a. Xét \(\Delta HBA\) và \(\Delta ABC\) có:
\(\widehat{B}\left(chung\right)\)
\(\widehat{BHA}=\widehat{BAC}\left(=90^0\right)\)
Do đó: \(\Delta HBA\infty\Delta ABC\left(g-g\right)\)
b. Vì \(\Delta ABC\) vuông tại A
=> \(AB^2+AC^2=BC^2\)
hay \(6^2+8^2=BC^2\)
=> \(\sqrt{BC}=\sqrt{100}\)
=> BC = 10cm
Vì \(\Delta HBA\infty\Delta ABC\left(cmt\right)\)
=> \(\dfrac{AH}{AC}=\dfrac{AB}{BC}\)
hay \(\dfrac{AH}{8}=\dfrac{6}{10}\)
=> AH = 4,8 cm
Vì \(\Delta ABH\) vuông tại H
=> \(BH^2+AH^2=AB^2\)
hay \(BH^2=6-4,8\)
=> BH = 1,2 cm
c. Xét \(\Delta ABC\) và \(\Delta HAC\) có:
\(\widehat{BAC}=\widehat{AHC}\left(=90^0\right)\)
\(\widehat{C}\left(chung\right)\)
Do đó: \(\Delta ABC\infty\Delta HAC\left(g-g\right)\)
Mà \(\Delta HBA\infty\Delta ABC\left(cmt\right)\)
=> \(\Delta HAC\infty\Delta HBA\)
=> \(\dfrac{AH}{HB}=\dfrac{HC}{AH}\)
hay \(AH^2=HB.HC\)
