The internal bisector of angle A in triangle ABC meets BC at D. If AB = 10 cm, AC = 15 cm and BC = 20 cm, find BD (in cm).

The bisector result, which follows from BPT applied to a line through $C$ parallel to $AD$, gives $\frac{BD}{DC}=\frac{AB}{AC}=\frac{10}{15}=\frac{2}{3}$. So $BC=20$ cm splits into 2 + 3 = 5 parts of 4 cm, and $BD=2$ parts = 8 cm. The value 12 is $DC$, obtained by matching $BD$ with the wrong side; 10 comes from treating $AD$ as a median and halving $BC$; 4 is just one part.