Giải thích các bước giải:

Xét $\Delta AHE,\Delta AHB$ có:

$\widehat{AHE}=\widehat{AHB}(=90^o)$

$\widehat{HAE}=90^o-\widehat{HAB}=\widehat{HBA}$

$\to \Delta HAE\sim\Delta HBA(g.g)$

$\to \dfrac{AE}{BA}=\dfrac{HA}{HB}$

$\to \dfrac{AF}{BC}=\dfrac{HA}{HB}$

Mà $\widehat{HAF}=\widehat{HAB}=90^o-\widehat{HBA}=\widehat{HBC}$

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

$\to \widehat{AHF}=\widehat{BHC}$

$\to \widehat{FHC}=\widehat{FHB}+\widehat{BHC}=\widehat{FHB}+\widehat{FHA}=\widehat{AHB}=90^o$