Proof: By Euclid
(related to Proposition: 4.02: Inscribing in Circle Triangle Equiangular with Given Angles)
 Let $GH$ have been drawn touching circle $ABC$ at $A$.^{1} And Let (angle) $HAC$, equal to angle $DEF$, have been constructed on the straight line $AH$ at the point $A$ on it, and (angle) $GAB$, equal to [angle] $DFE$, on the straight line $AG$ at the point $A$ on it [Prop. 1.23].
 And let $BC$ have been joined.
 Therefore, since some straight line $AH$ touches the circle $ABC$, and the straight line $AC$ has been drawn across (the circle) from the point of contact $A$, (angle) $HAC$ is thus equal to the angle $ABC$ in the alternate segment of the circle [Prop. 3.32].
 But, $HAC$ is equal to $DEF$.
 Thus, angle $ABC$ is also equal to $DEF$.
 So, for the same (reasons), $ACB$ is also equal to $DFE$.
 Thus, the remaining (angle) $BAC$ is equal to the remaining (angle) $EDF$ [Prop. 1.32].
 Thus, triangle $ABC$ is equiangular^{2} with triangle $DEF$, and has been inscribed in circle $ABC$].
 Thus, a triangle, equiangular with the given triangle, has been inscribed in the given circle.
 (Which is) the very thing it was required to do.
∎
Thank you to the contributors under CC BYSA 4.0!
 Github:

 nonGithub:
 @Fitzpatrick
References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes