Giải thích các bước giải:

a.Xét $\Delta EAF,\Delta EDB$ có:
Chung $\hat E$

$\widehat{EAF}=\widehat{EDB}(=90^o)$

$\to \Delta EAF\sim\Delta EDB(g.g)$

$\to \dfrac{EA}{ED}=\dfrac{EF}{EB}$

$\to EA.EB=EF.ED$

b.Xét $\Delta EBF,\Delta EDA$ có:

Chung $\hat E$

$\dfrac{EA}{ED}=\dfrac{EF}{EB}$

$\to \Delta EBF\sim\Delta EAD(c.g.c)$

$\to \widehat{EBF}=\widehat{EDA}$

$\to \widehat{ABF}=\widehat{ADE}$

c.Từ b $\to \widehat{EFB}=\widehat{EAD}$

$\to \widehat{DFB}=180^o-\widehat{EFB}=180^o-\widehat{EAD}=\widehat{DAB}=\dfrac12\widehat{BAC}=45^o$

$\to \Delta DBF$ vuông cân tại $D$