a) Xét \(\Delta ABC\) vuông tại A có :
\(BC^2=AB^2+AC^2\)
\(25=9+AC^2\)
\(AC^2=16\)
AC = 4 cm
Xét \(\Delta ABC\) có : AD là phân giác
=> \(\dfrac{AD}{AB}=\dfrac{DC}{BC}=\dfrac{AD+DC}{AB+BC}=\dfrac{AC}{AB+BC}=\dfrac{4}{3+5}=\dfrac{1}{2}\)
=> \(\dfrac{AD}{AB}=\dfrac{1}{2}\Leftrightarrow AD=\dfrac{1}{2}.3=1,5cm\)
b)Xét \(\Delta ABCvà\Delta HDC\) có:
\(\widehat{C}\) chung
\(\widehat{BAC}=\widehat{DHC}\)
=> \(\Delta ABC\sim\Delta HDC\)
=> \(\dfrac{CA}{CH}=\dfrac{CB}{CD}\Leftrightarrow CH.CB=CD.CA\)
a) Xét \(\Delta ABC\) vuông tại A:
\(BC^2=AB^2+AC^2\left(Pytago\right).\\ \Rightarrow5^2=3^2+AC^2.\\ \Rightarrow AC=4\left(cm\right).\)
Xét \(\Delta ABC:\)
BD là phân giác \(\widehat{ABC}\) (gt).
\(\Rightarrow\) \(\dfrac{AD}{CD}=\dfrac{AB}{BC}\) (Tính chất phân giác).
\(\Rightarrow\dfrac{AD}{CD+AD}=\dfrac{AB}{BC+AB}.\\ \Rightarrow\dfrac{AD}{AC}=\dfrac{AB}{BC+AB}.\)
Thay: \(\dfrac{AD}{4}=\dfrac{3}{5+3}.\)
\(\Rightarrow AD=\dfrac{3}{2}\left(cm\right).\)
b) Xét \(\Delta ABC\) và \(\Delta HDC:\)
\(\widehat{C}chung.\)
\(\widehat{A}=\widehat{H}\left(=90^o\right).\)
\(\Rightarrow\Delta ABC\sim\Delta HDC\left(g-g\right).\\ \Rightarrow\dfrac{CB}{CD}=\dfrac{CA}{CH}.\\ \Rightarrow CH.CB=CD.CA.\)
