Giải thích các bước giải:

a.Xét $\Delta ABH,\Delta ABC$ có: 

Chung $\hat B$

$\widehat{AHB}=\widehat{BAC}$

$\to \Delta BAH\sim\Delta BCA(g.g)$

$\to \dfrac{BA}{BC}=\dfrac{BH}{BA}$

$\to BA^2=BH.BC$

b.Xét $\Delta AHB,\Delta ACH$ có:

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

$\widehat{BAH}=\widehat{ACB}=\widehat{ACH}$

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

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

$\to HA^2=HB.HC$

c.Xét $\Delta ABI,\Delta BCK$ có:

$\widehat{ABI}=\widehat{KBC}$ vì $BK$ là phân giác $\hat B$

$\widehat{BAK}=\widehat{BKC}(=90^o)$

$\to\Delta ABI\sim\Delta KBC(g.g)$

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

$\to BA.BC=KB.BI$

Xét $\Delta AIB,\Delta KIC$ có:

$\widehat{AIB}=\widehat{KIC}$

$\widehat{BAI}=\widehat{IKC}$

$\to \Delta ABI\sim\Delta KCI(g.g)$

$\to \dfrac{AI}{KI}=\dfrac{IB}{CI}$

$\to IA.IC=IB.IK$

$\to AB.BC-AI.CI=KB.BI-BI.IK=BI^2$S