Proof: By Euclid
(related to Proposition: 4.03: Circumscribing about Circle Triangle Equiangular with Given Angles)
 Let $EF$ have been produced in each direction to points $G$ and $H$.
 And let the center $K$ of circle $ABC$ have been found [Prop. 3.1].
 And let the straight line $KB$ have been drawn, at random, across ($ABC$).
 And Let (angle) $BKA$, equal to angle $DEG$, have been constructed on the straight line $KB$ at the point $K$ on it, and (angle) $BKC$, equal to $DFH$ [Prop. 1.23].
 And let the (straight lines) $LAM$, $MBN$, and $NCL$ have been drawn through the points $A$, $B$, and $C$ (respectively), touching the circle $ABC$.^{1}
 And since $LM$, $MN$, and $NL$ touch circle $ABC$ at points $A$, $B$, and $C$ (respectively), and $KA$, $KB$, and $KC$ are joined from the center $K$ to points $A$, $B$, and $C$ (respectively), the angles at points $A$, $B$, and $C$ are thus right angles [Prop. 3.18].
 And since the (sum of the) four angles of quadrilateral $AMBK$ is equal to four right angles, inasmuch as $AMBK$ (can) also (be) divided into two triangles [Prop. 1.32], and angles $KAM$ and $KBM$ are (both) right angles, the (sum of the) remaining (angles), $AKB$ and $AMB$, is thus equal to two right angles.
 And $DEG$ and $DEF$ is also equal to two right angles [Prop. 1.13].
 Thus, $AKB$ and $AMB$ is equal to $DEG$ and $DEF$, of which $AKB$ is equal to $DEG$.
 Thus, the remainder $AMB$ is equal to the remainder $DEF$.
 So, similarly, it can be shown that $LNB$ is also equal to $DFE$.
 Thus, the remaining (angle) $MLN$ is also equal to the [remaining] (angle) $EDF$ [Prop. 1.32].
 Thus, triangle $LMN$ is equiangular with triangle $DEF$.
 And it has been drawn around circle $ABC$.
 Thus, a triangle, equiangular^{2} with the given triangle, has been circumscribed about the given circle.
 (Which is) the very thing it was required to do.^{2}
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes