Giải thích các bước giải:

Xét $\Delta BHA,\Delta BAC$ có:

Chung $\hat B$

$\widehat{BHA}=\widehat{BAC}(=90^o)$

$\to \Delta BHA\sim\Delta BAC(g.g)$

$\to \dfrac{BH}{BA}=\dfrac{BA}{BC}$

$\to BA^2=BH.BC$

Xét $\Delta BHF,\Delta BEC$ có:

Chung $\hat B$

$\widehat{BHF}=\widehat{BEC}(=90^o)$

$\to \Delta BHF\sim\Delta BEC(g.g)$

$\to \dfrac{BH}{BE}=\dfrac{ BF}{BC}$

$\to BF.BE=BH.BC$

$\to BF.BE=BA^2$

$\to \dfrac{BA}{BF}=\dfrac{BE}{BA}$

$\to \Delta BFA\sim\Delta BAE(c.g.c)$

$\to \widehat{BAF}=\widehat{BEA}$