smc

If it works for a parallelogram $ABCD$, it should also work for a unit square, with $A(0,0),B(0,1),C(1,1),D(1,0)$. We are given that $E$ is the midpoint of $BD$, so $E(0.5,0.5)$. If $F$ is on $DA$, then $F(x,0)$. We note that $DF=1-x$ and $DA=1$, so $DF=\frac{1}{3}DA$ means $1-x=\frac{1}{3}$, or $x=\frac{2}{3}$, and hence $F(\frac{2}{3},0)$. We note that $△DFE$ has a base $DF$ that is $\frac{1}{3}$ and an altitude from $E$ to $DF$ that is $\frac{1}{2}$. Therefore, $[DEF]=\frac{1}{2}bh=\frac{1}{2}\cdot \frac{1}{3}\cdot \frac{1}{2}=\frac{1}{12}$. Quadrilateral $ABEF$ can be split into $△ABE$ and $△AEF$. The first triangle is $\frac{1}{4}$ of the unit square cut diagonally, so $[ABE]=\frac{1}{4}$. The second triangle has base $AF$ that is $\frac{2}{3}$ and height $E$ to $AF$ that is $\frac{1}{2}$. Therefore, $[AEF]=\frac{1}{2}bh=\frac{1}{2}\cdot \frac{2}{3}\cdot \frac{1}{2}=\frac{1}{6}$. The entire quadrilateral $ABEF$ has area $\frac{1}{4}+\frac{1}{6}=\frac{5}{12}$. This is $5$ times larger than the area of $△DFE$, so the ratio is $1:5$, or $\boxed{1:5}$.