Proof: By Euclid
(related to Proposition: 3.36: Tangent Secant Theorem)
 $[D]CA$ is surely either through the center, or not.
 Let it first of all be through the center, and let $F$ be the center of circle $ABC$, and let $FB$ have been joined.
 Thus, (angle) $FBD$ is a right angle [Prop. 3.18].
 And since straight line $AC$ is cut in half at $F$, let $CD$ have been added to it.
 Thus, the (rectangle contained) by $AD$ and $DC$ plus the (square) on $FC$ is equal to the (square) on $FD$ [Prop. 2.6].
 And $FC$ (is) equal to $FB$.
 Thus, the (rectangle contained) by $AD$ and $DC$ plus the (square) on $FB$ is equal to the (square) on $FD$.
 And the (square) on $FD$ is equal to the (sum of the squares) on $FB$ and $BD$ [Prop. 1.47].
 Thus, the (rectangle contained) by $AD$ and $DC$ plus the (square) on $FB$ is equal to the (sum of the squares) on $FB$ and $BD$.
 Let the (square) on $FB$ have been subtracted from both.
 Thus, the remaining (rectangle contained) by $AD$ and $DC$ is equal to the (square) on the tangent $DB$.
 And so let $DCA$ not be through the center of circle $ABC$, and let the center $E$ have been found, and let $EF$ have been drawn from $E$, perpendicular to $AC$ [Prop. 1.12].
 And let $EB$, $EC$, and $ED$ have been joined.
 (Angle) $EBD$ (is) thus a right angle [Prop. 3.18].
 And since some straight line, $EF$, through the center, cuts some (other) straight line, $AC$, not through the center, at right angles, it also cuts it in half [Prop. 3.3].
 Thus, $AF$ is equal to $FC$.
 And since the straight line $AC$ is cut in half at point $F$, let $CD$ have been added to it.
 Thus, the (rectangle contained) by $AD$ and $DC$ plus the (square) on $FC$ is equal to the (square) on $FD$ [Prop. 2.6].
 Let the (square) on $FE$ have been added to both.
 Thus, the (rectangle contained) by $AD$ and $DC$ plus the (sum of the squares) on $CF$ and $FE$ is equal to the (sum of the squares) on $FD$ and $FE$.
 But the (square) on $EC$ is equal to the (sum of the squares) on $CF$ and $FE$.
 For [angle] $EFC$ [is] a right angle [Prop. 1.47].
 And the (square) on $ED$ is equal to the (sum of the squares) on $DF$ and $FE$ [Prop. 1.47].
 Thus, the (rectangle contained) by $AD$ and $DC$ plus the (square) on $EC$ is equal to the (square) on $ED$.
 And $EC$ (is) equal to $EB$.
 Thus, the (rectangle contained) by $AD$ and $DC$ plus the (square) on $EB$ is equal to the (square) on $ED$.
 And the (sum of the squares) on $EB$ and $BD$ is equal to the (square) on $ED$.
 For $EBD$ (is) a right angle [Prop. 1.47].
 Thus, the (rectangle contained) by $AD$ and $DC$ plus the (square) on $EB$ is equal to the (sum of the squares) on $EB$ and $BD$.
 Let the (square) on $EB$ have been subtracted from both.
 Thus, the remaining (rectangle contained) by $AD$ and $DC$ is equal to the (square) on $BD$.
 Thus, if some point is taken outside a circle, and two straight lines radiate from it towards the circle, and (one) of them cuts the circle, and (the other) touches (it), then 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, will be equal to the square on the tangent (line).
 (Which is) the very thing it was required to show.^{1}
∎
Thank you to the contributors under CC BYSA 4.0!
 Github:

 nonGithub:
 @Fitzpatrick
References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes