Rysunek pomocniczy:

---

Z twierdzenia o dwusiecznej:

$\frac{\left|CD\right|}{\left|BD\right|}=\frac{\left|AC\right|}{\left|AB\right|}$  

$\frac{\left|CD\right|}{\left|BC\right|-\left|CD\right|}=\frac{\left|AC\right|}{\left|AB\right|}$  

$\frac{\left|CD\right|}{9-\left|CD\right|}=\frac{8}{10}$  

$\frac{\left|CD\right|}{9-\left|CD\right|}=\frac{4}{5}$  

Mnożymy na krzyż.

$5\left|CD\right|=4\left(9-\left|CD\right|\right)$  

$5\left|CD\right|=36-4\left|CD\right|$  

$9\left|CD\right|=36\mid :9$  

$\left|CD\right|=4$  

Wówczas:

$\left|BD\right|=\left|BC\right|-\left|CD\right|=9-4=5$  

---

Odp. |BD|=5, |CD|=4.