Giải thích các bước giải:
a.Xét $\Delta DHB,\Delta EHA$ có:
$\widehat{BHD}=\widehat{AHE}$
$\widehat{BDH}=\widehat{AEH}(=90^o)$
$\to \Delta DHB\sim\Delta EHA(g.g)$
b.Ta có:
$AH.BC+BH.AC+CH.AB$
$=(AD-HD).BC+(BE-HE).AC+(CF-CH).AB$
$=(AD.BC+BE.AC+CF.AB)-(HD.BC+HE.AC+CH.AB)$
$=2(\dfrac12AD.BC+\dfrac12BE.AC+\dfrac12CF.AB)-2(\dfrac12HD.BC+\dfrac12HE.AC+\dfrac12CH.AB)$
$=2(S_{ABC}+S_{ABC}+S_{ABC})-2(S_{HBC}+S_{HAC}+S_{HAB})$
$=6S_{ABC}-2S_{ABC}$
$=4S_{ABC}$