Question

 

cho tam giac abc vuong tai a, duong cao ah

a. chung minh tam giac hba dong dang voi tam giac abc

b. chung minh ah^2 =hb* hc

c. tia phan giac cua goc ahc cat ac tai d . chung minh \(\frac{hb}{hc}\)=\(\frac{ad^2}{dc^2}\)

 

 

Answer 1

a) Xét \(\Delta HBA\)và \(\Delta ABC\)có :

\(\widehat{AHB}=\widehat{BAC}=90^o;\widehat{B}\left(chung\right)\)

\(\Rightarrow\)\(\Delta HBA\)\(\approx\)\(\Delta ABC\)( g.g )

b) Xét \(\Delta HBA\)và \(\Delta HAC\)có :

\(\widehat{AHB}=\widehat{AHC}=90^o\)

\(\widehat{BAH}=\widehat{ACH}\left(cung-phu-\widehat{B}\right)\)

\(\Rightarrow\Delta HBA\approx\Delta HAC\left(g.g\right)\)

\(\Rightarrow\frac{BH}{AH}=\frac{AH}{HC}\Rightarrow AH^2=BH.HC\)