a)Xét \(\Delta ABC\) vuông tại A có:
\(BC^2=AC^2+AB^2\)
\(BC^2=64+36\)
\(BC^2=100\)
BC=10cm
Xét \(\Delta ABC\) có: AD là phân giác của\(\widehat{BAC}\)
=> \(\dfrac{BD}{AB}=\dfrac{DC}{AC}=\dfrac{BD+DC}{AB+AC}=\dfrac{BC}{AB+AC}=\dfrac{10}{6+8}=\dfrac{5}{7}\)
=> \(\dfrac{BD}{AB}=\dfrac{5}{7}\Leftrightarrow\dfrac{BD}{6}=\dfrac{5}{7}\Rightarrow BD=\dfrac{5}{7}.6\approx4,3\) cm
b)
Xét \(\Delta HBA\) và \(\Delta ABC\) có:
\(\widehat{B}\) chung
\(\widehat{BHA}=\widehat{BAC}\left(=90^o\right)\)
=> \(\Delta HBA\sim\)\(\Delta ABC\) (g-g)
=> \(\dfrac{AH}{AC}=\dfrac{AB}{BC}\Leftrightarrow\dfrac{AH}{8}=\dfrac{6}{10}\Rightarrow AH=\dfrac{3}{5}.8=4,8cm\)
\(\dfrac{HB}{AB}=\dfrac{AB}{BC}\Leftrightarrow\dfrac{HB}{6}=\dfrac{6}{10}\Rightarrow HB=\dfrac{3}{5}.6=3,6cm\)
c) Có : \(\dfrac{HB}{AB}=\dfrac{AB}{BC}\) ( \(\Delta HBA\sim\)\(\Delta ABC\) )
=> \(AB^2=HB.BC\)
