[tex]\it Th.\ bisec. \Rightarrow\ \dfrac{DB}{DC}=\dfrac{AB}{AC}\Rightarrow\ \dfrac{DB}{DC}=\dfrac{5}{7}\Rightarrow\ \dfrac{DB}{DC+DB}=\dfrac{5}{7+5} \Rightarrow \\ \\ \\ \Rightarrow \dfrac{DB}{BC}=\dfrac{5}{12} \Rightarrow \dfrac{DB}{6}=\dfrac{5}{12} \Rightarrow DB=\dfrac{\ 5\cdot6^{(6}}{12}=\dfrac{5}{2}=2,5\ cm[/tex]
[tex]\it DC=BC-DB=6-2,5=3,5\ cm[/tex]
