Giải thích các bước giải:
a.Xét $\Delta BFH,\Delta AHE$ có:
$\widehat{FBH}=\widehat{ABH}=90^o-\widehat{BAH}=\widehat{HAE}$
$\widehat{FHB}=90^o-\widehat{EHC}=\widehat{AHE}$ vì $AH\perp BC, HE\perp HF$
$\to \Delta BHF\sim\Delta AHE(g.g)$
Gọi $AF\cap HE=D$
Xét $\Delta DHF, \Delta DAE$ có:
Chung $\hat D$
$\widehat{DHF}=\widehat{DAE}(=90^o)$
$\to\Delta DFH\sim\Delta DEA(g.g)$
$\to\dfrac{DF}{DE}=\dfrac{DH}{DA}$
$\to\dfrac{DF}{DH}=\dfrac{DE}{DA}$
Mà $\widehat{ADH}=\widehat{FDE}$
$\to\Delta DHA\sim\Delta DFE(c.g.c)$
$\to \widehat{DAH}=\widehat{FED}$
$\to \widehat{BAH}=\widehat{FEH}$
