Giải thích các bước giải:

a.Xét $\Delta AHB,\Delta ABC$ có:

Chung $\hat B$

$\widehat{AHB}=\widehat{BAC}(=90^o)$ 

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

Tương tự chứng minh được: $\Delta HAC\sim\Delta ABC(g.g)$

$\to\Delta HBA\sim\Delta HAC(\sim\Delta ABC)$

$\to \dfrac{BA}{BC}=\dfrac{BH}{BA},\dfrac{CA}{CH}=\dfrac{CB}{CA}, \dfrac{HA}{HB}=\dfrac{HC}{HA}$

$\to AB^2=BH\cdot BC, CA^2=CH\cdot CB, HA^2=HB\cdot HC$

b.Xét $\Delta BAD,\Delta CBI$ có:

$\widehat{ABD}=\widehat{IBC}$ vì $BI$ là phân giác $\hat B$

$\widehat{BAD}=\widehat{BIC}(=90^o)$

$\to \Delta BAD\sim\Delta BIC(g.g)$

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

$\to BA\cdot BC=BI\cdot BD$

c.Xét $\Delta BAM,\Delta BCD$ có:

$\widehat{ABM}=\widehat{DBC}$

$\widehat{BAM}=\widehat{BAH}=90^o-\widehat{HAC}=\widehat{ACH}=\widehat{BCD}$

$\to \Delta BAM\sim\Delta BCD(g.g)$

$\to \dfrac{BA}{BC}=\dfrac{AM}{CD}$

$\to BA\cdot CD=AM\cdot BC$

d.Ta có: $\Delta ABC$ vuông tại $A, AH\perp BC$

$\to S_{ABC}=\dfrac12AH\cdot BC=\dfrac12AB\cdot AC$

$\to AH\cdot BC=AB\cdot AC$

$\to (AH\cdot BC)^2=(AB\cdot AC)^2$

$\to AH^2\cdot BC^2=AB^2\cdot AC^2$

$\to \dfrac{BC^2}{AB^2\cdot AC^2}=\dfrac1{AH^2}$

$\to \dfrac{AB^2+AC^2}{AB^2\cdot AC^2}=\dfrac1{AH^2}$

$\to \dfrac1{AC^2}+\dfrac1{AB^2}=\dfrac1{AH^2}$