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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)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"