Giải thích các bước giải:

a.Xét $\Delta HAB,\Delta ABC$ có:

Chung $\hat B$

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

$\to \Delta BHA\sim\Delta BAC(g.g)$

b.Xét $\Delta HAB,\Delta HAC$ có:

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

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

$\to \Delta HAB\sim\Delta HCA(g.g)$

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

$\to AH^2=HB.HC$

c.Từ a $\to \dfrac{BH}{BA}=\dfrac{BA}{BC}$

$\to BA^2=BH.BC$

Vì $BH=BK$

$\to BK^2=BH.BC$

Xét $\Delta BDC,\Delta BHI$ có:

Chung $\hat B$

$\widehat{BDC}=\widehat{BHI}(=90^o)$

$\to \Delta BDC\sim\Delta BHI(g.g)$

$\to \dfrac{BD}{BH}=\dfrac{BC}{BI}$

$\to BD.BI=BH.BC=BK^2$

$\to \dfrac{BD}{BK}=\dfrac{BK}{BI}$

$\to \Delta BDK\sim\Delta BKI(c.g.c)$

$\to \widehat{BKI}=\widehat{BDK}=90^o$