a) \(BD=BA+AD=4+5=9\left(cm\right).\)
Ta có: \(\dfrac{BC}{BD}=\dfrac{6}{9}=\dfrac{2}{3}.\\ \dfrac{AB}{CB}=\dfrac{4}{6}=\dfrac{2}{3}.\\ \Rightarrow\dfrac{BC}{BD}=\dfrac{AB}{CB}. \)
Xét \(\Delta ABC\) và \(\Delta CBD:\)
\(\dfrac{BC}{BD}=\dfrac{AB}{CB}\left(cmt\right).\\ \)
\(\widehat{B}chung.\)
\(\Rightarrow\Delta ABC\sim\Delta CBD\left(c-g-c\right).\)
\(\Rightarrow\dfrac{AC}{CD}=\dfrac{AB}{CB}\) (2 cạnh tương ứng).
\(\Rightarrow\dfrac{AC}{7}=\dfrac{2}{3}.\\ \Rightarrow AC=\dfrac{14}{3}\left(cm\right).\)
c) Xét \(\Delta BDC:\)
BF là phân giác \(\widehat{B}\left(gt\right).\)
\(\Rightarrow\dfrac{CF}{FD}=\dfrac{BC}{BD}\) (Tính chất đường phân giác).
Xét \(\Delta ABC:\)
BE là phân giác \(\widehat{B}\left(gt\right).\)
\(\Rightarrow\dfrac{AB}{CB}=\dfrac{AE}{EC}\) (Tính chất đường phân giác).
Mà \(\left\{{}\begin{matrix}\dfrac{AB}{CB}=\dfrac{BC}{BD}\left(cmt\right).\\\dfrac{CF}{FD}=\dfrac{BC}{BD}\left(cmt\right).\end{matrix}\right.\)
\(\Rightarrow\dfrac{AE}{EC}=\dfrac{CF}{FD}.\\ \Rightarrow AE.FD=CF.EC.\)
