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 $\widehat{AID}=\widehat{EIF}$

$\to \Delta IAD\sim\Delta IEF(c.g.c)$

$\to \widehat{IFE}=\widehat{IDA}=\widehat{BDA}=\widehat{BCA}$

$\to EF//BC$