Question

cho tam giác ABC nhọn.AD,BE,CF là các đường cao . Gọi H là trực tâm.Cmr

a) HA/AD+HB/BE+HC/CF 

b)\(cos^2A+cos^2B+cos^2C< 1\)

Answer 1

a/

Ta có : \(\frac{HD}{AD}=\frac{S_{BHC}}{S_{ABC}}\) ; \(\frac{HE}{BE}=\frac{S_{AHC}}{S_{ABC}}\) ; \(\frac{HF}{FC}=\frac{S_{AHB}}{S_{ABC}}\)

\(\Rightarrow\frac{HD}{AD}+\frac{HE}{BE}+\frac{HF}{FC}=\frac{S_{BHC}+S_{AHC}+S_{AHB}}{S_{ABC}}=\frac{S_{ABC}}{S_{ABC}}=1\)

Ta có : \(1-\frac{HA}{AD}=\frac{HD}{AD}\) ; \(1-\frac{HB}{BE}=\frac{HE}{BE}\) ; \(1-\frac{HC}{CF}=\frac{HF}{CF}\)

Suy ra \(1-\frac{HA}{AD}+1-\frac{HB}{BE}+1-\frac{HC}{CF}=1\)

\(\Rightarrow\frac{HA}{AD}+\frac{BH}{BE}+\frac{CH}{CF}=2\)

Answer 2

b) cm: cos²A + cos²B + cos²C <1

xet tg BFC va tg BDA co:

 BFC=BDA=90^O (GT)

BCF=BAD(cung phu voi FBD)

=> tg BFC dong dang tg BDA(g.g)

=>BF/BD=BC/BA

xet tg BDF va tg BAC co :

ABC: goc chung

BF/BD=BC/BA(cmt)

=>tg BDF dong dang tg BAC(c.g.c)

=> S_BDF/S_BAC=(DB/AB)²

ma tg ABD vuong tai D => cosB=DB/AB(ti so luong giac cua goc nhon)

=> S_BDF/S_ABC=cos²A

tuong tu S_CDE/S_CAB=cos²C

=>cos²A+cos²B+cos²C =(S_BDF+S_AEF+S_CDE)/S_ABC

ma S_BDF+S_AEF+S_CDE=S_ABC-S_DEF<S_ABC

=>cos²A+cos²B+cos²C<1