Proof: By Euclid

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 equiangular² 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 BY-SA 4.0!

Presumably, by finding the center of $ABC$ [Prop. 3.1], drawing a straight line between the center and point $A$, and then drawing $GH$ through point $A$, at right angles to the aforementioned straight line [Prop. 1.11] (translator's note). ↩
I.e. a triangle with given angles (editor's note). ↩