a) Xét \(\Delta ABC\) có :
\(\widehat{BAC}=90^o\left(gt\right)\)
\(\Rightarrow BC^2=AB^2+AC^2\) (Đlý Pitago)
=> \(BC=\sqrt{AB^2+AC^2}=\sqrt{6^2+8^2}=10\left(cm\right)\)
b) Xét \(\Delta ABC,\Delta HBA\) có :
\(\left\{{}\begin{matrix}\widehat{B}:Chung\\\widehat{BAC}=\widehat{BHA}=90^o\end{matrix}\right.\)
=> \(\Delta ABC\sim\Delta HBA\left(g.g\right)\)
c) Vì \(\Delta ABC\sim\Delta HBA\left(câu-b\right)=>\dfrac{AB}{HB}=\dfrac{BC}{BA}\Rightarrow AB^2=BH.BC\)
=> \(6^2=BH.10=>BH=\dfrac{36}{10}=3,6\left(cm\right)\)
Ta có : \(D\in BC\Rightarrow BC=BH+HC=>HC=10-3,6=6,4\left(cm\right)\)
d) Xét \(\Delta ABC\) có :
AD là đường phân giác trong tam giác
=> \(\dfrac{DB}{DC}=\dfrac{AB}{AC}=\dfrac{6}{8}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{DB}{BC-DB}=\dfrac{3}{4}=>\dfrac{DB}{10-DB}=\dfrac{3}{4}\)
\(=>3\left(10-DB\right)=4DB=>7DB=30=>DB=\dfrac{30}{7}\left(cm\right)\)
