Proof: By Euclid
(related to Proposition: 3.33: Construction of Segment on Given Line Admitting Given Angle)
 So the [angle] $C$ is surely either acute, a right angle, or obtuse.
 First of all, let it be acute.
 And, as in the first diagram (from the left), Let (angle) $BAD$, equal to angle $C$, have been constructed on the straight line $AB$, at the point. A (on it) [Prop. 1.23].
 Thus, $BAD$ is also acute.
 Let $AE$ have been drawn, at right angles to $DA$ [Prop. 1.11].
 And let $AB$ have been cut in half at $F$ [Prop. 1.10].
 And let $FG$ have been drawn from point $F$, at right angles to $AB$ [Prop. 1.11].
 And let $GB$ have been joined.
 And since $AF$ is equal to $FB$, and $FG$ (is) common, the two (straight lines) $AF$, $FG$ are equal to the two (straight lines) $BF$, $FG$ (respectively).
 And angle $AFG$ (is) equal to [angle] $BFG$.
 Thus, the base $AG$ is equal to the base $BG$ [Prop. 1.4].
 Thus, the circle drawn with center $G$, and radius $GA$, will also go through $B$ (as well as $A$).
 Let it have been drawn, and let it be (denoted) $ABE$.
 And let $EB$ have been joined.
 Therefore, since $AD$ is at the extremity of diameter $AE$, (namely, point) $A$, at right angles to $AE$, the (straight line) $AD$ thus touches the circle $ABE$ [Prop. 3.16 corr.] .
 Therefore, since some straight line $AD$ touches the circle $ABE$, and some (other) straight line $AB$ has been drawn across from the point of contact $A$ into circle $ABE$, angle $DAB$ is thus equal to the angle $AEB$ in the alternate segment of the circle [Prop. 3.32].
 But, $DAB$ is equal to $C$.
 Thus, angle $C$ is also equal to $AEB$.
 Thus, a segment $AEB$ of a circle, accepting the angle $AEB$ (which is) equal to the given (angle) $C$, has been drawn on the given straight line $AB$.
 And so let $C$ be a right angle.
 And let it again be necessary to draw a segment of a circle on $AB$, accepting an angle equal to the right[angle] $C$.
 Let the (angle) $BAD$ [again] have been constructed, equal to the right angle $C$ [Prop. 1.23], as in the second diagram (from the left).
 And let $AB$ have been cut in half at $F$ [Prop. 1.10].
 And let the circle $AEB$ have been drawn with center $F$, and radius either $FA$ or $FB$.
 Thus, the straight line $AD$ touches the circle $ABE$, on account of the angle at $A$ being a right angle [Prop. 3.16 corr.] .
 And angle $BAD$ is equal to the angle in segment $AEB$.
 For (the latter angle), being in a semicircle, is also a right angle [Prop. 3.31].
 But, $BAD$ is also equal to $C$.
 Thus, the (angle) in (segment) $AEB$ is also equal to $C$.
 Thus, a segment $AEB$ of a circle, accepting an angle equal to $C$, has again been drawn on $AB$.
 And so let (angle) $C$ be obtuse.
 And Let (angle) $BAD$, equal to ($C$), have been constructed on the straight line $AB$, at the point $A$ (on it) [Prop. 1.23], as in the third diagram (from the left).
 And let $AE$ have been drawn, at right angles to $AD$ [Prop. 1.11].
 And let $AB$ have again been cut in half at $F$ [Prop. 1.10].
 And let $FG$ have been drawn, at right angles to $AB$ [Prop. 1.10].
 And let $GB$ have been joined.
 And again, since $AF$ is equal to $FB$, and $FG$ (is) common, the two (straight lines) $AF$, $FG$ are equal to the two (straight lines) $BF$, $FG$ (respectively).
 And angle $AFG$ (is) equal to angle $BFG$.
 Thus, the base $AG$ is equal to the base $BG$ [Prop. 1.4].
 Thus, a circle of center $G$, and radius $GA$, being drawn, will also go through $B$ (as well as $A$).
 Let it go like $AEB$ (in the third diagram from the left).
 And since $AD$ is at right angles to the diameter $AE$, at its extremity, $AD$ thus touches circle $AEB$ [Prop. 3.16 corr.] .
 And $AB$ has been drawn across (the circle) from the point of contact $A$.
 Thus, angle $BAD$ is equal to the angle constructed in the alternate segment $AHB$ of the circle [Prop. 3.32].
 But, angle $BAD$ is equal to $C$.
 Thus, the angle in segment $AHB$ is also equal to $C$.
 Thus, a segment $AHB$ of a circle, accepting an angle equal to $C$, has been drawn on the given straight line $AB$.
 (Which is) the very thing it was required to do.^{1}
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes