Is AD the bisector of angle A in triangle ABC, given that D lies on BC with BD = 10 cm and DC = 6 cm, and that AB = 15 cm and AC = 9 cm?

$\frac{BD}{DC}=\frac{10}{6}=\frac{5}{3}$ and $\frac{AB}{AC}=\frac{15}{9}=\frac{5}{3}$. The two ratios agree, and the converse of the bisector proportionality (itself a BPT consequence) then forces $AD$ to bisect angle A. The second option misreads the arithmetic; the third confuses "larger part" with "bisector"; the fourth overlooks that the side lengths alone settle the question.