Giải thích các bước giải:

a.Xét $\Delta CDI,\Delta CIA$ có:

Chung $\hat I$

$\widehat{BCI}=\widehat{BCx}=\dfrac12\hat A=\widehat{IAC}$ vì $AD$ là phân giác $\hat A$ 

$\to\Delta ICD\sim\Delta IAC(g.g)$
$\to\dfrac{IC}{IA}=\dfrac{ID}{IC}$

$\to CI^2=DI.AI$

b.Xét $\Delta ABD,\Delta ACI$ có:

$\widehat{BAD}=\widehat{IAC}$

$\widehat{ABD}=180^o-\widehat{BAD}-\widehat{ADB}=180^o-\dfrac12\hat A-\widehat{IDC}=180^o-\widehat{DCI}-\widehat{CDI}=\widehat{DIC}=\widehat{AIC}$

$\to\Delta ABD\sim\Delta AIC(g.g)$

c.Ta có:

$\Delta ABD\sim\Delta AIC$

$\Delta AIC\sim\Delta  CID$

$\to\Delta ABD\sim\Delta CID$

$\to \dfrac{BD}{ID}=\dfrac{AD}{CD}$

$\to \dfrac{BD}{AD}=\dfrac{ID}{CD}$

Mà $\widehat{BDI}=\widehat{ADC}$

$\to\Delta DBI\sim\Delta DAC(c.g.c)$

$\to\widehat{IBD}=\widehat{DAC}$

$\to \widehat{IBC}=\widehat{DAC}=\dfrac12\hat A=\widehat{BCI}$

$\to\Delta IBC$ cân tại $I$

d.Ta có $\dfrac{BD}{ID}=\dfrac{AD}{CD}\to DB.DC=DA.DI$

Từ câu b

$\to \dfrac{AB}{AI}=\dfrac{AD}{AC}\to AB.AC=AI.AD$

$\to AB.AC-DB.DC=AI.AD-DA.DI=AD(AI-DI)=AD^2$

$\to đpmc$