Giải thích các bước giải:

a.Xét $\Delta HBF,\Delta HCE$ có:

$\hat F=\hat E(=90^o)$

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

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

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

Mà $\widehat{EHF}=\widehat{BHC}$

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

b.Xét $\Delta CDH,\Delta CFB$ có:

Chung $\hat C$

$\widehat{CDH}=\widehat{CFB}(=90^o)$

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

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

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

$\to \widehat{FDC}=\widehat{BHC}=\widehat{FHE}$