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