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In a triangle $$ABC$$, $$D$$, $$E$$ are points on the line segments $$AB$$ and $$AC$$ respectively. $$DE$$ is parallel to $$BC$$. If $$\frac{AB}{AC} = \frac{6}{7}$$ and $$\frac{AD}{DB} = \frac{3}{2}$$ then $$\frac{AC}{AE} =$$
Since $$DE \parallel BC$$, by the basic proportionality theorem $$\frac{AE}{EC} = \frac{AD}{DB} = \frac{3}{2}$$. So $$\frac{AE}{AC} = \frac{3}{5}$$ and $$\frac{AC}{AE} = \frac{5}{3}$$. The ratio $$\frac{AB}{AC}$$ is not needed.
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