finding the center of a circle (elementary geometry)

Based on the described construction and using the OP's notations, we have the following figure:

It is enough to demonstrate that $DG=r$, the radius of the circle.

By the construction, the triangle $ABD$ is equilateral and then $AB=BD=AF$. $BDG$ is an isosceles triangle because the triangles $ABD$ and $ADG$ are congruent. $\angle AGD = \angle BGD$ because / and $AB=AD=AF$ (See the Wiki article called Inscribed angle)

Now, connecting $A$ and $F$ with the center $O$ we get the green triangle $AFO$. To show that $DG=FO=r$, it is enough to prove that the triangles $BDG$ and $AFO$ are congruent.

We have already seen that the bases of these triangles equal. What remains is to show that $\angle AOF = \angle BGD$.

This is easy. Just look at the figure and you will see that the $\angle BDG =2\times\angle AGD = \angle AOF$. (See the Wiki article called Inscribed angle.)

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Finally, this is how to use the construction at stake to find the center of the circle:

1. Construct $E$, $F$, and $D$ and then $G$ as described.
2. The straight $ED$ and the circle centered at $G$ through $D$ will meet in $O$.

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Note

This construction is not that bad compared to the most frequently used one in the case of which you construct two bisecting perpendiculars. In the case of this present construction it is enough to draw the line through $F$ and $D$ which already exist after constructing the first bisecting perpendicular.