Let $$E$$ be the midpoint of the side $$\overline{BC}$$ of $$\triangle ABC$$ and let $$F$$ be chosen in $$\overline{AC}$$ so that $$AC = 3FC$$. Determine the ratio of the areas, $$\triangle ABEF \colon \triangle EFC$$.

Triangles $$EFC$$ and $$ABC$$ share the angle $$C$$, so $$\frac{[EFC]}{[ABC]} = \frac{FC}{AC} \cdot \frac{EC}{BC} = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$$. The quadrilateral $$ABEF$$ is the rest of the triangle, so its area is $$\frac{5}{6}[ABC]$$. Hence $$[ABEF] \colon [EFC] = 5 \colon 1$$, i.e. the ratio is 5.