Ta có : \(\frac{S_{BHC}+S_{HAC}+S_{HAB}}{S_{ABC}}=\frac{S_{ABC}}{S_{ABC}}=1\)
hay \(\frac{HD}{AD}+\frac{HE}{BE}+\frac{HF}{CF}=1\)
\(\Rightarrow\left(1-\frac{HD}{AD}\right)+\left(1-\frac{HD}{BD}\right)+\left(1-\frac{HF}{CF}\right)=2\)
\(\Rightarrow\frac{AH}{AD}+\frac{BH}{BE}+\frac{CH}{CF}=2\)
Ta có đpcm.