Proof: By Euclid
(related to Proposition: 3.37: Converse of Tangent Secant Theorem)
 For let $DE$ have been drawn touching $ABC$ [Prop. 3.17], and let the center of the circle $ABC$ have been found, and let it be (at) $F$.
 And let $FE$, $FB$, and $FD$ have been joined.
 (Angle) $FED$ is thus a right angle [Prop. 3.18].
 And since $DE$ touches circle $ABC$, and $DCA$ cuts (it), the (rectangle contained) by $AD$ and $DC$ is thus equal to the (square) on $DE$ [Prop. 3.36].
 And the (rectangle contained) by $AD$ and $DC$ was also equal to the (square) on $DB$.
 Thus, the (square) on $DE$ is equal to the (square) on $DB$.
 Thus, $DE$ (is) equal to $DB$.
 And $FE$ is also equal to $FB$.
 So the two (straight lines) $DE$, $EF$ are equal to the two (straight lines) $DB$, $BF$ (respectively).
 And their base, $FD$, is common.
 Thus, angle $DEF$ is equal to angle $DBF$ [Prop. 1.8].
 And $DEF$ (is) a right angle.
 Thus, $DBF$ (is) also a right angle.
 And $FB$ produced is a diameter, And a (straight line) drawn at right angles to a diameter of a circle, at its extremity, touches the circle [Prop. 3.16 corr.] .
 Thus, $DB$ touches circle $ABC$.
 Similarly, (the same thing) can be shown, even if the center happens to be on $AC$.
 Thus, if some point is taken outside a circle, and two straight lines radiate from the point towards the circle, and one of them cuts the circle, and the (other) meets (it), and the (rectangle contained) by the whole (straight line) cutting (the circle), and the (part of it) cut off outside (the circle), between the point and the convex circumference, is equal to the (square) on the (straight line) meeting (the circle), then the (straight line) meeting (the circle) will touch the circle.
 (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