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In $$\triangle ABC$$, $$P$$ is a point on the line segment $$BC$$. A line $$l$$ parallel to the side $$BC$$ intersects the line segments $$AB$$, $$AP$$ and $$AC$$ in $$D$$, $$E$$, $$F$$, respectively, such that $$AD = 2019\,DB$$. Then $$\frac{AE \cdot FC}{AF \cdot EP}$$ is equal to
Since $$DE \parallel BP$$, in $$\triangle ABP$$ we get $$\frac{AE}{EP} = \frac{AD}{DB} = 2019$$. Since $$DF \parallel BC$$, in $$\triangle ABC$$ we get $$\frac{AF}{FC} = \frac{AD}{DB} = 2019$$. Hence $$\frac{AE \cdot FC}{AF \cdot EP} = \frac{AE}{EP} \cdot \frac{FC}{AF} = 2019 \times \frac{1}{2019} = 1$$.
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