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In a triangle $$ABC$$, $$D$$ and $$E$$ are points on the sides $$AB$$ and $$AC$$ such that the line segment $$DE$$ is parallel to the side $$BC$$. $$F$$ is a point on the line segment $$AD$$ such that $$FE$$ is parallel to the line segment $$DC$$. If $$AF = 4$$, $$FD = 6$$ then find $$DB$$.
Correct Answer: 15
In triangle $$ADC$$, $$FE \parallel DC$$, so $$\frac{AE}{EC} = \frac{AF}{FD} = \frac{4}{6}$$. In triangle $$ABC$$, $$DE \parallel BC$$, so $$\frac{AD}{DB} = \frac{AE}{EC} = \frac{2}{3}$$. With $$AD = 4 + 6 = 10$$, $$DB = \frac{3}{2} \times 10 = 15$$.
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