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 $ABC$.
- 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 with the given triangle, has been circumscribed about the given circle.
- (Which is) the very thing it was required to do.
Thank you to the contributors under CC BY-SA 4.0!
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"