Giải thích các bước giải:

a.Xét $\Delta AHB,\Delta AHC$ có:

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

$\widehat{HAB}=90^o-\hat B=\hat C$

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

$\to \dfrac{AH}{HC}=\dfrac{HB}{HA}$

$\to AH^2=HB\cdot HC$

b.Xét $\Delta ABC, \Delta HAB$  có:

Chung $\hat B$

$\hat A=\hat H(=90^o)$

$\to \Delta ABC\sim\Delta HBA(g.g)$

$\to \dfrac{AB}{HB}=\dfrac{BC}{AB}$

$\to AB^2=BH\cdot BC$

c.Xét $\Delta BKC, \Delta BHS$ có:

Chung $\hat B$

$\hat K=\hat H(=90^o)$

$\to \Delta BKC\sim\Delta BHS(g.g)$

$\to \dfrac{BK}{BH}=\dfrac{BC}{BS}$

$\to BK\cdot BS=BH\cdot BC=BA^2$

Ta có: $BA=BE\to BE^2=BK\cdot BS$

$\to \dfrac{BE}{BK}=\dfrac{BS}{BE}$

Mà $\widehat{KBE}=\widehat{SBE}$

$\to \Delta BKE\sim\Delta BES(c.g.c)$

$\to \widehat{BES}=\widehat{BKE}=90^o$

$\to BE\perp SE$