Giải thích các bước giải:

a.Xét $\Delta ABC,\Delta HBA$ có:

Chung $\hat B$

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

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

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

Vì $BI$ là phân giác $\hat B$

$\to\dfrac{IA}{IH}=\dfrac{BA}{BH}$

$\to IA\cdot BH=IH\cdot BA$

c.Ta có $BD$ là phân giác $\hat B$

$\to\dfrac{AD}{DC}=\dfrac{BA}{BC}$

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

Từ câu b $\to IA\cdot BH=IH\cdot BA\to \dfrac{BH}{BA}=\dfrac{IH}{IA}$

$\to \dfrac{HI}{IA}=\dfrac{AD}{DC}$

d.Kẻ $CE\perp BD$

Xét $\Delta ABD,\Delta EBC$ có:

$\widehat{ABD}=\widehat{EBC}$

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

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

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

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

Xét $\Delta DAB,\Delta DEC$ có:

$\widehat{ADB}=\widehat{EDC}$

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

$\to\Delta DAB\sim\Delta DEC(g.g)$

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

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

$\to BA\cdot BC-DA\cdot DC=BD\cdot BE-DB\cdot DE=BD(BE-DE)=BD\cdot DB=BD^2$

$\to BD^2<BA\cdot BC$