Bạn kiểm tra lại đề nhé:
Chứng minh: \(\frac{HE}{AA'}+\frac{HE}{BB'}+\frac{HF}{CC'}=2\)
Ta có:
\(\frac{HA'}{AA'}=\frac{S\left(HBC\right)}{S\left(ABC\right)}\); \(\frac{HB'}{BB'}=\frac{S\left(HAC\right)}{S\left(ABC\right)}\); \(\frac{HC'}{CC'}=\frac{S\left(BHA\right)}{S\left(ABC\right)}\)
=> \(\frac{HA'}{AA'}+\frac{HB'}{BB'}+\frac{HC'}{CC'}=\frac{S\left(HAB\right)+S\left(HAC\right)+S\left(HBC\right)}{S\left(ABC\right)}=1\)
=> \(\frac{2HA'}{AA'}+\frac{2HB'}{BB'}+\frac{2HC'}{CC'}=2\)
Lại có: E; D; F lần lượt đối xứng với H qua BC; AC; AB
=> HE = 2HA'; HD = 2HC'; HF = 2HB'
=> \(\frac{HE}{AA'}+\frac{HE}{BB'}+\frac{HF}{CC'}=2\)