Proof: By Euclid
(related to Proposition: 3.32: Angles made by Chord with Tangent)
 That is to say, that angle $FBD$ is equal to the angle constructed in segment $BAD$, and angle $EBD$ is equal to the angle constructed in segment $DCB$.
 For let $BA$ have been drawn from $B$, at right angles to $EF$ [Prop. 1.11].
 And let the point $C$ have been taken at random on the circumference $BD$.
 And let $AD$, $DC$, and $CB$ have been joined.
 And since some straight line $EF$ touches the circle $ABCD$ at point $B$, and $BA$ has been drawn from the point of contact, at right angles to the tangent, the center of circle $ABCD$ is thus on $BA$ [Prop. 3.19].
 Thus, $BA$ is a diameter of circle $ABCD$.
 Thus, angle $ADB$, being in a semicircle, is a right angle [Prop. 3.31].
 Thus, the remaining angles (of triangle $ADB$) $BAD$ and $ABD$ are equal to one right angle [Prop. 1.32].
 And $ABF$ is also a right angle.
 Thus, $ABF$ is equal to $BAD$ and $ABD$.
 Let $ABD$ have been subtracted from both.
 Thus, the remaining angle $DBF$ is equal to the angle $BAD$ in the alternate segment of the circle.
 And since $ABCD$ is a quadrilateral in a circle, (the sum of) its opposite angles is equal to two right angles [Prop. 3.22].
 And $DBF$ and $DBE$ is also equal to two right angles [Prop. 1.13].
 Thus, $DBF$ and $DBE$ is equal to $BAD$ and $BCD$, of which $BAD$ was shown (to be) equal to $DBF$.
 Thus, the remaining (angle) $DBE$ is equal to the angle $DCB$ in the alternate segment $DCB$ of the circle.
 Thus, if some straight line touches a circle, and some (other) straight line is drawn across, from the point of contact into the circle, cutting the circle (in two), then those angles the (straight line) makes with the tangent will be equal to the angles in the alternate segments of 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