Cho tam giác \(ABC\) có các đường cao \(AA',BB',CC'\), trực tâm \(H\). Khi đó, \(\dfrac{HA'}{AA'}+\dfrac{HB'}{BB'}+\dfrac{HC'}{CC'}=\)?
\(\dfrac{1}{2}.\)\(1\).\(2.\)\(4.\)Hướng dẫn giải:
Ta có \(S_{ABC}=S_{ABH}+S_{BCH}+S_{ACH}\)
\(\Rightarrow\dfrac{S_{ABH}}{S_{ABC}}+\dfrac{S_{BCH}}{S_{ABC}}+\dfrac{S_{ACH}}{S_{ABC}}=1\)
\(\Rightarrow\dfrac{HC'.AB}{CC'.AB}+\dfrac{HA'.BC}{AA'.BC}+\dfrac{HB'.AC}{BB'.AC}=1\)
\(\Rightarrow\dfrac{HA'}{AA'}+\dfrac{HB'}{BB'}+\dfrac{HC'}{CC'}=1.\)