM}(=90^o)$

$\to\Delta DHN\sim\Delta DAM(g.g)$

$\to \dfrac{DH}{DA}=\dfrac{DN}{DM}$

$\to \dfrac{DH}{DN}=\dfrac{DA}{DM}$

Mà $\widehat{ADH}=\widehat{NDM}$

$\to \Delta DAH\sim\Delta DMN(c.g.c)$

$\to \widehat{DMN}=\widehat{DAH}$

$\to \widehat{HMN}=\widehat{HAC}$

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

$\to \Delta HMN\sim\Delta HAC(g.g)$

$\to\dfrac{HM}{HA}=\dfrac{HN}{HC}$

$\to \dfrac{HM}{HN}=\dfrac{HA}{HC}$

Xét $\Delta HME,\Delta HNF$ có:

$\widehat{MHE}=\widehat{MHN}-\widehat{EHN}=90^o-\widehat{EHN}=\widehat{EHF}-\widehat{EHN}=\widehat{NHF}$

$\widehat{EMH}=\widehat{AMD}=90^o-\hat D=\widehat{HND}=\widehat{HNF}$

$\to \Delta HME\sim\Delta HNF(g.g)$

$\to \dfrac{MF}{NF}=\dfrac{HM}{HN}=\dfrac{HA}{HC}$ không đổi