Proof: By Euclid
(related to Proposition: 1.09: Bisecting an Angle)
 Let the point $D$ have been taken at random on $AB$, and let $AE$, equal to $AD$, have been cut off from $AC$ [Prop. 1.3], and let $DE$ have been joined.
 And let the equilateral triangle $DEF$ have been constructed upon $DE$ [Prop. 1.1], and let $AF$ have been joined.
 I say that the angle $BAC$ has been cut in half by the straight line $AF$.
 For since $AD$ is equal to $AE$, and $AF$ is common, the two (straight lines) $DA$, $AF$ are equal to the two (straight lines) $EA$, $AF$, respectively.
 And the base $DF$ is equal to the base $EF$.
 Thus, angle $DAF$ is equal to angle $EAF$ [Prop. 1.8].
 Thus, the given rectilinear angle $BAC$ has been cut in half by the straight line $AF$.
 (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"