Giải thích các bước giải:

a.Xét $\Delta BDA, \Delta BFC$ có:
Chung $\hat B$

$\widehat{BDA}=\widehat{BFC}(=90^o)$

$\to \Delta BDA\sim\Delta BFC(g.g)$

$\to\dfrac{BD}{BF}=\dfrac{BA}{BC}$

$\to BF.BA=BD.BC$

 $\to \dfrac{BF}{BC}=\dfrac{BF}{BA}$

Mà $\widehat{FBD}=\widehat{ABC}$

$\to \Delta BDF\sim\Delta BAC(c.g.c)$

$\to \widehat{BFD}=\widehat{ACB}$

b.Xét $\Delta HFB,\Delta HEC$ có:

$\widehat{FHB}=\widehat{EHC}$

$\widehat{HFB}=\widehat{HEC}(=90^o)$

$\to \Delta HBF\sim\Delta HCE(g.g)$

$\to\dfrac{HB}{HC}=\dfrac{HF}{HE}$

$\to HB.HE=HC.HF$

$\to \dfrac{HE}{HC}=\dfrac{HF}{HB}$

Mà $\widehat{FHE}=\widehat{BHC}$

$\to \Delta HFE\sim\Delta HBC(c.g.c)$

$\to \widehat{FEH}=\widehat{HCB}$

$\to \widehat{FEB}=\widehat{FCB}$

c.Xét $\Delta CHD, \Delta CFB$ có:

Chung $\hat C$

$\widehat{HDC}=\widehat{BFC}(=90^o)$

$\to \Delta CDH\sim\Delta CFB(g.g)$

$\to\dfrac{CH}{CB}=\dfrac{CD}{CF}$

$\to CH.CF=CD.CB$

$\to BF.BA+CH.CF=BD.BC+CD.CB=BC^2$

d.Từ câu b $\to \widehat{EFH}=\widehat{HBC}$

Xét $\Delta CDF, \Delta CHB$ có:

Chung $\hat C$

$CH.CF=CD.CB\to \dfrac{CH}{CB}=\dfrac{CD}{CF}$

$\to \Delta CDF\sim\Delta CHB(c.g.c)$

$\to \widehat{CFD}=\widehat{HBD}=\widehat{HBC}=\widehat{HFE}$

$\to FC$ là phân giác $\widehat{EFD}$

$\to \widehat{EFD}=2\widehat{CFD}=2\widehat{EBD}=2\widehat{EBO}=\widehat{EOC}$

Vì $\Delta EBC$ cân tại $E, O$ là trung điểm $BC\to OE=OB=OC$

$\to 180^o-\widehat{DFE}=180^o-\widehat{EOC}$

$\to \widehat{IFD}=\widehat{EOI}$

Mà $\widehat{FID}=\widehat{EIO}$

$\to\Delta IFD\sim\Delta IOE(g.g)$

$\to\dfrac{IF}{IO}=\dfrac{ID}{IE}$

$\to ID.IO=IE.IF$