Proof: By Euclid
(related to Proposition: 4.04: Inscribing Circle in Triangle)
- Let the angles $ABC$ and $ACB$ have been cut in half by the straight lines $BD$ and $CD$ (respectively) [Prop. 1.9], and let them meet one another at point $D$, and let $DE$, $DF$, and $DG$ have been drawn from point $D$, perpendicular to the straight lines $AB$, $BC$, and $CA$ (respectively) [Prop. 1.12].
- And since angle $ABD$ is equal to $CBD$, and the right angle $BED$ is also equal to the right angle $BFD$, $EBD$ and $FBD$ are thus two triangles having two angles equal to two angles, and one side equal to one side - the (one) subtending one of the equal angles (which is) common to the (triangles) - (namely), $BD$.
- Thus, they will also have the remaining sides equal to the (corresponding) remaining sides [Prop. 1.26].
- Thus, $DE$ (is) equal to $DF$.
- So, for the same (reasons), $DG$ is also equal to $DF$.
- Thus, the three straight lines $DE$, $DF$, and $DG$ are equal to one another.
- Thus, the circle drawn with center $D$, and radius one of $E$, $F$, or $G$, will also go through the remaining points, and will touch the straight lines $AB$, $BC$, and $CA$, on account of the angles at $E$, $F$, and $G$ being right angles.
- For if it cuts (one of) them then it will be a (straight line) drawn at right angles to a diameter of the circle, from its extremity, falling inside the circle.
- The very thing was shown (to be) absurd [Prop. 3.16].
- Thus, the circle drawn with center $D$, and radius one of $E$, $F$, or $G$, does not cut the straight lines $AB$, $BC$, and $CA$.
- Thus, it will touch them and will be the circle inscribed in triangle $ABC$.
- Let it have been (so) inscribed, like $FGE$ (in the figure).
- Thus, the circle $EFG$ has been inscribed in the given triangle $ABC$.
- (Which is) the very thing it was required to do.
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"