Let $ABCD$ be the given 4-gon, with $E$, $F$, $G$, $H$ as the midpoints of $AB$, $BC$, $CD$, $DA$ respectively. Draw the diagonals $AC$ and $BD$.

In triangle $ABD$, $E$ and $H$ are midpoints, so $EH \parallel BD$. In triangle $BCD$, $F$ and $G$ are midpoints, so $FG \parallel BD$. Thus, $EH \parallel FG$. Similarly, in triangle $ABC$, $EF \parallel AC$, and in triangle $ADC$, $HG \parallel AC$. Thus, $EF \parallel HG$. Since opposite sides are parallel, $EFGH$ is a parallelogram.

For the areas, the four corner triangles formed are $\Delta AEH$, $\Delta EBF$, $\Delta FCG$, and $\Delta GDH$. By the properties of triangles and midpoints,

$$\text{Area}(\Delta AEH) = \frac{1}{4}\,\text{Area}(\Delta ABD)$$

and

$$\text{Area}(\Delta CFG) = \frac{1}{4}\,\text{Area}(\Delta CBD).$$

Similarly,

$$\text{Area}(\Delta EBF) = \frac{1}{4}\,\text{Area}(\Delta ABC)$$

and

$$\text{Area}(\Delta HDG) = \frac{1}{4}\,\text{Area}(\Delta ACD).$$

Adding these:

$$\text{Area}(\Delta AEH) + \text{Area}(\Delta CFG) + \text{Area}(\Delta EBF) + \text{Area}(\Delta HDG) = \frac{1}{4}\left[\text{Area}(\Delta ABD) + \text{Area}(\Delta CBD)\right] + \frac{1}{4}\left[\text{Area}(\Delta ABC) + \text{Area}(\Delta ACD)\right].$$

Since $\text{Area}(\Delta ABD) + \text{Area}(\Delta CBD) = \text{Area}(ABCD)$ and $\text{Area}(\Delta ABC) + \text{Area}(\Delta ACD) = \text{Area}(ABCD)$, the sum is

$$\frac{1}{2}\,\text{Area}(ABCD).$$

The area of parallelogram $EFGH$ is the remaining portion:

$$\text{Area}(ABCD) - \frac{1}{2}\,\text{Area}(ABCD) = \frac{1}{2}\,\text{Area}(ABCD).$$

∴ The parallelogram's area is half that of the 4-gon.