Giải thích các bước giải:
d. Ta có: \(\dfrac{HD}{AD}+\dfrac{HE}{BE}+\dfrac{HF}{CF}\\ =\dfrac{2HD\cdot BC}{2AD\cdot BC}+\dfrac{2HE\cdot AC}{2EB\cdot AC}+\dfrac{2AB\cdot HF}{2CF\cdot AB}\\=\dfrac{S_{BHC}}{S_{ABC}}+\dfrac{S_{HAC}}{S_{ABC}}+\dfrac{S_{AHF}}{S_{ABC}}\\ =\dfrac{S_{BHC}+S_{AHC}+S_{AHF}}{S_{ABC}}\\ =\dfrac{S_{ABC}}{S_{ABC}}=1\)
Vậy ta có đpcm