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A B C A' B' C' H Ta có : \(\dfrac{HA'}{AA'}=\dfrac{S_{HBC}}{S_{ABC}}\)( Vì có chung đáy BC nên tỉ số hai đường cao cũng bằng tỉ số hai diện tích) ( * )
Tương tự , ta cũng có :
\(\dfrac{HB'}{BB'}=\dfrac{S_{HCA}}{S_{ABC}}\) (**)
\(\dfrac{HC'}{CC'}=\dfrac{S_{HAB}}{S_{ABC}}\) (***)
Từ : ( * ; ** ; ***) =>\(\dfrac{HA'}{AA'}+\dfrac{HB'}{BB'}+\dfrac{HC'}{CC'}=\dfrac{S_{HAC}+S_{HAB}+S_{HBC}}{S_{ABC}}\)
\(=\dfrac{S_{ABC}}{S_{ABC}}=1\left(đpcm\right)\)
Lời giải:
Ta thấy:
\(\left\{\begin{matrix} S_{HBC}=\frac{HA'.BC}{2}\\ S_{ABC}=\frac{AA'.BC}{2}\end{matrix}\right.\Rightarrow \frac{S_{HBC}}{S_{ABC}}=\frac{HA'}{AA'}(*)\)
\(\left\{\begin{matrix} S_{HAC}=\frac{HB'.AC}{2}\\ S_{ABC}=\frac{BB'.AC}{2}\end{matrix}\right.\Rightarrow \frac{S_{HAC}}{S_{ABC}}=\frac{HB'}{BB'}(**)\)
\(\left\{\begin{matrix} S_{HAB}=\frac{HC'.AB}{2}\\ S_{ABC}=\frac{CC'.AB}{2}\end{matrix}\right.\) \(\Rightarrow \frac{S_{HAB}}{S_{ABC}}=\frac{HC'}{CC'}(***)\)
Từ \((*); (**); (***)\Rightarrow \frac{HA'}{AA'}+\frac{HB'}{BB'}+\frac{HC'}{CC'}=\frac{S_{HBC}+S_{HCA}+S_{HAB}}{S_{ABC}}=\frac{S_{ABC}}{S_{ABC}}=1\)
Ta có : \(\frac{HA'}{AA'}=\frac{S_{HBC}}{S_{ABC}};\frac{HB'}{AB'}=\frac{S_{HAC}}{S_{ABC}};\frac{HC'}{AC'}=\frac{S_{HAB}}{S_{ABC}}\)
nên \(\frac{HA'}{AA'}+\frac{HB'}{BB'}+\frac{HC'}{CC'}=\frac{S_{HBC}+S_{HAB}+S_{HAC}}{S_{ABC}}=\frac{S_{ABC}}{S_{ABC}}=1\)
Vậy \(\frac{HA'}{AA'}+\frac{HB'}{BB'}+\frac{HC'}{CC'}=1\)
Ta có:
\(\dfrac{HA'}{AA'}+\dfrac{HB'}{BB'}+\dfrac{HC'}{CC'}\)
\(\dfrac{HA'.BC}{AA'.BC}+\dfrac{HB'.AC}{BB'.AC}+\dfrac{HC'.AB}{CC'.AB}\)
\(\dfrac{S_{BHC}}{S_{ABC}}+\dfrac{S_{AHC}}{S_{ABC}}+\dfrac{S_{AHB}}{S_{ABC}}=\dfrac{S_{ABC}}{S_{ABC}}=1\)