In $$\triangle ABC$$, $$AB = 4$$, $$AC = 6$$, $$BC = 9$$. If $$D$$ is a point on $$\overline{BC}$$ such that $$3BD = 2DC$$ then $$\frac{\angle BAD}{\angle CAD}$$ is

Solution

$$3BD = 2DC$$ gives $$\frac{BD}{DC} = \frac{2}{3}$$, and $$\frac{AB}{AC} = \frac{4}{6} = \frac{2}{3}$$. Since $$D$$ divides $$BC$$ in the ratio of the adjacent sides, by the converse of the angle bisector theorem $$AD$$ bisects $$\angle A$$. So $$\angle BAD = \angle CAD$$ and the ratio is 1.

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