Solution. The set of points of a parabola in the plane consists of those points that are equidistant from a given point ($F$, the focus) and a given line (the directrix) that does not pass through the point.

Draw a tangent to the circle at point $D$, which we will denote as $v$. Since $F C=D C$, according to the definition of a parabola, $v$ is the directrix.

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According to the problem statement, the following needs to be proven: $F D \cdot F E=D E^{2}$. The triangle $F B D$ is a right triangle by Thales' theorem, so applying the leg theorem to it: $F B^{2}=F D \cdot F E$. The two equations can be matched, and then we only need to prove the following: $D E=F B$.

To do this, drop a perpendicular from point $B$ to the directrix, thus obtaining point $G$. We know that every point on the parabola is equidistant from the focus and the directrix, so $F B=B G$. We also know that $B G=D E$, since they are opposite sides of the same rectangle. Therefore: $F B=B G=D E$, and thus $D E=F B$. This proves the statement.