Proof: By Euclid

For (if) not then, if possible, let $G$ (be the center of the circle), and let $GA$, $GD$, and $GB$ have been joined.
And since $AD$ is equal to $DB$, and $DG$ (is) common, the two (straight lines) $AD$, $DG$ are equal to the two (straight lines) $BD$, $DG$,¹ respectively.
And the base $GA$ is equal to the base $GB$.
For (they are both) radii.
Thus, angle $ADG$ is equal to angle $GDB$ [Prop. 1.8].
And when a straight line stood upon (another) straight line make adjacent angles (which are) equal to one another, each of the equal angles is a right angle [Def. 1.10] .
Thus, $GDB$ is a right angle.
And $FDB$ is also a right angle.
Thus, $FDB$ (is) equal to $GDB$, the greater to the lesser.
The very thing is impossible.
Thus, (point) $G$ is not the center of the circle $ABC$.
So, similarly, we can show that neither is any other (point) except $F$.
Thus, point $F$ is the center of the [circle] $ABC$.

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The Greek text has "$GD$, $DB$", which is obviously a mistake (translator's note). ↩