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In $$\triangle ABC$$, $$AB = x$$, $$AC = y$$. If $$D$$ is a point on the side $$BC$$ such that $$AD$$ is the internal angle bisector of the angle $$A$$ then $$BC \colon BD$$ is
By the angle bisector theorem $$\frac{BD}{DC} = \frac{AB}{AC} = \frac{x}{y}$$. So $$BD = \frac{x}{x + y} \cdot BC$$, giving $$BC \colon BD = (x + y) \colon x$$.
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