Đáp án: $\dfrac{HA}{AA'}+\dfrac{HB}{BB'}+\dfrac{HC}{CC'}=2$

 

Giải thích các bước giải:

Ta có:

${{S}_{\Delta HBC}}=\dfrac{1}{2}HA'.BC$

${{S}_{\Delta ABC}}=\dfrac{1}{2}AA'.BC$

$\Rightarrow \dfrac{{{S}_{\Delta HBC}}}{{{S}_{\Delta ABC}}}=\dfrac{HA'}{AA'}$

$\Rightarrow \dfrac{{{S}_{\Delta HBC}}}{{{S}_{\Delta ABC}}}=\dfrac{HA'+HA}{AA'}-\dfrac{HA}{AA'}$

$\Rightarrow \dfrac{{{S}_{\Delta HBC}}}{{{S}_{\Delta ABC}}}=\dfrac{AA'}{AA'}-\dfrac{HA}{AA'}$

$\Rightarrow \dfrac{HA}{AA'}=1-\dfrac{{{S}_{\Delta HBC}}}{{{S}_{\Delta ABC}}}$

Chứng minh tương tự:

$\dfrac{HB}{BB'}=1-\dfrac{{{S}_{\Delta HCA}}}{{{S}_{\Delta ABC}}}$

$\dfrac{HC}{CC'}=1-\dfrac{{{S}_{\Delta HAB}}}{{{S}_{\Delta ABC}}}$

$\Rightarrow \dfrac{HA}{AA'}+\dfrac{HB}{BB'}+\dfrac{HC}{CC'}=3-\left( \dfrac{{{S}_{\Delta HBC}}+{{S}_{\Delta HCA}}+{{S}_{\Delta HAB}}}{{{S}_{\Delta ABC}}} \right)$

$\Rightarrow \dfrac{HA}{AA'}+\dfrac{HB}{BB'}+\dfrac{HC}{CC'}=3-\dfrac{{{S}_{\Delta ABC}}}{{{S}_{\Delta ABC}}}$

$\Rightarrow \dfrac{HA}{AA'}+\dfrac{HB}{BB'}+\dfrac{HC}{CC'}=3-1$

$\Rightarrow \dfrac{HA}{AA'}+\dfrac{HB}{BB'}+\dfrac{HC}{CC'}=2$