Giải thích các bước giải:

a.Xét $\Delta ABE,\Delta ACF$ có:

Chung $\hat A$

$\widehat{AEB}=\widehat{AFC}(=90^o)$

$\to\Delta ABE\sim\Delta ACF(g.g)$

$\to \dfrac{AB}{AC}=\dfrac{AE}{AF}$

$\to AE\cdot AC=AF\cdot AB$

b.Xét $\Delta DBH,\Delta DAC$ có:

$\widehat{HDB}=\widehat{ADC}(=90^o)$

$\widehat{HBD}=90^o-\widehat{BHD}=90^o-\widehat{AHE}=\widehat{HAE}=\widehat{DAC}$

$\to \Delta DBH\sim\Delta DAC(g.g)$

$\to \dfrac{DB}{DA}=\dfrac{DH}{DC}$

$\to DB\cdot DC=DH\cdot DA$

c.Ta có: $IH\perp MN$

$\to \widehat{BHI}=\widehat{BHM}+\widehat{MHI}=\widehat{BHM}+90^o=\widehat{NHE}+90^o=\widehat{NHE}+\widehat{NEH}=\widehat{ANH}$

Xét $\Delta AHN,\Delta BHI$ có:

$\widehat{HAN}=\widehat{HAE}=\widehat{HBD}=\widehat{HBI}$

$\widehat{ANH}=\widehat{BHI}$

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

$\to \dfrac{HN}{IH}=\dfrac{AH}{BI}$

$\to HN=\dfrac{AH\cdot IH}{BI}$

Tương tự chứng minh được $\Delta AHM\sim\Delta CIH(g.g)$

$\to \dfrac{HM}{IH}=\dfrac{AH}{CI}$

$\to HM=\dfrac{AH\cdot IH}{IC}$

Do $I$ là trung điểm $BC\to IB=IC$

$\to \dfrac{AH\cdot IH}{IB}=\dfrac{AH\cdot IH}{IC}$

$\to HM=HN$

$\to H$ là trung điểm $MN$