Proof: By Euclid
(related to Proposition: 3.18: Radius at Right Angle to Tangent)
 For if not, let $FG$ have been drawn from $F$, perpendicular to $DE$ [Prop. 1.12].
 Therefore, since angle $FGC$ is a right angle, (angle) $FCG$ is thus acute [Prop. 1.17].
 And the greater angle is subtended by the greater side [Prop. 1.19].
 Thus, $FC$ (is) greater than $FG$.
 And $FC$ (is) equal to $FB$.
 Thus, $FB$ (is) also greater than $FG$, the lesser than the greater.
 The very thing is impossible.
 Thus, $FG$ is not perpendicular to $DE$.
 So, similarly, we can show that neither (is) any other (straight line) except $FC$.
 Thus, $FC$ is perpendicular to $DE$.
 Thus, if some straight line touches a circle, and some (other) straight line is joined from the center (of the circle) to the point of contact, then the (straight line) so joined will be perpendicular to the tangent.
 (Which is) the very thing it was required to show.^{1}
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes