Proof: By Euclid
(related to Proposition: 3.19: Right Angle to Tangent of Circle Goes Through Center)
 For (if) not, if possible, let $F$ be (the center of the circle), and let $CF$ have been joined.
 Therefore, since some straight line $DE$ touches the circle $ABC$, and $FC$ has been joined from the center to the point of contact, $FC$ is thus perpendicular to $DE$ [Prop. 3.18].
 Thus, $FCE$ is a right angle.
 And $ACE$ is also a right angle.
 Thus, $FCE$ is equal to $ACE$, the lesser to the greater.
 The very thing is impossible.
 Thus, $F$ is not the center of circle $ABC$.
 So, similarly, we can show that neither is any (point) other (than one) on $AC.$ Thus, if some straight line touches a circle, and a straight line is drawn from the point of contact, at right angles to the tangent, then the center (of the circle) will be on the (straight line) so drawn.
 (Which is) the very thing it was required to show.
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"