Vì AD là tia phân giác góc A nên theo tính chất đường phân giác ta có:
\(\dfrac{AB}{AC}=\dfrac{BD}{CD}\Rightarrow\dfrac{AB}{AC+AB}=\dfrac{BD}{BD+CD}\)hay\(\dfrac{5}{5+7}=\dfrac{BD}{6}\)
\(\Rightarrow BD=5.6:12=2,5\)
theo dinh luat tia phan giac ta co :
\(\dfrac{AB}{AC}\)=\(\dfrac{BD}{CD}\Rightarrow\) \(\dfrac{BD}{CD}=\dfrac{5}{7}\)\(\Rightarrow\dfrac{BD}{BC-BD}=\dfrac{5}{7}\Rightarrow\dfrac{BD}{6-BD}=\dfrac{5}{7}\Rightarrow7.BD=30-5.BD\Rightarrow BD=2,5\left(cm\right)\)
ta có AD là phân giác góc B\(\rightarrow\)
\(\dfrac{AB}{AC}=\dfrac{DB}{DC}\Rightarrow\dfrac{AB}{AC+AB}=\dfrac{DB}{DC+DB}\)
\(\dfrac{5}{12}=\dfrac{DB}{6}\Rightarrow DB=\dfrac{30}{12}=2,5\)
