[tex]\it Th.\ bisec. \Rightarrow \dfrac{BD}{CD}=\dfrac{AB}{AC} \Rightarrow \dfrac{BD}{CD}=\dfrac{\ 9^{(3}}{15} \Rightarrow\dfrac{BD}{CD}=\dfrac{3}{5} \Rightarrow\dfrac{BD}{CD}=0,6 \Rightarrow\\ \\ \\ \Rightarrow BD=0,6\cdot CD\ \ \ \ \ (1)\\ \\ BD+CD=BC \Rightarrow BD+CD=20 \Rightarrow BD=20-CD\ \ \ \ \ \ (2)[/tex]
[tex]\it (1),\ (2) \Rightarrow 0,6CD=20-CD\bigg|_{\cdot5} \Rightarrow 3CD=100-5CD\bigg|_{+5CD} \Rightarrow\\ \\ \\ \Rightarrow8CD=100\bigg|_{:4} \Rightarrow 2CD=25\bigg|_{:2} \Rightarrow CD=12,5\ cm\\ \\ \\ BD=BC-CD=20-12,5=7,5\ cm[/tex]
