Kẻ $AE// BC, E\in FK$
$⇒\widehat{AEH}=\widehat{HFC}=\widehat{FHC}=\widehat{AHE}$
$⇒\Delta AEH ⊥ A$
`⇒AE=AH`
`⇒AE=FI`
Ta có :
$\widehat{KAE}=\widehat{KIF}, \widehat{AEK}=\widehat{KFI}$
$⇒\Delta KAE=\Delta KIF(g.c.g)$
`⇒ AK = KI`
`⇒ ĐPCM`
Xin hay nhất !