Giải thích các bước giải:
a.Ta có $\Delta ABC$ vuông tại $A\to BC=\sqrt{AB^2+AC^2}=25$
Mà $BI$ là phân giác $\hat B$
$\to \dfrac{IA}{IC}=\dfrac{BA}{BC}=\dfrac35$
$\to \dfrac{IA}{IA+IC}=\dfrac3{3+5}$
$\to \dfrac{AI}{AC}=\dfrac38$
$\to AI=\dfrac38AC=\dfrac{15}{2}$
$\to BI=\sqrt{AB^2+AI^2}=\dfrac{15\sqrt{5}}{2}$
b.Xét $\Delta AIB,\Delta CID$ có:
$\widehat{AIB}=\widehat{CID}$(đối đỉnh)
$\widehat{BAI}=\widehat{CDI}(=90^o)$
$\to \Delta ABI\sim\Delta DCI(g.g)$
$\to \dfrac{IA}{ID}=\dfrac{IB}{IC}$
$\to \dfrac{IA}{IB}=\dfrac{ID}{IC}$
Mà $\widehat{AID}=\widehat{BIC}$
$\to \Delta AID\sim\Delta BIC(c.g.c)$
$\to \widehat{ADI}=\widehat{ICB}$
$\to \widehat{ADB}=\widehat{ACB}$
c.Ta có $AE//CD, CD\perp BD\to AE\perp DB$
$DF//AB, AB\perp AC\to DF\perp AC$
$\to \widehat{AEI}=\widehat{DFI}(=90^o)$
Mà $\widehat{AIE}=\widehat{DIF}$
$\to \Delta AIE\sim\Delta DIF(g.g)$
$\to\dfrac{IA}{ID}=\dfrac{IE}{IF}$
$\to \dfrac{IA}{IE}=\dfrac{ID}{IF}$
Mà $\widehat{AID}=\widehat{EIF}$
$\to \Delta IAD\sim\Delta IEF(c.g.c)$
$\to \widehat{IFE}=\widehat{IDA}=\widehat{BDA}=\widehat{BCA}$
$\to EF//BC$
