Proof: By Euclid

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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