A problem by Efren Medallo

Let $\overline {DF} = \overline {BC} = x$ and $\overline {BF} = y$ . By virtue of similar triangles,

$\frac{\overline{BF}}{\overline{DF}}= \frac{\overline{BC}}{\overline{AC}}$

$\frac{y}{x} = \frac{x}{8}$

$y = \frac {x^{2}}{8}$

Since it is known that $\Delta ABC \cong \Delta DEF$ , then polygon $ADEC$ has an area of $\Delta ABC + \Delta DEF - \Delta DBF$ . That is,

$\frac{8x}{2} + \frac{8x}{2} - \frac{xy}{2} = 28$

or

$\frac {x(16-y)}{2} = 28$

substituting the first equation to the second yields

$x^{3} - 128x + 448 = 0$

whose zeroes are

$x=-2\left(1+\sqrt{29}\right),\:x=4,\:x=2\left(\sqrt{29}-1\right)$

The first value of $x$ is a negative value, and the third value of $x$ is greater than 8, thus the only plausible value would be $x = 4$ .

Using $x=4$ gives a value of $y=2$ . From the figure, $\overline{FC} = x-y$ , thus the answer is $\boxed{2}$ .