Bài 3: Cho $\triangle ABC$ có 3 góc nhọn, các đường cao AA‟ , BB‟ , CC‟ và trực tâm H.
Tính tổng: $\frac{HA{}'}{AA{}'}+\frac{HB{}'}{BB{}'}+\frac{HC{}'}{CC{}'}$ .
Bài Làm:
Ta có : $\frac{S_{HBC}}{S_{ABC}}=\frac{\frac{1}{2}HA{}'.BC}{\frac{1}{2}AA{}'.BC}=\frac{HA{}'}{AA{}'}$ (1)
$\frac{S_{HAB}}{S_{ABC}}=\frac{\frac{1}{2}HC{}'.AB}{\frac{1}{2}CC{}'.AB}=\frac{HC{}'}{CC{}'}$ (2)
$\frac{S_{HAC}}{S_{ABC}}=\frac{\frac{1}{2}HB{}'.AC}{\frac{1}{2}BB{}'.AC}=\frac{HB{}'}{BB{}'}$ (3)
Cộng (1) + (2) + (3) theo vế ta được :
$\frac{HA{}'}{AA{}'}+\frac{HB{}'}{BB{}'}+\frac{HC{}'}{CC{}'}=\frac{S_{HBC}+S_{HAB}+S_{HAC}}{S_{ABC}}$
<=> $\frac{S_{ABC}}{S_{ABC}}=1$ .
Vậy $\frac{HA{}'}{AA{}'}+\frac{HB{}'}{BB{}'}+\frac{HC{}'}{CC{}'}=1$ .