Chứng minh !!!!!!

E

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Ta có: $\dfrac{{{S_{BHC}}}}{{{S_{ABC}}}} = \dfrac{{\frac{1}{2}HD.BC}}{{\dfrac{1}{2}AD.BC}} = \dfrac{{HD}}{{AD}}$
Tương tự, có: $\dfrac{{{S_{AHB}}}}{{{S_{ABC}}}} = \dfrac{{HF}}{{CF}};\dfrac{{{S_{AHC}}}}{{{S_{ABC}}}} = \dfrac{{HE}}{{BE}}$
$ \to \dfrac{{HD}}{{AD}} + \dfrac{{HE}}{{BE}} + \dfrac{{HF}}{{CF}} = \dfrac{{{S_{BHC}}}}{{{S_{ABC}}}} + \dfrac{{{S_{AHC}}}}{{{S_{ABC}}}} + \dfrac{{{S_{AHB}}}}{{{S_{ABC}}}} = \dfrac{{{S_{BHC}} + {S_{AHC}} + {S_{AHB}}}}{{{S_{ABC}}}} = 1$
 
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