Proof: By Euclid
(related to Proposition: 3.17: Construction of Tangent from Point to Circle)
 For let the center $E$ of the circle have been found [Prop. 3.1], and let $AE$ have been joined.
 And let (the circle) $AFG$ have been drawn with center $E$ and radius $EA$.
 And let $DF$ have been drawn from from (point) $D$, at right angles to $EA$ [Prop. 1.11].
 And let $EF$ and $AB$ have been joined.
 I say that the (straight line) $AB$ has been drawn from point $A$ touching circle $BCD$.
 For since $E$ is the center of circles $BCD$ and $AFG$, $EA$ is thus equal to $EF$, and $ED$ to $EB$.
 So the two (straight lines) $AE$, $EB$ are equal to the two (straight lines) $FE$, $ED$ (respectively).
 And they contain a common angle at $E$.
 Thus, the base $DF$ is equal to the base $AB$, and triangle $DEF$ is equal to triangle $EBA$, and the remaining angles (are equal) to the (corresponding) remaining angles [Prop. 1.4].
 Thus, (angle) $EDF$ (is) equal to $EBA$.
 And $EDF$ (is) a right angle.
 Thus, $EBA$ (is) also a right angle.
 And $EB$ is a radius.
 And a (straight line) drawn at right angles to the diameter of a circle, from its extremity, touches the circle [Prop. 3.16 corr.] .
 Thus, $AB$ touches circle $BCD$.
 Thus, the straight line $AB$ has been drawn touching the given circle $BCD$ from the given point $A$.
 (Which is) the very thing it was required to do.
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"