b) có \(ED\perp AC;AB\perp AC\Rightarrow ED//AB\)
\(\Rightarrow\widehat{EDF}=\widehat{FAB}\left(slt\right)\); \(\widehat{FED}=\widehat{FBA}\left(slt\right)\)
Xét \(\Delta EDF\) và \(\Delta BAF\) có :
\(\widehat{EDF}=\widehat{FAB}\left(cmt\right)\)
\(\widehat{FED}=\widehat{FBA}\left(cmt\right)\)
\(\Rightarrow\)\(\Delta EDF\) \(\sim\) \(\Delta BAF\) ( gg)
\(\Rightarrow\frac{FD}{FA}=\frac{FE}{FB}\Leftrightarrow FA.FE=FB.FD\)
a) Xét \(\Delta ABC\) vuông tại A:
\(\Rightarrow BC^2=AB^2+AC^2\)
\(\Leftrightarrow BC^2=42^2+56^2\)
\(\Leftrightarrow BC=70cm\)
Xét \(\Delta ABC\) có AD là phân giác
\(\Rightarrow\frac{AB}{AC}=\frac{BD}{CD}\Leftrightarrow\frac{AB}{AB+AC}=\frac{BD}{BD+CD}\Leftrightarrow\frac{AB}{AB+AC}=\frac{BD}{BC}\Leftrightarrow\frac{42}{42+56}=\frac{BD}{70}\Leftrightarrow BD=\frac{42+70}{42+56}=\frac{8}{7}\)
Vậy BD = \(\frac{8}{7}cm\)
Có BD + DC = BC
\(\Rightarrow DC=BC-BD=70-\frac{8}{7}=\frac{482}{7}cm\)